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Who is our nearest planetary neighbor, on average?


Santa Claus flies to the South PoleSeven Spheres of Unequal Mass, a weighing problem with a twistDescribe a large integerFast Mental Calculation of $7.5^7$Math in Space (without the help of celebrities)Find the value of $bigstar$: Puzzle 8 - InequalityWho drinks beer while running anyway?A Crucial DeliveryRanking And AverageHow long will my money last at roulette?













16












$begingroup$


Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.



Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.




What is the closest planet to Earth on average?







mathematics physics







share|improve this question











$endgroup$







  • 2




    $begingroup$
    Though this is a somewhat mathematical puzzle, the answer is a little surprising...
    $endgroup$
    – Dr Xorile
    yesterday






  • 5




    $begingroup$
    I assume you mean the closest planet besides Earth?
    $endgroup$
    – Deusovi♦
    yesterday






  • 2




    $begingroup$
    Didn't I see someone promoting their paper on this today?
    $endgroup$
    – Jay
    yesterday






  • 1




    $begingroup$
    @Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
    $endgroup$
    – yo'
    16 hours ago















16












$begingroup$


Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.



Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.




What is the closest planet to Earth on average?







mathematics physics







share|improve this question











$endgroup$







  • 2




    $begingroup$
    Though this is a somewhat mathematical puzzle, the answer is a little surprising...
    $endgroup$
    – Dr Xorile
    yesterday






  • 5




    $begingroup$
    I assume you mean the closest planet besides Earth?
    $endgroup$
    – Deusovi♦
    yesterday






  • 2




    $begingroup$
    Didn't I see someone promoting their paper on this today?
    $endgroup$
    – Jay
    yesterday






  • 1




    $begingroup$
    @Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
    $endgroup$
    – yo'
    16 hours ago













16












16








16


2



$begingroup$


Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.



Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.




What is the closest planet to Earth on average?







mathematics physics







share|improve this question











$endgroup$




Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.



Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.




What is the closest planet to Earth on average?







mathematics physics




mathematics physics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 14 mins ago









smci

35229




35229










asked yesterday









Dr XorileDr Xorile

13.6k32674




13.6k32674







  • 2




    $begingroup$
    Though this is a somewhat mathematical puzzle, the answer is a little surprising...
    $endgroup$
    – Dr Xorile
    yesterday






  • 5




    $begingroup$
    I assume you mean the closest planet besides Earth?
    $endgroup$
    – Deusovi♦
    yesterday






  • 2




    $begingroup$
    Didn't I see someone promoting their paper on this today?
    $endgroup$
    – Jay
    yesterday






  • 1




    $begingroup$
    @Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
    $endgroup$
    – yo'
    16 hours ago












  • 2




    $begingroup$
    Though this is a somewhat mathematical puzzle, the answer is a little surprising...
    $endgroup$
    – Dr Xorile
    yesterday






  • 5




    $begingroup$
    I assume you mean the closest planet besides Earth?
    $endgroup$
    – Deusovi♦
    yesterday






  • 2




    $begingroup$
    Didn't I see someone promoting their paper on this today?
    $endgroup$
    – Jay
    yesterday






  • 1




    $begingroup$
    @Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
    $endgroup$
    – yo'
    16 hours ago







2




2




$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
yesterday




$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
yesterday




5




5




$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
yesterday




$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
yesterday




2




2




$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
yesterday




$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
yesterday




1




1




$begingroup$
@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
16 hours ago




$begingroup$
@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
16 hours ago










6 Answers
6






active

oldest

votes


















17












$begingroup$

The closest planet to Earth on average is:




Mercury




The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!




As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.

By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.

Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.




Earth to Mars:




$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$




Earth to Venus:




$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$




Earth to Mercury:




$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$







share|improve this answer









$endgroup$












  • $begingroup$
    The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
    $endgroup$
    – darksky
    19 hours ago











  • $begingroup$
    @darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
    $endgroup$
    – Amorydai
    16 hours ago







  • 1




    $begingroup$
    @darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
    $endgroup$
    – Dr Xorile
    16 hours ago






  • 1




    $begingroup$
    @Amorydai, how did you calculate the integrals?
    $endgroup$
    – Dr Xorile
    14 hours ago


















10












$begingroup$

I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.




First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.

Orbits




Now, notice that:




I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.




So:




Let's plot the distance away from the Earth over time, which looks vaguely like this:

Orbital graph




Here, you should note:




The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).


Also, Mars is clearly out of the question. Goodbye Mars.




So:




It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.


The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.




So my guess is that:




Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)




NB: Click on images for slightly higher quality if they're a bit fuzzy.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    Love this answer. Super helpful for gaining the intuition.
    $endgroup$
    – Dr Xorile
    14 hours ago


















7












$begingroup$

The answer:




Mercury




Reasoning:




The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.







share|improve this answer









$endgroup$








  • 3




    $begingroup$
    Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
    $endgroup$
    – Soltius
    20 hours ago











  • $begingroup$
    By this logic, the average closest planet to our moon is also Mercury.
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    13 hours ago











  • $begingroup$
    @BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
    $endgroup$
    – Matthew Barber
    9 hours ago










  • $begingroup$
    @Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
    $endgroup$
    – Matthew Barber
    8 hours ago


















4












$begingroup$

It's




Mercury.




Because:




Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).


As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.







share|improve this answer











$endgroup$












  • $begingroup$
    Why is it continuous and monotonic? And how did you deal with Mars?
    $endgroup$
    – boboquack
    yesterday










  • $begingroup$
    @boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
    $endgroup$
    – JonMark Perry
    23 hours ago











  • $begingroup$
    Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
    $endgroup$
    – Gareth McCaughan♦
    22 hours ago










  • $begingroup$
    @GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
    $endgroup$
    – JonMark Perry
    21 hours ago











  • $begingroup$
    The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
    $endgroup$
    – Gareth McCaughan♦
    21 hours ago


















3












$begingroup$


I'd say the closest planet to Earth is Earth with an average distance of 0







share|improve this answer










New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$








  • 2




    $begingroup$
    Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
    $endgroup$
    – Santana Afton
    8 hours ago











  • $begingroup$
    Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
    $endgroup$
    – yo'
    6 hours ago


















2












$begingroup$

Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.



So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.



Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)



The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.



I suspect there may be an easier more purely geometrical way to do this.






share|improve this answer









$endgroup$












  • $begingroup$
    @boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
    $endgroup$
    – Dr Xorile
    14 hours ago











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6 Answers
6






active

oldest

votes








6 Answers
6






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$

The closest planet to Earth on average is:




Mercury




The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!




As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.

By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.

Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.




Earth to Mars:




$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$




Earth to Venus:




$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$




Earth to Mercury:




$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$







share|improve this answer









$endgroup$












  • $begingroup$
    The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
    $endgroup$
    – darksky
    19 hours ago











  • $begingroup$
    @darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
    $endgroup$
    – Amorydai
    16 hours ago







  • 1




    $begingroup$
    @darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
    $endgroup$
    – Dr Xorile
    16 hours ago






  • 1




    $begingroup$
    @Amorydai, how did you calculate the integrals?
    $endgroup$
    – Dr Xorile
    14 hours ago















17












$begingroup$

The closest planet to Earth on average is:




Mercury




The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!




As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.

By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.

Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.




Earth to Mars:




$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$




Earth to Venus:




$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$




Earth to Mercury:




$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$







share|improve this answer









$endgroup$












  • $begingroup$
    The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
    $endgroup$
    – darksky
    19 hours ago











  • $begingroup$
    @darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
    $endgroup$
    – Amorydai
    16 hours ago







  • 1




    $begingroup$
    @darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
    $endgroup$
    – Dr Xorile
    16 hours ago






  • 1




    $begingroup$
    @Amorydai, how did you calculate the integrals?
    $endgroup$
    – Dr Xorile
    14 hours ago













17












17








17





$begingroup$

The closest planet to Earth on average is:




Mercury




The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!




As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.

By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.

Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.




Earth to Mars:




$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$




Earth to Venus:




$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$




Earth to Mercury:




$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$







share|improve this answer









$endgroup$



The closest planet to Earth on average is:




Mercury




The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!




As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.

By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.

Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.




Earth to Mars:




$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$




Earth to Venus:




$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$




Earth to Mercury:




$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$








share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









AmorydaiAmorydai

1,16514




1,16514











  • $begingroup$
    The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
    $endgroup$
    – darksky
    19 hours ago











  • $begingroup$
    @darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
    $endgroup$
    – Amorydai
    16 hours ago







  • 1




    $begingroup$
    @darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
    $endgroup$
    – Dr Xorile
    16 hours ago






  • 1




    $begingroup$
    @Amorydai, how did you calculate the integrals?
    $endgroup$
    – Dr Xorile
    14 hours ago
















  • $begingroup$
    The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
    $endgroup$
    – darksky
    19 hours ago











  • $begingroup$
    @darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
    $endgroup$
    – Amorydai
    16 hours ago







  • 1




    $begingroup$
    @darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
    $endgroup$
    – Dr Xorile
    16 hours ago






  • 1




    $begingroup$
    @Amorydai, how did you calculate the integrals?
    $endgroup$
    – Dr Xorile
    14 hours ago















$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
19 hours ago





$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
19 hours ago













$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
16 hours ago





$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
16 hours ago





1




1




$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
16 hours ago




$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
16 hours ago












$begingroup$
@darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
$endgroup$
– Dr Xorile
16 hours ago




$begingroup$
@darksky, that's what the question asked. A circular orbit implies constant speed and the lack of resonances makes this the correct assumption. A PhD candidate recently published a more detailed result showing this to be true even with more accurate assumptions
$endgroup$
– Dr Xorile
16 hours ago




1




1




$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
14 hours ago




$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
14 hours ago











10












$begingroup$

I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.




First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.

Orbits




Now, notice that:




I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.




So:




Let's plot the distance away from the Earth over time, which looks vaguely like this:

Orbital graph




Here, you should note:




The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).


Also, Mars is clearly out of the question. Goodbye Mars.




So:




It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.


The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.




So my guess is that:




Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)




NB: Click on images for slightly higher quality if they're a bit fuzzy.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    Love this answer. Super helpful for gaining the intuition.
    $endgroup$
    – Dr Xorile
    14 hours ago















10












$begingroup$

I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.




First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.

Orbits




Now, notice that:




I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.




So:




Let's plot the distance away from the Earth over time, which looks vaguely like this:

Orbital graph




Here, you should note:




The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).


Also, Mars is clearly out of the question. Goodbye Mars.




So:




It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.


The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.




So my guess is that:




Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)




NB: Click on images for slightly higher quality if they're a bit fuzzy.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    Love this answer. Super helpful for gaining the intuition.
    $endgroup$
    – Dr Xorile
    14 hours ago













10












10








10





$begingroup$

I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.




First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.

Orbits




Now, notice that:




I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.




So:




Let's plot the distance away from the Earth over time, which looks vaguely like this:

Orbital graph




Here, you should note:




The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).


Also, Mars is clearly out of the question. Goodbye Mars.




So:




It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.


The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.




So my guess is that:




Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)




NB: Click on images for slightly higher quality if they're a bit fuzzy.






share|improve this answer











$endgroup$



I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.




First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.

Orbits




Now, notice that:




I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.




So:




Let's plot the distance away from the Earth over time, which looks vaguely like this:

Orbital graph




Here, you should note:




The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).


Also, Mars is clearly out of the question. Goodbye Mars.




So:




It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.


The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.




So my guess is that:




Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)




NB: Click on images for slightly higher quality if they're a bit fuzzy.







share|improve this answer














share|improve this answer



share|improve this answer








edited 20 hours ago

























answered yesterday









boboquackboboquack

15.7k149119




15.7k149119







  • 1




    $begingroup$
    Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    Love this answer. Super helpful for gaining the intuition.
    $endgroup$
    – Dr Xorile
    14 hours ago












  • 1




    $begingroup$
    Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    Love this answer. Super helpful for gaining the intuition.
    $endgroup$
    – Dr Xorile
    14 hours ago







1




1




$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
16 hours ago




$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
16 hours ago












$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
14 hours ago




$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
14 hours ago











7












$begingroup$

The answer:




Mercury




Reasoning:




The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.







share|improve this answer









$endgroup$








  • 3




    $begingroup$
    Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
    $endgroup$
    – Soltius
    20 hours ago











  • $begingroup$
    By this logic, the average closest planet to our moon is also Mercury.
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    13 hours ago











  • $begingroup$
    @BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
    $endgroup$
    – Matthew Barber
    9 hours ago










  • $begingroup$
    @Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
    $endgroup$
    – Matthew Barber
    8 hours ago















7












$begingroup$

The answer:




Mercury




Reasoning:




The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.







share|improve this answer









$endgroup$








  • 3




    $begingroup$
    Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
    $endgroup$
    – Soltius
    20 hours ago











  • $begingroup$
    By this logic, the average closest planet to our moon is also Mercury.
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    13 hours ago











  • $begingroup$
    @BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
    $endgroup$
    – Matthew Barber
    9 hours ago










  • $begingroup$
    @Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
    $endgroup$
    – Matthew Barber
    8 hours ago













7












7








7





$begingroup$

The answer:




Mercury




Reasoning:




The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.







share|improve this answer









$endgroup$



The answer:




Mercury




Reasoning:




The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.








share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









Matthew BarberMatthew Barber

5073




5073







  • 3




    $begingroup$
    Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
    $endgroup$
    – Soltius
    20 hours ago











  • $begingroup$
    By this logic, the average closest planet to our moon is also Mercury.
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    13 hours ago











  • $begingroup$
    @BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
    $endgroup$
    – Matthew Barber
    9 hours ago










  • $begingroup$
    @Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
    $endgroup$
    – Matthew Barber
    8 hours ago












  • 3




    $begingroup$
    Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
    $endgroup$
    – Soltius
    20 hours ago











  • $begingroup$
    By this logic, the average closest planet to our moon is also Mercury.
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    13 hours ago











  • $begingroup$
    @BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
    $endgroup$
    – Matthew Barber
    9 hours ago










  • $begingroup$
    @Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
    $endgroup$
    – Matthew Barber
    8 hours ago







3




3




$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
20 hours ago





$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
20 hours ago













$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
13 hours ago





$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
13 hours ago













$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
9 hours ago




$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
9 hours ago












$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
8 hours ago




$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
8 hours ago











4












$begingroup$

It's




Mercury.




Because:




Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).


As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.







share|improve this answer











$endgroup$












  • $begingroup$
    Why is it continuous and monotonic? And how did you deal with Mars?
    $endgroup$
    – boboquack
    yesterday










  • $begingroup$
    @boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
    $endgroup$
    – JonMark Perry
    23 hours ago











  • $begingroup$
    Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
    $endgroup$
    – Gareth McCaughan♦
    22 hours ago










  • $begingroup$
    @GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
    $endgroup$
    – JonMark Perry
    21 hours ago











  • $begingroup$
    The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
    $endgroup$
    – Gareth McCaughan♦
    21 hours ago















4












$begingroup$

It's




Mercury.




Because:




Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).


As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.







share|improve this answer











$endgroup$












  • $begingroup$
    Why is it continuous and monotonic? And how did you deal with Mars?
    $endgroup$
    – boboquack
    yesterday










  • $begingroup$
    @boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
    $endgroup$
    – JonMark Perry
    23 hours ago











  • $begingroup$
    Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
    $endgroup$
    – Gareth McCaughan♦
    22 hours ago










  • $begingroup$
    @GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
    $endgroup$
    – JonMark Perry
    21 hours ago











  • $begingroup$
    The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
    $endgroup$
    – Gareth McCaughan♦
    21 hours ago













4












4








4





$begingroup$

It's




Mercury.




Because:




Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).


As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.







share|improve this answer











$endgroup$



It's




Mercury.




Because:




Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).


As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.








share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered yesterday









JonMark PerryJonMark Perry

20.2k64098




20.2k64098











  • $begingroup$
    Why is it continuous and monotonic? And how did you deal with Mars?
    $endgroup$
    – boboquack
    yesterday










  • $begingroup$
    @boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
    $endgroup$
    – JonMark Perry
    23 hours ago











  • $begingroup$
    Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
    $endgroup$
    – Gareth McCaughan♦
    22 hours ago










  • $begingroup$
    @GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
    $endgroup$
    – JonMark Perry
    21 hours ago











  • $begingroup$
    The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
    $endgroup$
    – Gareth McCaughan♦
    21 hours ago
















  • $begingroup$
    Why is it continuous and monotonic? And how did you deal with Mars?
    $endgroup$
    – boboquack
    yesterday










  • $begingroup$
    @boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
    $endgroup$
    – JonMark Perry
    23 hours ago











  • $begingroup$
    Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
    $endgroup$
    – Gareth McCaughan♦
    22 hours ago










  • $begingroup$
    @GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
    $endgroup$
    – JonMark Perry
    21 hours ago











  • $begingroup$
    The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
    $endgroup$
    – Gareth McCaughan♦
    21 hours ago















$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
yesterday




$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
yesterday












$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
23 hours ago





$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
23 hours ago













$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
22 hours ago




$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
22 hours ago












$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
21 hours ago





$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
21 hours ago













$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
21 hours ago




$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
21 hours ago











3












$begingroup$


I'd say the closest planet to Earth is Earth with an average distance of 0







share|improve this answer










New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$








  • 2




    $begingroup$
    Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
    $endgroup$
    – Santana Afton
    8 hours ago











  • $begingroup$
    Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
    $endgroup$
    – yo'
    6 hours ago















3












$begingroup$


I'd say the closest planet to Earth is Earth with an average distance of 0







share|improve this answer










New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$








  • 2




    $begingroup$
    Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
    $endgroup$
    – Santana Afton
    8 hours ago











  • $begingroup$
    Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
    $endgroup$
    – yo'
    6 hours ago













3












3








3





$begingroup$


I'd say the closest planet to Earth is Earth with an average distance of 0







share|improve this answer










New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




I'd say the closest planet to Earth is Earth with an average distance of 0








share|improve this answer










New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this answer



share|improve this answer








edited 20 hours ago









Ahmed Ashour

976313




976313






New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 21 hours ago









user57862user57862

311




311




New contributor




user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user57862 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 2




    $begingroup$
    Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
    $endgroup$
    – Santana Afton
    8 hours ago











  • $begingroup$
    Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
    $endgroup$
    – yo'
    6 hours ago












  • 2




    $begingroup$
    Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
    $endgroup$
    – yo'
    16 hours ago










  • $begingroup$
    @yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
    $endgroup$
    – Santana Afton
    8 hours ago











  • $begingroup$
    Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
    $endgroup$
    – yo'
    6 hours ago







2




2




$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
16 hours ago




$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
16 hours ago












$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
8 hours ago





$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
8 hours ago













$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
6 hours ago




$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
6 hours ago











2












$begingroup$

Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.



So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.



Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)



The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.



I suspect there may be an easier more purely geometrical way to do this.






share|improve this answer









$endgroup$












  • $begingroup$
    @boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
    $endgroup$
    – Dr Xorile
    14 hours ago
















2












$begingroup$

Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.



So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.



Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)



The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.



I suspect there may be an easier more purely geometrical way to do this.






share|improve this answer









$endgroup$












  • $begingroup$
    @boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
    $endgroup$
    – Dr Xorile
    14 hours ago














2












2








2





$begingroup$

Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.



So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.



Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)



The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.



I suspect there may be an easier more purely geometrical way to do this.






share|improve this answer









$endgroup$



Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.



So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.



Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)



The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.



I suspect there may be an easier more purely geometrical way to do this.







share|improve this answer












share|improve this answer



share|improve this answer










answered 19 hours ago









Gareth McCaughan♦Gareth McCaughan

64.5k3164253




64.5k3164253











  • $begingroup$
    @boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
    $endgroup$
    – Dr Xorile
    14 hours ago

















  • $begingroup$
    @boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
    $endgroup$
    – Dr Xorile
    14 hours ago
















$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
14 hours ago





$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
14 hours ago


















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