Kdtyjb

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.

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:

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.

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.

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.

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

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 $frac2pi(1+r)E(frac{4r}{1+r^2})$ 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 $frac{r+u}{sqrt{(r+u)^2+v^2}}+frac{r-u}{sqrt{(r-u)^2+v^2}}$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=frac{p}{sqrt{p^2+q^2}}=costan^{-1}frac qp$. 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 $frac{partial g}{partial r}>0$, which means that $frac{partial f}{partial r}>0$, which means that $frac{partialint f}{partial r}>0$, 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.