AU - R, or 149,591,621.9 km.

Since the South Pole is twice as far away (the diameter of the earth is twice the radius, or R x 2), that distance would be 1 AU + (R x 2), or 149,610,756.2 km.

To figure the percentages, we divide. First, the radius of the earth is divided by 1 AU, like so: R / (1 AU) x 100. This shows us that the North Pole would be 0.00426349% closer to the sun. The South Pole, again, is twice as far away, or .00852698% farther away from the sun.

In terms of percentages, there is not that great of a difference—but it is enough to give us winter and summer each year, with a tilt not nearly as great as the 90 degrees postulated in this problem!

You can think of this as a geometry question disguised as an astronomy question!For these purposes, we can assume the Earth is a sphere. (Technically it's a slightly-lumpy oblate spheroid, but it's actually very close to a sphere.)If we imagine the Earth titled even further on its axis (about four times as much as its present 23 degrees, all the way to 90 degrees) so that the North Pole is pointed directly at the Sun and the South Pole is pointed directly away, the distance between them is simply the diameter of that sphere.We're given the radius is 6378.1 km (ridiculously precise by the way; the variation in Earth's radius is more than the 100 meter precision this is giving us, as anyone in Colorado will attest); the diameter is simply twice that, 12,756.2 km.If we compare this to the size of 1 AU, given to us as 149,598,000 km, we can see that the ratio is 0.000085; that is, the North Pole is only 0.0085% closer to the Sun than the South Pole. The difference in temperature, however, would be quite large; in this extreme scenario the North Pole would always been daylight and the South Pole would always be in darkness, so they could easily differ in temperature by 100 C.