The Earth's radius is 6378.1 km. Its average distance from the Sun is 1 AU=149,598,000 km. Because the Earth is tilted, one of its poles will be closer to the sun than the other. The question is...
The Earth's radius is 6378.1 km. Its average distance from the Sun is 1 AU=149,598,000 km. Because the Earth is tilted, one of its poles will be closer to the sun than the other. The question is "how much closer?". If the Earth was tilted on its side in its orbit so that the North Pole pointed directed to the Sun, how much closer would the North Pole be to the Sun than the South Pole? What percentage of the Earth's total distance from the Sun is this?
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.