- 12 Apr '06 01:31 / 1 editKirby, our intrepid lunar explorer is driving his solar powered rover on the equator of the moon and notices his solar cells are going bad. He finds he has to keep the cells aimed right at the sun full bore or the power drops below the level required for his life support equiptment. He finds he has to turn off his radios, lasers and such and is alone on the surface of the moon at local noon time, with everything but life support turned off he can keep going but he has to keep at local noon all the time, 24/7 or his life support will run out of power.

He knows he can make it to his base but its all the way round the other side of the moon, now he can drive and steer around major barriers ok but has to be close to local noon all the time. So how fast does he have to drive to keep himself at local noon and how long does it take to get back to his base halfway round the moon?

So how fast in Km/hr is that? - 12 Apr '06 07:44

Where is the base situated? The latitude is very important for solving this problem. And where is he himself at? At the antipode of the base perhaps?*Originally posted by sonhouse***Kirby, our intrepid lunar explorer is driving his solar powered rover on the equator of the moon and notices his solar cells are going bad. He finds he has to keep the cells aimed right at the sun full bore or the power drops below the level required for his life support equiptment. He finds he has to turn off his radios, lasers and such and is alone on the ...[text shortened]... long does it take to get back to his base halfway round the moon?**

So how fast in Km/hr is that?

There is no need for him to be at local moon noon at all. It's only important for him to have the sun over the horizon, that’s enough. The moon has no atmosphere that block sun light as her on Good ol' Earth where the sun becomes fainter near the horizon.

If he is near to the lunar north pole or south pole (depending of the season) he have eternal sun shine and do not have to move at all. He just have to stay put, mace a call to the base to come and pick him up when they have the time.

A day on the moon is some 14 (earth) days long and the night is as long (not considering any seasonal effect), this is an important clue to the solution of the problem.

He should prefer going on the maria which is relatively flat and straight forward, and avoid the more rocky regions of the lunar surface.

So the most efficient route from where he is to the base is, with this circumstances, not possible to calculate, the same with his velocity. - 12 Apr '06 23:26

The fact that you are on the moon gives you a definite advantage but I should have said the solar panels are just like the mars rover, they can't be pointed they are just lying flat on the top so to get the max power he has to DRIVE the rover to keep at local noon. Both he and the base are on lunar equator so if he just drives in either direction around the equator, he will get there eventually, mountains aside.*Originally posted by FabianFnas***Where is the base situated? The latitude is very important for solving this problem. And where is he himself at? At the antipode of the base perhaps?**

There is no need for him to be at local moon noon at all. It's only important for him to have the sun over the horizon, that’s enough. The moon has no atmosphere that block sun light as her on Good ol' Ea ...[text shortened]... o the base is, with this circumstances, not possible to calculate, the same with his velocity. - 13 Apr '06 07:21

Okay, both base and current position is at mutual antipodes and on the equator.*Originally posted by sonhouse***The fact that you are on the moon gives you a definite advantage but I should have said the solar panels are just like the mars rover, they can't be pointed they are just lying flat on the top so to get the max power he has to DRIVE the rover to keep at local noon. Both he and the base are on lunar equator so if he just drives in either direction around the equator, he will get there eventually, mountains aside.**

Let’s also assume that the lunar equator is parallell with the ecliptic (it is not in reality) to make the math easier.

Well, we know the circumference of the moon, (we use half of that), we know the time of one lunar revolution around its own axis (we use half of that), then we divide the one with the other and we get the speed.

Does it matter if you travel eastward or westward? Distance is the same? - 13 Apr '06 12:52

Distance would be the same but of course only one direction keeps up with the sun.*Originally posted by FabianFnas***Okay, both base and current position is at mutual antipodes and on the equator.**

Let’s also assume that the lunar equator is parallell with the ecliptic (it is not in reality) to make the math easier.

Well, we know the circumference of the moon, (we use half of that), we know the time of one lunar revolution around its own axis (we use half of that), ...[text shortened]... nd we get the speed.

Does it matter if you travel eastward or westward? Distance is the same?