3 ms·

I think it actually might be the best spot because the far side of the moon is always positioned facing away from earth. Is there anywhere else in the solar sy

I think it actually might be the best spot because the far side of the moon is always positioned facing away from earth. Is there anywhere else in the solar system that can claim that?

Supermancho 3y ago
> Is there anywhere else in the solar system that can claim that? I know Mercury shares this. Not super useful, except at the boundary.
- ben_w 3y ago
  I'm 90% sure you're mixing up two other things, about planets in the goldilocks zone of red dwarfs being tidally locked to their host star? Mercury definitely isn't tidally locked to the sun, and its sidereal day isn't a whole integer fraction of Earth's year. I'm not 100% certain without doing some algebra, but don't think it's even possible for one planet to be tidally locked to any planet in a different orbit, be that Mercury to Earth or otherwise, owing to the distance not being constant while angular momentum has to be. (If two planets share an orbit, there are an infinite number of trivial solutions).
  - Supermancho 3y ago
    Indeed, you are correct. https://www.quora.com/Is-Mercury-tidally-locked-so-that-one-side-always-faces-the-Sun https://www.quora.com/Is-Mercury-tidally-locked-so-that-one-... There is more info about what is and is not: https://en.wikipedia.org/wiki/Tidal_locking https://en.wikipedia.org/wiki/Tidal_locking
ben_w 3y ago
Always facing away from Earth, but not always facing away from the sun. Not my field, so I don't know how important interference from solar RF is. If you're building on this scale, why not put two on opposite sides of, say, Deimos? The delta-v between there and LEO is less than the delta-v between the lunar surface and LEO: https://upload.wikimedia.org/wikipedia/commons/9/93/Solar_system_delta_v_map.svg https://upload.wikimedia.org/wikipedia/commons/9/93/Solar_sy...
- pdabbadabba 3y ago
  Wow! TIL. At first I thought you must be reading the map wrong, but you're right. Landing on Deimos does indeed seem to require less delta-V than landing on our own moon, assuming that this map is correct. And that seems to hold even without aerobraking. 14917 ms-1 [Moon] v. 14520 ms-1 [Deimos], if I'm reading it right.
  - kyleyeats 3y ago
    It's because Deimos basically doesn't have a gravity well to overcome. You're just getting alongside it in orbit. It's not really landing at all.
- marcosdumay 3y ago
  I don't know if it's possible to make this map to scale, but it's hard to read right now.
deleted 3y ago
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kyleyeats 3y ago
There's an orbit that's always dark to Earth: https://en.wikipedia.org/wiki/Counter-Earth https://en.wikipedia.org/wiki/Counter-Earth A telescope in that orbit would never see earthlight, like one on the moon.
- bequanna 3y ago
  Communication with earth in that orbit would be a bit of a pain tho, right?
  - kyleyeats 3y ago
    Some might see that as a benefit! But yeah you'd want a satellite orbiting in the L4/L5 Lagrange point of the Earth/Sun system. It would seem stationary to both Earth and Counter-Earth/L3.

I think it actually might be the best spot because the far side of the moon is always positioned facing away from earth. Is there anywhere else in the solar system that can claim that?

Supermancho 3y ago

> Is there anywhere else in the solar system that can claim that? I know Mercury shares this. Not super useful, except at the boundary.

ben_w 3y ago
I'm 90% sure you're mixing up two other things, about planets in the goldilocks zone of red dwarfs being tidally locked to their host star? Mercury definitely isn't tidally locked to the sun, and its sidereal day isn't a whole integer fraction of Earth's year. I'm not 100% certain without doing some algebra, but don't think it's even possible for one planet to be tidally locked to any planet in a different orbit, be that Mercury to Earth or otherwise, owing to the distance not being constant while angular momentum has to be. (If two planets share an orbit, there are an infinite number of trivial solutions).
- Supermancho 3y ago
  Indeed, you are correct. https://www.quora.com/Is-Mercury-tidally-locked-so-that-one-side-always-faces-the-Sun https://www.quora.com/Is-Mercury-tidally-locked-so-that-one-... There is more info about what is and is not: https://en.wikipedia.org/wiki/Tidal_locking https://en.wikipedia.org/wiki/Tidal_locking

ben_w 3y ago

Always facing away from Earth, but not always facing away from the sun. Not my field, so I don't know how important interference from solar RF is. If you're building on this scale, why not put two on opposite sides of, say, Deimos? The delta-v between there and LEO is less than the delta-v between the lunar surface and LEO: https://upload.wikimedia.org/wikipedia/commons/9/93/Solar_system_delta_v_map.svg https://upload.wikimedia.org/wikipedia/commons/9/93/Solar_sy...

pdabbadabba 3y ago
Wow! TIL. At first I thought you must be reading the map wrong, but you're right. Landing on Deimos does indeed seem to require less delta-V than landing on our own moon, assuming that this map is correct. And that seems to hold even without aerobraking. 14917 ms-1 [Moon] v. 14520 ms-1 [Deimos], if I'm reading it right.
- kyleyeats 3y ago
  It's because Deimos basically doesn't have a gravity well to overcome. You're just getting alongside it in orbit. It's not really landing at all.
marcosdumay 3y ago
I don't know if it's possible to make this map to scale, but it's hard to read right now.

deleted 3y ago

[deleted]

kyleyeats 3y ago

There's an orbit that's always dark to Earth: https://en.wikipedia.org/wiki/Counter-Earth https://en.wikipedia.org/wiki/Counter-Earth A telescope in that orbit would never see earthlight, like one on the moon.

bequanna 3y ago
Communication with earth in that orbit would be a bit of a pain tho, right?
- kyleyeats 3y ago
  Some might see that as a benefit! But yeah you'd want a satellite orbiting in the L4/L5 Lagrange point of the Earth/Sun system. It would seem stationary to both Earth and Counter-Earth/L3.