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Reading this made me wonder a few physics questions. - If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle yo

Reading this made me wonder a few physics questions. - If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle you shoot it? - The escape velocity of the moon is 2.38km/s. Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it? Or is the required velocity higher at smaller angles?

ArchReaper 7y ago
1) No 2) * Escape velocity is the velocity required to counter the force of gravity, ie 'away' or 'out' from the center of the planet, opposite the force of gravity. * Edit: Looks like I'm wrong, guess I slept that day in physics
- teraflop 7y ago
  Escape velocity is actually the same in any direction, not just directly upward. This is true because in a two-body system, whether or not a trajectory escapes to infinity depends only on whether the sign of specific orbital energy[1] is positive or negative. Specific orbital energy is a function of kinetic energy and potential energy, but is independent of the direction of motion. [1]: https://en.wikipedia.org/wiki/Specific_orbital_energy https://en.wikipedia.org/wiki/Specific_orbital_energy
gnode 7y ago
> If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle you shoot it? In most cases no. Although the orbit of the bullet is cyclical (below escape velocity), unless you shoot perfectly level with the surface, its orbit will intersect the moon. Obviously if you shoot slightly down, it'll hit the ground. If you shoot slightly up, it'll hit the ground behind you on its return. > Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it? Yes. Although if it hits the moon first, it'll slow down and not escape (unless you were to shoot through the moon).
- smellf 7y ago
  Edit: I was wrong, never mind.
  - chowells 7y ago
    Bullets are sealed. They don't use the atmosphere as an oxidizer. Doesn't mean they fire well in a vacuum, because there are other factors involved. But you could certainly design a firing system to work perfectly fine in a vacuum.
    - metacyclic 7y ago
      They are capable of being fired underwater.
- vortico 7y ago
  - Ah, I see. Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit? A slightly different question, but if you shoot a bullet perpendicular to the moon's surface at any speed (as long as it won't hit the ground), it will always hit you in the back after one orbit, right? - I suppose if you think about it in terms of kinetic energy, this makes sense.
  - gnode 7y ago
    I realize that I made an assumption in my first answer, that the moon is not rotating. A rotation may cause you to move out of the orbital path of the bullet before it gets back to you. If you shot from the poles though, you would remain in the bullet's orbit. That said, in ideal conditions, even with rotation, the bullet will hit you eventually, but not for a long time on average, given the scale of the moon. In reality it's orbit would probably destabilise due to collisions with dust and perturbations from other gravitational bodies (the Earth, Sun, etc. create an n-body situation which slowly changes the orbit over time). > Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit? The velocity must be great enough to stay above the surface, and not high enough to escape. Additionally, you must shoot horizontally, like I mentioned before, and not rotate out of the orbit (e.g. by being at the poles, or at the rotational equator, shooting along the equator).

ArchReaper 7y ago

1) No 2) * Escape velocity is the velocity required to counter the force of gravity, ie 'away' or 'out' from the center of the planet, opposite the force of gravity. * Edit: Looks like I'm wrong, guess I slept that day in physics

teraflop 7y ago
Escape velocity is actually the same in any direction, not just directly upward. This is true because in a two-body system, whether or not a trajectory escapes to infinity depends only on whether the sign of specific orbital energy[1] is positive or negative. Specific orbital energy is a function of kinetic energy and potential energy, but is independent of the direction of motion. [1]: https://en.wikipedia.org/wiki/Specific_orbital_energy https://en.wikipedia.org/wiki/Specific_orbital_energy

gnode 7y ago

> If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle you shoot it? In most cases no. Although the orbit of the bullet is cyclical (below escape velocity), unless you shoot perfectly level with the surface, its orbit will intersect the moon. Obviously if you shoot slightly down, it'll hit the ground. If you shoot slightly up, it'll hit the ground behind you on its return. > Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it? Yes. Although if it hits the moon first, it'll slow down and not escape (unless you were to shoot through the moon).

smellf 7y ago
Edit: I was wrong, never mind.
- chowells 7y ago
  Bullets are sealed. They don't use the atmosphere as an oxidizer. Doesn't mean they fire well in a vacuum, because there are other factors involved. But you could certainly design a firing system to work perfectly fine in a vacuum.
  - metacyclic 7y ago
    They are capable of being fired underwater.
vortico 7y ago
- Ah, I see. Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit? A slightly different question, but if you shoot a bullet perpendicular to the moon's surface at any speed (as long as it won't hit the ground), it will always hit you in the back after one orbit, right? - I suppose if you think about it in terms of kinetic energy, this makes sense.
- gnode 7y ago
  I realize that I made an assumption in my first answer, that the moon is not rotating. A rotation may cause you to move out of the orbital path of the bullet before it gets back to you. If you shot from the poles though, you would remain in the bullet's orbit. That said, in ideal conditions, even with rotation, the bullet will hit you eventually, but not for a long time on average, given the scale of the moon. In reality it's orbit would probably destabilise due to collisions with dust and perturbations from other gravitational bodies (the Earth, Sun, etc. create an n-body situation which slowly changes the orbit over time). > Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit? The velocity must be great enough to stay above the surface, and not high enough to escape. Additionally, you must shoot horizontally, like I mentioned before, and not rotate out of the orbit (e.g. by being at the poles, or at the rotational equator, shooting along the equator).