I guess I could do the math myself but what fun would that be when I can get the great minds of this message board to do it for me.
Using a gun and ammunition that is commercially available in the United States, can I hit the Earth when shooting from the surface of the Moon?
Would I aim directly at the Earth or should I aim somewhere to the right (or maybe it’s the left) of the Earth’s path as it moves across the Moon’s “sky”? Does the Moon have a sky?
Even if not commercially available, is there a bullet shape and composition that could reach the solid surface of the Earth instead of just becoming a small meteor due to the friction from the atmosphere?
Just to be clear, if this were possible, I would only do it if I could be ensured of hitting a safe target without endangering anyone.
Even if you could somehow calculate exactly where to aim in the sky in order to hit a target a quarter million miles away several million years from now, it’d almost certainly burn up in the atmosphere.
Perhaps if you managed to get near the L1 Earth/Moon Lagrange point with a bullet you might reduce the necessary muzzle velocity by a fraction; once past the L1 point the Earth’s gravity would take over. But I don’t think this would be much of a reduction.
That’s what I was thinking. I would only have to get to L1. Am I thinking right? (Probably not if I am coming up with stupid questions like this early in the morning,)
At L1 there is no net gravity between Earth & Moon. If you were standing (floating) there you could throw a rock with enough velocity to hit the Earth. Or at least hit the Earth’s atmosphere. Your problem is aim.
As mentioned above, from the lunar surface your problem is lunar escape velocity. Which you’re not going to achieve with a commercial firearm. Or even with artillery.
I don’t think I have to achieve the escape velocity. I am not trying to escape the Moon’s gravitational pull. I am just trying to get far enough where the Earth’s gravitational pull overwhelms the Moon’s gravitational pull, i.e., just past L1.
Although the L1 point is on the order of 60-80K km out from the Moon. By which point you’re a long way towards the top of the gravity well. My implication being (and without doing all the math), that although you don’t need to get to 100% of lunar escape velocity to fire a projectile to there, you do need to get kinda close to that fast. Regardless of whether there’s the Earth out there as a backstop and as something distorting the Moon’s gravitational field.
Yes you are. The appropriate formulation of the question is what velocity is required from the surface of the moon to reach a height of roughly 60,000 km, or about 16% of the distance from the moon to the earth where the L1 point is located. Without doing the math, that doesn’t seem to be much less than escape velocity.
Various equations including one for maximum height as a function of velocity, launch angle, and gravity can be found here.
Should work better than normal. A conventional firearm doesn’t rely on atmospheric air to oxidize its propellant, so it should be able to launch the bullet out of the barrel without any trouble. And since there’s no atmospheric air to push out of the way, you’d actually end up with a slightly higher muzzle velocity in a vacuum than you would at sea level on earth.
According to my quick calculation (which might be totally wrong since I haven’t had my coffee yet) a high-powered rifle with a muzzle velocity of 1200 m/s fired from the moon would reach an impressive 442 km before it started to fall back. This is, however, rather well short of the L1 point some 60,000 km away!
Did you account for the fact that the acceleration due to gravity is from two bodies (the Earth and the Moon). Most simple formulas for calculating maximum height from velocity assume just one source of gravity. As the bullet approaches L1 the (negative) acceleration is influenced by both the Earth and the Moon.
I am too busy learning all about drafting indemnity clauses in contracts this morning to do the math. And I really don’t feel like doing math this morning anyway.
ETA: If its only about 442 km then maybe the influence of the Earth’s gravity could be safely ignored.
Same as what happens to you when you fire a gun anywhere else. You’ll get a momentum equal to the mass of the bullet multiplied by the speed of the bullet (but in the opposite direction). But since your mass is much greater than that of the bullet, the resulting velocity for your body will be much lower.