Please help with physics....im retarded.....?

Question

In Robert Heinlein's The Moon Is a Harsh Mistress, the colonial inhabitants of the moon threaten to launch rocks down onto the Earth if they are not given independence (or at least representation). Assuming that a rail gun could launch a rock of mass m at 2.90 times the lunar escape speed, calculate the speed of the rock as it enters the Earth's atmosphere.

_____km/h

like i said....im retarded....the answer is suppose to be km/s

Answer 1

This is not easy. First of all, the rail gun could not be aimed directly at the earth because you would miss, at least initially. The rock may enter a decaying orbit and eventually hit... I'm not sure. Like shooting at a moving animal, you would have to aim backwards relative to the direction the moon is moving in order to hit the constantly moving earth (moving relative to the moon, that is). Okay, that's probably too complicated.

Let's assume the moon is hanging in the sky, not orbiting, and you shoot the rail gun directly at the earth. What are the forces acting on the rock? Gravity from the moon, gravity from the earth. The rock slows down initially because of the proximity of the moon, then speeds up because of the larger gravitational attraction of the earth.

For gravity,

F = G * m1 * m2 / r²

Call the mass of the moon Mm and the mass of the earth Me (7.30 E22 kg and 5.97 E24 kg, respectively). For simplicity, let's assume the rock weighs 1 kg (the answer is the same regardless of the mass of the rock). Then:

F = G * (Me * (1 kg) / (D - x)² - Mm * (1 kg)/x²)

where x = distance from the rock to the moon and D = distance from the earth to the moon. Obviously, this force changes as the rock moves.

F = ma, m = 1 kg so F * kg = a

V = Vo + at = Vo + (F * kg)t

Escape velocity from the moon = 2.4 km/s, so initial velocity is 2.4 * 2.9 = 6.96 km/s = 6960 m/s

By the way, G = 6.673 E-11 m3/kg-s²

Right as the rock leaves the surface of the moon (1735 km from the center), the force acting on it from the moon is about 1.61 N, and the force from the earth is about 0.003N. By the time the rock reaches the earth, the force is around 9.8N.

Now there is probably some way to do this using calculus, but I don't have the energy to research it right now, so I did it using numerical methods (an Excel spreadsheet), assuming 5000-km increments rather than continuous motion, so the answer I got was a rough approximation. I calculate that the velocity when it hits earth's atmosphere would be about 13.7 km/s. The trip would take about 17.5 hours, at an average velocity of about 6 km/s. Approximately.

Reference: http://nssdc.gsfc.nasa.gov/planetary/factsheet/index.html