Rotational/gravitational physics question.?

Question

I have this problem for extra credit, one of 35, and this is the last one i need. i think i am my own worse enemy on this one because everything i try that makes sense to me doesn't work. I only have 2 submissions left. Thanks!

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 2.50 times the lunar escape speed, calculate the speed of the rock as it enters the Earth's atmosphere.

Answer 1

That's an interesting problem based on a great novel. Start by looking up the lunar escape velocity, which I found to be 2380 m/s. That makes our initial velocity, v_0 = 2.5*2380 = 5950. With the rock's mass m, the initial kinetic energy is 0.5*m*(v_0)^2 = 0.5*m*5950^2 = 17,701,250*m.

Now consider the change in gravitational potential energy. On a scale as large as the distance from the Earth to the Moon, we can't use good old mg(Δh), which is mass times gravitational acceleration times change in height (Δh). We have to use the formula -GMm/r1 - (-GMm/r1), where G is the gravitational constant G = 6.67 x 10^-11 m^3/(kg-s^2), M is the mass of the Earth, M = 5.97 x 10^24 kg, m is the mass of the rock as given, r1 is the distance between the centers of mass of the bodies (the rock and the Earth) as it is launched, and r2 is the distance as the rock collides with the atmosphere. If the Earth and Moon have their centers of mass a distance d apart, with lunar radius Rm and Earth's atmosphere radius Re, r1 would be d - Rm and r2 would be Re. The average value of d is 384,403,000 m and Rm is 1,737,103 m. Re is a little more complicated. The physical radius of Earth is 6,372,797 m, but the atmosphere has no definite limit. Let's use 120,000 m, the point at which the atmosphere affects re-entry. That makes Re = 6,492,797. So r1 = 384,403,000 - 1,737,103 = 382,665,897 m and r2 = 6,492,797 m.

The result is that the change in gravitational potential energy is -GMm/r1 - (-GMm/r2) = (GMm)(-1/r1 + 1/r2) = (6.67 x 10^-11)(5.97 x 10^24)m(-1/382,665,897 + 1/6,492,797) = 60,288,755*m. This accounts only for the loss of gravitational potential energy with respect to the Earth. I am neglecting the gain in potential energy with respect to the Moon, but you might want to calculate that value and subtract it from the value I just got. In that case, M is the mass of the moon instead of the mass of the Earth, r1 is Rm and r2 is d - Re.

That means that the final total kinetic energy will be the original kinetic energy plus the loss in gravitational potential energy with respect to the Earth, or 17,701,250*m + 60,288,755*m = 77,990,005*m. The final kinetic energy is also equal to 0.5*m*(v_f)^2, where v_f is the final velocity. 0.5*m*(v_f)^2 = 77,990,005*m ==> (v_f)^2 = 155,980,010 = v_f = 12,489 m/s, more than twice the original velocity. If you account for the gain of gravitational potential energy with respect to the Moon, your final result will be smaller than this, but should still be considerably larger than the launch velocity.

Answer 2

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