## Moon fall speed

Using the JPL impact calculator I get:
Energy:10^13 Megatons
Transient crater: 2660 miles diameter with 939 miles depth
Final crater: 7920 miles diameter with 3.16 miles depth

Paradoxically, the Earth loses negligible mass, nor are orbit or inclination altered appreciably (despite having inexplicably transformed from a sphere to a lopsided dumbbell)

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 Quote by Saladsamurai Wouldn't it be reasonable to treat the Moon and Earth as spheres of uniform density for something like this. I mean, whether it landed on land or sea, I'll be calling out of work that day.
Whether you go to work or not that day, to compute the force you still need to provide some sort of math that describes the rate of decceleration.

 Quote by Saladsamurai And I think the malls will be closed. Except for Starbuck's.
That's a good tip. When the Moon drops from the sky, run for Starbucks :)

 Quote by DaveC426913 Is this counting the fact that, at the Moon's distance, g is nowhere near 9.8m/s^2?
Yes.

9.8 km/s^2 has nothing to do with 9.8 km/s, unless the fall time were 1 second. The Moon's initial acceleration towards Earth would be very small (G*6e24*1.0123/384000000^2 = 0.0000027 km/s^2).

As I watched the simulation, I kept thinking "Janus is wrong... Janus is wrong". The Moon was moving way too slow, even after it crossed geosychronous altitude, for me to think it had a chance to build to 9.8. But as it got close to Earth's surface, Earth's gravity started pulling big time. At surface-to-surface contact, it was ~9.8- 9.9.

The numerical method relies on the facts that:

* The Moon starts with 0.0 km/s of velocity relative to Earth and at a distance (D) of either lunar perigee, lunar apogee, or somewhere inbetween..
* And that acceleration due to Earth's gravity at that moment will be G(M+m)/D^2.
* And that after delta t (t=time), this acceleration will produce a velocity of v(t)
* And that after delta t, this velocity will produce a change of position of d(v,delta t).
* And that acceleration due to Earths gravity at the new distance will be G(M+m)^newD^2
* And that as delta t is made smaller and smaller, the simulated answer will asymptotically approach the true answer
* And that a delta t of 1 second is sufficient enough to get an answer correct to > 0.1 km/s, while spending less than 2 minutes of computer time.
* And if you make the computer repeat this process again and again until D < (Earth Radii + Moon Radii), you will have numerically computed v and t.