|Sep11-07, 10:31 PM||#18|
Moon fall speed
Using the JPL impact calculator I get:
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)
|Sep12-07, 02:37 AM||#19|
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.
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