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#19
Sep1207, 02:37 AM

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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 surfacetosurface 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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