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Mathematics
General Math
Correctness of Ball-Dropping Problem Answer
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[QUOTE="skeeter, post: 6776022, member: 1088"] Worked this out long ago and arrived at the same conclusion. I'm used to using $r(t)$ instead of $y(t)$. $r'' = -\dfrac{GM_e}{R_e^3} \cdot r$ $r'' = -w^2 \cdot r \implies \omega = \sqrt{\dfrac{GM_e}{R_e^3}}$ $r = R_e \cos(\omega t)$ $r' = -R_e \omega \sin(\omega t)$ $r'' = -R_e \omega^2 \cos(\omega t)$ what's interesting is the period of motion ... $T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{R_e^3}{GM_e}} \implies T^2 = \dfrac{4\pi^2 R_e^3}{GM_e}$ ... Kepler's law of harmony. The time it takes from ball drop to return from the tunnel is the same as if the ball were orbiting the Earth close to its surface (no air resistance, of course). [/QUOTE]
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Mathematics
General Math
Correctness of Ball-Dropping Problem Answer
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