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Here's a hypothetical situation. There's an object, at rest, 500 km above the surface of Earth, with some force preventing it from accelerating downwards. The force is removed and the object immediately accelerates at 8.44 m/s^2 downwards. By the time it reaches the surface, it has an acceleration of 9.81 m/s^2. Assuming negligible air resistance, how would you calculate the final velocity of the object, at the instant before crashing?

Jake
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    I'm not sure this is a great fit here. More importantly, what have you done yourself to try to answer the question? Please [edit] the question to show what you have tried. – user Mar 12 '15 at 15:10

1 Answers1

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Specific energy conservation. If it falls from r2 to r1...

$$ E = KE + GE = \frac{1}{2} v^2 - \frac{GM}{r} \\ E_1 = E_2 \\ -\frac{GM}{r_2} = \frac{1}{2} v^2 - \frac{ GM}{r_1} \\ v = \sqrt{ 2 GM \left( \frac{1}{r_1} - \frac{1}{r_2} \right) } $$

For small distances, Taylor approx of 1/r for r2 about r1...

$$ \frac{1}{r_2} \approx \frac{1}{r_1}-\frac{1}{r_1^2} (r_2-r_1) \\ v \approx \sqrt{ 2 g (r_2 - r_1) } $$

You should recognize this as the usual form that comes from (1/2 m v^2 = m g h). This shows that the first expression for v is the proper generalization of the constant-gravity model which is taught first in physics classes.

The specific energy conservation is one the first-principles from which the laws of orbital mechanics are derived.

AlanSE
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