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Formulas for classical gravity

Gravity at Altitude

Gravity is an inverse square law. Therefore the force of gravity at two different distances from the same mass can be found by:

{$$ \large \left(\frac{g_1}{g_2} \right) = \left(\frac{R^2_2}{R^2_1} \right) $$}

Example: At what height is the force of gravity half of what it is at the surface of the Earth?

The center of gravity of Earth is taken to be the center of the Earth. We want the gravity to be half, so:

{$$ \large \left(\frac{g_1}{ {g_1\over{2}} } \right) = 2 = \left(\frac{R^2_2}{R^2_1} \right) $$}

If we say the surface of the Earth is 1 Earth radius from the center of mass then clearly the height we want is {$ \sqrt{2} \text{ Earth radiuses} $}.

{$$ \large \begin{align} 2 &= \left(\frac{R^2_2}{1} \right) \cr 2 &= R^2_2 \cr \sqrt{2} &= R_2 \end{align} $$}

The radius of the Earth is 6371 km (that is an average because the Earth is not a perfect sphere). So the answer is about 9010 km less 6372 km = 2638 km which is the height about the surface.


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