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If so, that means gravity is only 9.8 m/s^2 at the surface of the earth?

Qmechanic
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meaningless
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2 Answers2

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Yes, $F=mg$ can be derived from applying $F=\frac{G m_1 m_2}{r^2}$ to the surface of the Earth. The acceleration felt by an object of mass $m$ dropped a distance $R_{\rm Earth}$ from the Earth's center (where $R_{\rm Earth}$ is the Earth's radius) is (see, eg, Wolfram Alpha)

$$ a = \frac{F}{m} = \frac{G M_{\rm Earth}}{R_{\rm Earth}^2} = \frac{\left(6.67 \times 10^{-11} {\rm N\ m^2\ kg^{-2}}\right)\left(5.97 \times 10^{24} {\rm kg}\right)}{\left(6.37 \times 10^6 {\rm m}\right)^2} = 9.81 {\rm m\ s^{-2}} = g $$ It's an interesting exercise to calculate the acceleration due to gravity on, say, the moon and the sun, by adjusting $M$ and $R$ in the above formula appropriately.

Andrew
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Yes, the GM/r^2 is basically g, which is 9.81 m/s^2 near surface of Earth. When further up it's drops as r increases. The m is the small mass.

Username
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