How do you derive $g = \frac{G m}{r^{2}}$ $g=(Gm)/r^2$ and $V = - \frac{G m}{r}$ $V=-(Gm)/r$ ?

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How do you derive $g = \frac{G m}{r^{2}}$ $g=(Gm)/r^2$ and $V = - \frac{G m}{r}$ $V=-(Gm)/r$ ?

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Answer 1 · 2018-01-28T10:45:50

Here you go....

Explanation:

Weight of an object on Earth = Gravitational force exerted on an object by the Earth.
Let us consider a body having mass "m".
$m g = G \cdot \frac{m M}{R^{2}}$ $mg=G*(mM)/R^2$
where,
g=acceleration due to gravity,
G=Gravitational constant,
M=Mass of the Earth,
and R=radius of the Earth.
Weight of the body (on Earth)=
mass $(m) \cdot$ $(m)*$ gravitational acceleration $(g)$ $(g)$
Cancel out the $m$ $m$ on both sides to get:
$g = G \cdot \frac{M}{R^{2}} = \frac{G M}{R^{2}}$ $g=G*M/R^2=(GM)/R^2$

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How do you derive $g = \frac{G m}{r^{2}}$ $g=(Gm)/r^2$ and $V = - \frac{G m}{r}$ $V=-(Gm)/r$ ?

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Explanation:

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