Given Newton's gravity formula, $F = G \cdot \frac{M \cdot m}{r^{2}}$ $F=G(Mm)/(r^2)$ , what are the SI units of the universal gravity constant, $G$ $G$ ?

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Given Newton's gravity formula, $F = G \cdot \frac{M \cdot m}{r^{2}}$ $F=G(Mm)/(r^2)$ , what are the SI units of the universal gravity constant, $G$ $G$ ?

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Answer 1 · 2016-03-19T12:16:05

$G$ $G$ has the SI units $\frac{m^{3}}{k g \cdot s^{2}}$ $m^3/(kg*s^2)$

Explanation:

$F$ $F$ is the force of gravity: $\frac{k g \cdot m}{s^{2}}$ $(kg*m)/s^2$

$M$ $M$ and $m$ $m$ are masses, in $k g$ $kg$

$r$ $r$ is the distance between the masses, in $m$ $m$ (meters)

This gives

$\frac{k g \cdot m}{s^{2}} = G \times \frac{k g^{2}}{m^{2}}$ $(kg*m)/s^2 = G xx (kg^2)/(m^2)$

$m^{2} \cdot \frac{k g \cdot m}{s^{2}} = G \times k g^{2}$ $m^2*(kg*m)/s^2 = G xx kg^2$
$m^{2} \cdot \frac{k g \cdot m}{s^{2}} \cdot \frac{1}{k^{2}} = G$ $m^2*(kg*m)/s^2*1/k^2 = G$
$m^2*(cancel(kg)*m)/s^2*1/k^cancel(2) = G$
$(m^3)/s^2*1/k = G$
$G = m^3/(kg*s^2)$

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Given Newton's gravity formula, $F = G \cdot \frac{M \cdot m}{r^{2}}$ $F=G(Mm)/(r^2)$ , what are the SI units of the universal gravity constant, $G$ $G$ ?

1 Answer

Explanation:

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Given Newton's gravity formula, F=G*(M*m)/(r^2)F=G⋅M⋅mr2F=G*(M*m)/(r^2), what are the SI units of the universal gravity constant, GGG?

1 Answer

Explanation:

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Given Newton's gravity formula, $F = G \cdot \frac{M \cdot m}{r^{2}}$ $F=G(Mm)/(r^2)$ , what are the SI units of the universal gravity constant, $G$ $G$ ?