Question #21bcd

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Question #21bcd

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Answer 1 · 2016-10-24T14:54:59

Yes

Explanation:

As per Law of Universal Gravitation the force of attraction between two bodies is given by the expression

$F_{G} = G \frac{M_{1} . M_{2}}{r^{2}}$ $F_G =G (M_1.M_2)/r^2$

Where $G$ $G$ is the proportionality constant.
It has the value $6.67408 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ $6.67408 xx 10^-11 m^3 kg^-1 s^-2$

If $M_{p}$ $M_p$ is the mass of planet and $R_{p}$ $R_p$ is its radius. The force acting on a mass $m$ $m$ having radius $r << R_{p}$ $r"<<"R_p$ located on the surface of the planet is

$F_{G} = G \frac{M_{p} . m}{R_{p}^{2}}$ $F_G =G (M_p.m)/R_p^2$
Comparing above with the expression between mass and weight
$W = m g$ $W=mg$
we get expression for gravity of the planet as
$g_{p} = G \frac{M_{p}}{R_{p}^{2}}$ $g_p =G (M_p)/R_p^2$
$\Rightarrow g_{p} \propto M_{p}$ $=>g_p propM_p$

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Question #21bcd

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