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If the mass of the earth is 80 times that of one of the moons of Jupiter and the diameter is double that of the moon and g on the surface of earth is $9.8 m / s^{2}$ . Then the value of g on the moon is

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0.49 {ms}^{- 2}

0.98 {ms}^{- 2}

4 {.9ms}^{- 2}

{49ms}^{- 2}

answer is A.

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Detailed Solution

Given:
$M_{e} = 80 x m, M_{m} = m$

$d_{e} = 2 d_{m}$

$r_{e} = 2 r, r_{m} = r$

To find:
$g_{e} = g; g_{m} = ?$

Formula:
$g = \frac{G M}{R^{2}}$

Solution:

$\frac{g_{m}}{g_{e}} = \frac{\frac{{GM}_{m}}{r_{m}^{2}}}{\frac{{GM}_{e}}{r_{e}^{2}}}$

$= \frac{M_{m} r_{e}^{2}}{M_{e} r_{m}^{2}}$

$= \frac{m \times 4 r^{2}}{80 x m \times r^{2}}$

$\frac{g_{m}}{g_{e}} = \frac{1}{20}$

$g_{m} = \frac{9.8}{20}$

$g_{m} = 0.49 {ms}^{- 2}$

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