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Q.

Weight of 1 kg becomes 1/6 on the moon. If the radius of the moon is 1.768 ´10⁶ m, then the mass of the moon will be (Take $G = 6 .67 \times 10^{- 11} N - m^{2} / {kg}^{2}$ )

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a

1.99 ´10³⁰ kg

b

7.56 ´10²² kg

c

5.98 ´10²⁴ kg

d

7.65 ´ 10²² kg

answer is D.

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Detailed Solution

$g_{m} = \frac{{GM}_{m}}{R_{m}^{2}} and g_{m} = \frac{g_{e}}{6} = \frac{9 .8}{6} m / s^{2} = 1 .63 m / s^{2}$

Substituting, $R_{m} = 1 .768 \times 10^{6} m, g_{m} = 1 .63 m / s^{2}$ and $G = 6 .67 \times 10^{- 11} N - m^{2} / {kg}^{2}$

We get $M_{m} = 7 .65 \times 10^{22} kg$

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Weight of 1 kg becomes 1/6 on the moon. If the radius of the moon is 1.768 ´106 m, then the mass of the moon will be (Take G=6.67×10−11N−m2/kg2)