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Is it true that the value of g on the moon is $\frac{1}{6}$ th of its value on the earth?

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True

False

answer is A.

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Detailed Solution

The above statement is true.
As we know, “The moon's mass is

\frac{1}{100}

times and

\frac{1}{4}

times the earth's radius”.
Since,

g_{moon} = \frac{G M_{m}}{{R_{m}}^{2}} = \frac{G M_{e}}{100 {(\frac{R_{e}}{4})}^{2}}

, Mm =

\frac{M_{e}}{100}

and Rm =

\frac{R_{e}}{4}

gmoon = moon’s gravity

{M_{m}}_{}

= mass of the moon

M_{e}

= mass of the earth

R_{m}

= radius of the moon

R_{e}

= radius of the earth

g_{moon} = \frac{16 G M_{e}}{100 R_{e}^{2}}

— eqn (1)

g_{earth} = \frac{G {M_{e}}_{}}{R_{e}^{2}}

— eqn (2)
On dividing eqn 1 by eqn 2, We get

g_{moon} = \frac{1}{6} g_{earth}

That's why the weight of any object on monon is

\frac{1}{6}

th as compared to the weight on the earth.

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