What will be the acceleration due to gravity on the surface of the moon, if its radius is $1 / 4^{t h}$ the radius of the earth and its mass is $1 / 80$ times the mass of the earth.

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Solution

The acceleration due to gravity is: $\frac{G M}{r^{2}}$ where G is constant

M affects the result by direct proportion and r by inverse square proportion.Reducing the radius by four, increases by 16 times.Reducing the mass of factor of 80 reduces the result by same factor.
The final result is $\frac{16}{80}$ = $\frac{1}{5}$ . Hence the surface acceleration on the moon is 0.2g.

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What will be the acceleration due to gravity on the surface of the moon, if its radius is 1/4th the radius of the earth and its mass is 1/80 times the mass of the earth.

What will be the acceleration due to gravity on the surface of the moon, if its radius is $1 / 4^{t h}$ the radius of the earth and its mass is $1 / 80$ times the mass of the earth.