A feather of mass $m$ and a hammer of mass $100 m$ are both released from rest from the same height on the surface of the moon. Mass of moon is $M$ and radius of moon is $R$ . Both feather and hammer are released simultaneously. What is the acceleration of the hammer?

\frac{m v^{2}}{r}

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\frac{G M}{R^{2}}

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\frac{G M m}{R^{2}}

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100 \frac{G M}{R^{2}}

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100 \frac{G M m}{R^{2}}

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Solution

The correct option is B $\frac{G M}{R^{2}}$
The acceleration of both the objects is the acceleration due to gravity at the surface of the moon= $g_{m o o n} = \frac{G M_{m o o n}}{R_{m o o n}^{2}} = \frac{G M}{R^{2}}$

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A feather of mass m and a hammer of mass 100m are both released from rest from the same height on the surface of the moon. Mass of moon is M and radius of moon is R. Both feather and hammer are released simultaneously. What is the acceleration of the hammer?

The correct option is B GMR2The acceleration of both the objects is the acceleration due to gravity at the surface of the moon=gmoon=GMmoonR2moon=GMR2

A feather of mass $m$ and a hammer of mass $100 m$ are both released from rest from the same height on the surface of the moon. Mass of moon is $M$ and radius of moon is $R$ . Both feather and hammer are released simultaneously. What is the acceleration of the hammer?

The correct option is B $\frac{G M}{R^{2}}$
The acceleration of both the objects is the acceleration due to gravity at the surface of the moon= $g_{m o o n} = \frac{G M_{m o o n}}{R_{m o o n}^{2}} = \frac{G M}{R^{2}}$