Question

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 the moon is$M$ and the radius of the moon is$R$. Both feather and hammer are released simultaneously. What is the acceleration of the hammer?

Answer 1

Hint: We would use the Newton’s law of Universal Gravitation to solve this question.
Formula used: Newton’s Law of universal Gravitation
$F = G\dfrac{{Mm}}{{{R^2}}}$
Force due to acceleration due to gravity,$F = mg$

Complete step by step answer:
According to the Newton’s law of universal gravitation, any two bodies attract each other by the force,
$F = G\dfrac{{Mm}}{{{R^2}}}$ . . . (1)
Where,
F is the force of attraction between them.
G is the universal gravitational constant$G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
$m$and$M$are the masses of respective bodies.
And R is the distance between them (calculated from their centers).
Let the hammer be released from a height h.
Then, from equation (1), we can write
$F = \dfrac{{GM100m}}{{{{(R + h)}^2}}}$ . . . (2)
Where,$100m$is the mass of the hammer.
Also, force applied on the hammer due to the gravitational field of moon will be
$F = 100m{g_m}$ . . . (3)
Where,${g_m}$is the acceleration due to gravity of moon.
Since, both the forces are equal, from equation (2) and (3), we can write
$100{g_m} = \dfrac{{GM100m}}{{{{(R + h)}^2}}}$
$ \Rightarrow {g_m} = \dfrac{{GM}}{{{{(R + h)}^2}}}$
This will be the acceleration of the human.

Note:Since, the distance between moon and hummer will be much less that the radius of the moon, we can write $\operatorname{Re} h \approx R.$
Then we will get
${g_m} = \dfrac{{GM}}{{{R^2}}}$
Further, if we want to write the above equation in terms of $g$ acceleration due to gravity on earth.
Then, we can write
$\dfrac{g}{6} = \dfrac{{GM}}{{{R^2}}}$
$ \Rightarrow g = \dfrac{{6GM}}{{{R^2}}}$
$(\therefore g = 6{g_m})$.