Question

If a person can jump $1\,{\text{m}}$ at the surface of the Earth, he will be able to jump 6 metres at the surface of the moon. Therefore, the ratio of moon’s acceleration due to gravity to Earth’s acceleration due to gravity, taken at the surfaces, would be
A. 6
B. 1/6
C. 0.6
D. 0.06

Answer 1

Use the third kinematic equation for the final velocity of an object. This kinematic equation gives the relation between the final velocity, initial velocity, acceleration and displacement of the object. Rewrite this equation for the person jumping and derive the relation between the acceleration due to gravity and the height for which the man can jump and solve it for the required ratio.

Formula used:
The third kinematic equation for final velocity of an object is given by
${v^2} = {u^2} + 2as$ …… (1)
Here, $v$ is the final velocity of the object, $u$ is initial velocity of the object, $a$ is acceleration of the object and $h$ is displacement of the object.

Complete step by step answer:
We have given that the person can jump $1\,{\text{m}}$ high on the surface of the Earth and the same person can jump $6\,{\text{m}}$ high on the surface of the moon.
${h_E} = 1\,{\text{m}}$
$\Rightarrow{h_M} = 6\,{\text{m}}$
We have asked to determine the ratio of the acceleration due to gravity on the surface of the Earth and moon.Let us first rewrite equation (1) for the person jumping.
${v^2} = {u^2} - 2gh$
When the person jumps on the surface of the Earth or moon, its final velocity $v$ becomes zero when he reaches the maximum attainable height. Hence, the above equation becomes
${u^2} = 2gh$
When the person jumps, its initial velocity $u$ is the same either on the surface of the Earth or moon. Thus, we can write that the acceleration due to gravity $g$ on the surface of any planet is inversely proportional to the height $h$ up to which the person can jump.
$g \propto \dfrac{1}{h}$
Let us rewrite the above relation for the acceleration due to gravity on the surface of the Earth and moon.
$\dfrac{{{g_E}}}{{{g_M}}} = \dfrac{{{h_M}}}{{{h_E}}}$
Here, ${g_E}$ and ${g_M}$ are the acceleration due to gravity on the surface of the Earth and moon respectively.
Substitute $6\,{\text{m}}$ for ${h_M}$ and $1\,{\text{m}}$ for ${h_E}$ in the above equation.
$\dfrac{{{g_E}}}{{{g_M}}} = \dfrac{{6\,{\text{m}}}}{{1\,{\text{m}}}}$
$\Rightarrow \dfrac{{{g_E}}}{{{g_M}}} = 6$
$\therefore \dfrac{{{g_M}}}{{{g_E}}} = \dfrac{1}{6}$
Therefore, the ratio of acceleration due to gravity on the surface of the moon and the Earth is $\dfrac{1}{6}$.

Hence, the correct option is B.

Note: One can also solve the same question by another method. One can use the law of conservation of energy for the person. The kinetic energy with which the person jumps from the surface is equal to the potential energy of the person and derives the same relation between the acceleration due to gravity and height up to which the person can jump and solve the rest by the same method.