Question

A high jumper can jump 2m on the earth. With the same effort, how high will he be able to jump on a planet whose density is $1/{3^{{\text{rd}}}}$ and radius is $1/{4^{{\text{th}}}}$ of the earth?
A. 12 m
B. 2 m
C. 8 m
D. 24 m

Answer 1

Recall the formula for acceleration due to gravity. Express the mass of the planet in terms of its density and volume. Use the expression for mass in the formula for acceleration due to gravity. Determine the acceleration due to gravity of the planet in terms of acceleration due to gravity of the earth. Use a kinematic equation to determine the height attained by the high jumper.

Formula used:
$g = \dfrac{{GM}}{{{R^2}}}$
Here, G is the gravitational constant and M is the mass of earth.
The mass of the sphere of radius R and density $\rho$ can be expressed as,
$M = \dfrac{4}{3}\pi {R^3}\rho$
${v^2} = {u^2} - 2gs$
Here, v is the final velocity, u is the initial velocity, and g is the acceleration due to gravity and s is the distance.

Complete step by step solution:
We know that a measure of how high we can jump on a certain planet depends on the acceleration due to gravity of that planet.
We assume the radius and density of the earth is R and $\rho$ respectively and the radius and density of that planet is $R'$ and $\rho '$.
We have the expression for acceleration due to gravity,
$g = \dfrac{{GM}}{{{R^2}}}$ …… (1)
Here, G is the gravitational constant and M is the mass of earth.
The mass of the earth of radius R and density $\rho$ can be expressed as,
$M = \dfrac{4}{3}\pi {R^3}\rho$ …… (2)
Using equation (2) in equation (1), we get,
$g = \dfrac{4}{3}G\pi R\rho$ …… (3)
Similarly, we can express the acceleration due to gravity of that planet as follows,
$g' = \dfrac{4}{3}G\pi R'\rho '$ …… (4)
We have given that the density of that planet is $1/{3^{{\text{rd}}}}$ of the earth and radius is $1/{4^{{\text{th}}}}$ of the radius of earth. Therefore, we can write,
$\rho ' = \dfrac{\rho }{3}$ and, $R' = \dfrac{R}{4}$
Using the above two equations in equation (4), we get,
$g' = \dfrac{4}{3}G\pi \left( {\dfrac{R}{4}} \right)\left( {\dfrac{\rho }{3}} \right)$
$\Rightarrow g' = \dfrac{{\dfrac{4}{3}G\pi R\rho }}{{12}}$
Using equation (3) in the above equation, we get,
$g' = \dfrac{g}{{12}}$ …… (5)
From the kinematic equation, we have,
${v^2} = {u^2} - 2gs$
Here, v is the final velocity, u is the initial velocity, and s is the height attained by the high jumper.
Since the final velocity at highest point is zero, we can write the above equation as follows,
${u^2} = 2gs$
$\Rightarrow s = \dfrac{{{u^2}}}{{2g}}$
Therefore, we can write,
$s \propto \dfrac{1}{g}$
We assume the height attained by the high jumper on earth is h and that of on the planet is$h'$. Therefore, we can write,
$\dfrac{{h'}}{h} = \dfrac{g}{{g'}}$
Using equation (5) in the above equation, we get,
$\dfrac{{h'}}{h} = \dfrac{g}{{\dfrac{g}{{12}}}}$
$\Rightarrow \dfrac{{h'}}{h} = 12$
Substituting 2 m for h in the above equation, we get,
$\dfrac{{h'}}{2} = 12$
$\Rightarrow h' = 24\,m$

So, the correct answer is “Option D”.

Note:
To answer this question, students should remember the formulae for acceleration due to gravity and density of the sphere. While using the kinematic equation, the sign of acceleration due to gravity should be taken positive for downward motion while negative for upward motion of the body.