Question

A high jumper can jump 2.0 m on earth, With the same effort how high will he be able to jump on a planet whose density is one-third and radius one-fourth those of the earth?
A. 4 m
B. 8 m
C. 18 m
D. 24 m

Answer 1

Hint: We are given the height, to which a high jumper can jump on the surface of earth. We are given the relation of density and radius of another planet with earth and are asked to find the height, to which the same jumper can jump on that planet’s surface. Since it is the same jumper and he takes the same effort on both the surfaces his potential energy on both surfaces will be the same. Hence we can equate the potential energy and solve for height on the other planet’s surface.

Formula used:
Potential energy, $PE=mgh$
Acceleration due to gravity, $g=\dfrac{GM}{{{R}^{2}}}$
Density, $\rho =\dfrac{M}{V}$
Volume, $V=\dfrac{4}{3}\pi {{R}^{3}}$

Complete answer:
In the question we are given the height up to which a jumper can jump on the surface of earth.
${{h}_{e}}=2m$, ‘${{h}_{e}}$’ is the height on earth.
We need to find the height up to which the same jumper can jump on the surface on another planet. The relation between density and radius of the planet and earth is given to us.
${{\rho }_{p}}=\dfrac{1}{3}{{\rho }_{e}}$, were ‘${{\rho }_{p}}$’ is the density of the planet and ‘${{\rho }_{e}}$’ is the density of earth.
${{R}_{p}}=\dfrac{1}{4}{{R}_{e}}$, were ‘${{R}_{p}}$’ is the radius of the planet and ‘${{R}_{e}}$’ is the radius of earth.
Since the jumper is jumping on the planet with the same effort he takes to jump on earth, the potential energy will be equal in both places, i.e.
Potential energy of the jumper on the surface of earth is equal to the potential energy of the jumper on the surface of the planet.
We know that potential energy, $PE=mgh$, where ‘m’ is the mass, ’g’ is acceleration due to gravity and ‘h’ is the height.
Therefore we can write,
$m{{g}_{e}}{{h}_{e}}=m{{g}_{p}}{{h}_{p}}$
Since mass of the jumper is constant,
$\Rightarrow {{g}_{e}}{{h}_{e}}={{g}_{p}}{{h}_{p}}$
From the above equation we can find the height, to which the jumper jumps on the planet as,
$\Rightarrow {{h}_{p}}=\dfrac{{{g}_{e}}{{h}_{e}}}{{{g}_{p}}}$
To solve this we need to find the acceleration due to gravity of earth’s surface and acceleration due to gravity of the planet’s surface.
We know the equation for acceleration due to gravity.
$g=\dfrac{GM}{{{R}^{2}}}$, where ‘G’ is gravitational constant, ‘N’ is mass and ‘R’ is radius.
From this we get the acceleration due to gravity of the earth’s surface as,
${{g}_{e}}=\dfrac{G{{M}_{e}}}{{{R}_{e}}^{2}}$
We know that, density
\[\rho =\dfrac{M}{V}\], where ‘M’ is mass and ‘V’ is volume.
Therefore, $\Rightarrow M=\rho V$
Hence we get acceleration due to gravity of the earth’s surface as,
${{g}_{e}}=\dfrac{G{{\rho }_{e}}{{V}_{e}}}{{{R}_{e}}^{2}}$
We have ${{V}_{e}}=\dfrac{4}{3}\pi {{R}_{e}}^{3}$, by substituting this in the above equation we get,
$\begin{align}
  & \Rightarrow {{g}_{e}}=\dfrac{G{{\rho }_{e}}\left( \dfrac{4}{3}\pi {{R}_{e}}^{3} \right)}{{{R}_{e}}^{2}} \\
 & \Rightarrow {{g}_{e}}=\dfrac{4}{3}\pi G{{\rho }_{e}}{{R}_{e}} \\
\end{align}$
Similarly we can find the acceleration due to gravity of the planet’s surface.
Since $g=\dfrac{GM}{{{R}^{2}}}$we have,
${{g}_{p}}=\dfrac{G{{M}_{p}}}{{{R}_{p}}^{2}}$
We know that ${{M}_{p}}={{\rho }_{p}}{{V}_{p}}$ and ${{V}_{p}}=\dfrac{4}{3}\pi {{R}_{p}}^{3}$
Therefore,
$\Rightarrow {{g}_{p}}=\dfrac{G{{\rho }_{p}}\left( \dfrac{4}{3}\pi {{R}_{p}}^{3} \right)}{{{R}_{p}}^{2}}$
$\Rightarrow {{g}_{p}}=\dfrac{4}{3}\pi G{{\rho }_{p}}{{R}_{p}}$
Now that we have acceleration due to gravity of earth’s surface and acceleration due to gravity of the planet’s surface, we can find the height, up to which the jumper can jump on the planet’s surface by substituting these in the equation,
${{h}_{p}}=\dfrac{{{g}_{e}}{{h}_{e}}}{{{g}_{p}}}$
\[\begin{align}
  & \Rightarrow {{h}_{p}}=\dfrac{\left( \dfrac{4}{3}\pi G{{\rho }_{e}}{{R}_{e}} \right){{h}_{e}}}{\left( \dfrac{4}{3}\pi G{{\rho }_{p}}{{R}_{p}} \right)} \\
 & \Rightarrow {{h}_{p}}=\left( \dfrac{{{\rho }_{e}}{{R}_{e}}{{h}_{e}}}{{{\rho }_{p}}{{R}_{p}}} \right) \\
\end{align}\]
Since ${{\rho }_{p}}=\dfrac{1}{3}{{\rho }_{e}}$ and ${{R}_{p}}=\dfrac{1}{4}{{R}_{e}}$, we can substitute this in the above equation
$\Rightarrow {{h}_{p}}=\left( \dfrac{{{\rho }_{e}}{{R}_{e}}{{h}_{e}}}{\left( \dfrac{1}{3}{{\rho }_{e}} \right)\left( \dfrac{1}{4}{{R}_{e}} \right)} \right)$
$\Rightarrow {{h}_{p}}=3\times 4\times {{h}_{e}}$
It is given in the question that, ${{h}_{e}}=2m$
$\begin{align}
  & \Rightarrow {{h}_{p}}=3\times 4\times 2 \\
 & \Rightarrow {{h}_{p}}=24m \\
\end{align}$
Therefore the jumper can jump up to a height of 24 m on the other planet.

Hence the correct answer is option D.

Note:
Acceleration due to gravity is the acceleration of a body due to the gravitational force.
Let us assume that there is a body of mass ‘m’ on the surface of earth.
Let ‘M’ be the mass of earth.
We know that the distance between the body and earth’s centre is the radius of earth ‘R’.
Now according the second law of Newton, we have
$F=\dfrac{GMm}{{{R}^{2}}}$
Since the object is on the earth’s surface there is a force acting on the body due to earth.
This force is given as, $F=ma$
But here the acceleration is acceleration due to gravity, ‘g’.
Therefore, $F=mg$
Now we can equate these two forces, thus we get
$mg=\dfrac{GMm}{{{R}^{2}}}$
$\Rightarrow g=\dfrac{GM}{{{R}^{2}}}$
Therefore we have acceleration due to gravity as, $g=\dfrac{GM}{{{R}^{2}}}$
From this it is clear that acceleration due to gravity of a body does not depend on the mass of that body.