Question

A man can jump to a height of 1.5 m on a planet A. What is the height he may be able to jump on another planet whose density and radius are respectively one- quarter and one- third that of planet A.
A. 1.5 m
B. 15 m
C. 18 m
D. 28 m

Answer 1

Hint: Gravity is a fundamental interaction between two objects having a certain mass. It is the universal force that pulls objects with mass toward one another. It is a form of mutual attraction force between the interacting objects. The force of this mutual attraction depends on the size of the objects and the square of the distance separating them. So far, as known in Physics, Gravitational pull is the weakest known force existing in nature. Work is said to be done when a force moves an object over some distance. Hence work done is force applied on a body multiplied with the displacement undertaken by the body as a result of the force.

Formulas used:
Work done(W)= mg x h
\[g = \left[ {G \times \rho \times \left( {\dfrac{4}{3}} \right)\pi {R^3}} \right]{R^2} \\ \Rightarrow g = \dfrac{{GM}}{{{R^2}}}\,or\,M = \dfrac{{g \times {R^2}}}{G} \]
Where, m= mass, g = acceleration due to gravity and h= height.

Complete step by step solution:
Work is said to be done when a force moves an object over some distance. Hence work done is the force applied to a body multiplied by the displacement undertaken by the body due to the force. First, we need to find the work done by that man for jumping 1.5 m height on planet A. In this case, as gravity is the force thus, work W = mg \[ \times \] h.
⇒ W = mg \[ \times \] 1.5

According to the question,
The density of the second planet, \[\rho ' = \dfrac{\rho }{4}\]
The radius of the second planet, \[R' = \dfrac{R}{3}\]
On planet A, \[g = \dfrac{{GM}}{{{R^2}}}\]
M = density x volume = \[\rho \times \left( {\dfrac{4}{3}} \right)\pi {R^3}\]
\[g = [G \times \rho \times \left( {\dfrac{4}{3}} \right)\pi {R^3}]{R^2}\]
\[\Rightarrow g = G \times \rho \times \left( {\dfrac{4}{3}} \right)\pi R\]
In the case of the other planet,
\[g' = G \times \rho ' \times \left( {\dfrac{4}{3}} \right)\pi R'\]
\[\Rightarrow g' = \left( {\dfrac{1}{{12}}} \right)[G \times \rho \times \left( {\dfrac{4}{3}} \right)\pi R]\]
\[\Rightarrow g'= \left( {\dfrac{1}{{12}}} \right)g\]

Solving further,
\[W{\text{ }} = {\text{ }}F\; \times S \\
\Rightarrow W{\text{ }} = {\text{ }}mgh \\ \]
\[\Rightarrow {H_{\max }} = \dfrac{{{u^2}}}{{2g}}\]
\[\Rightarrow {H_{\max }} \propto \dfrac{1}{g}\]
\[\Rightarrow \dfrac{{H{'_{\max }}}}{{{H_{\max }}}} = \dfrac{g}{{g'}} = 12\]
Further solving,
\[H{'_{\max }}\] \[ = 12 \times 1.5 = \]18 m
Hence, man can jump to 18m on the second planet.

Therefore, the correct answer is option C.

Note: We know, \[W{\text{ }} = {\text{ }}F\; \times S\], where w = work done, F = force and S = displacement of the object. When gravity is the force, \[F{\text{ }} = {\text{ m}} \times a\] is now \[F{\text{ }} = {\text{ m}} \times {\text{g}}\] where g stands for acceleration due to gravity. Hence, the equation becomes, \[W{\text{ }} = {\text{ }}mgh\].