Question

A planet has a mass \[\dfrac{1}{10}\] and radius \[\dfrac{1}{3}\] of that of earth. A person on earth can throw a stone to a height of 90m, then he will be able to throw the stone on the former planet to a height of –
A) 90m
B) 40m
C) 100m
D) 45m

Answer 1

Hint: We are given the mass and radius of a planet relative to earth. We can find the gravitational force and the acceleration due to gravity on this planet relative to earth and finally get the answer for the height at which the stone can be thrown.

Complete answer:
We know that the gravitational force is an essential force which acts on any object in a planet by virtue of which the objects weigh. The mass and the radius of the planet determines the effect of gravity on any object on its near surface. We know that the acceleration produced by the earth is \[9.8m{{s}^{-2}}\]. This acceleration is accountable for the downward movement or the limiting upward movement of an object thrown up in earth. Similarly, all masses have a particular gravitational force due to its mass and radius.
 Now, let us consider the given planet which has a mass and radius as \[\dfrac{1}{10}\text{th and }\dfrac{1}{3}th\] respectively as that of earth. We know that the gravitational force as derived from the Newton’s law of gravitation to be –
Where, F is the gravitational force acting on a body of mass ‘m’ due to earth’s mass ’M’ and radius ‘R’.
Now we know that,
\[\begin{align}
  & M'=\dfrac{M}{10} \\
 & \text{and,} \\
 & R'=\dfrac{R}{3} \\
\end{align}\]
Also, we know the relation of the radius and mass to the acceleration due to gravity as –
\[g=\dfrac{GM}{{{R}^{2}}}\]
We can substitute these in the force law to get the force on the unknown planet as –
\[\begin{align}
  & F'=\dfrac{GM'm}{{{R}^{'2}}} \\
 & \Rightarrow \text{ }F'=\dfrac{G\dfrac{M}{10}m}{{{(\dfrac{R}{3})}^{2}}} \\
 & \Rightarrow \ F'=\dfrac{9GMm}{10{{R}^{2}}} \\
 & \text{but, } \\
 & \text{we know that the acceleration due to gravity,} \\
 & g=\dfrac{GM}{{{R}^{2}}} \\
 & \Rightarrow \text{ }F'=\dfrac{9}{10}gm \\
\end{align}\]
Now, we know the acceleration due to gravity of the unknown planet as –
\[g'=\dfrac{9}{10}g\]
Now, let us compute the height attained by a stone which attains 90m on earth thrown by the same person. Now, according to the equations of motion –
\[{{v}^{2}}-{{u}^{2}}=2ah\]
Given an upward motion, the final velocity ‘v’ reduces to zero and the acceleration ‘a’ is the acceleration due to gravity against the motion.
\[\begin{align}
  & \Rightarrow 0-{{u}^{2}}=-2gh \\
 & \Rightarrow \text{ }u=\sqrt{2gh} \\
\end{align}\]
Now, let us apply this to both the planets as –
For earth –
\[\begin{align}
  & u=\sqrt{2gh} \\
 & \Rightarrow \text{ }u=\sqrt{2\times g\times 90}=\sqrt{180g}\text{ --(1)} \\
\end{align}\]
For the unknown planet –
\[\begin{align}
  & u=\sqrt{2g'h'} \\
 & \Rightarrow \text{ }u=\sqrt{2\times \dfrac{9}{10}g\times h'}=\sqrt{\dfrac{18}{10}gh'}\text{ --(2)} \\
\end{align}\]
Now, we know that the stone is thrown by the same person, i.e., the initial velocity ‘u’ will be the same. So, we can equate (1) and (2) –
\[\begin{align}
  & \sqrt{180g}=\sqrt{\dfrac{18}{10}gh'} \\
 & \Rightarrow \text{ 180}=\dfrac{18}{10}h' \\
 & \Rightarrow \text{ }h'=100m \\
\end{align}\]
So, the height attained by the stone on the new planet is 100m.

The correct answer is option C.

Note:
We can qualitatively relate the acceleration due to gravity by comparing the mass and the radius of the two given planets. We need to use the gravitational law when we have to get a value for some parameters like height, as we have done here.