Question

A uniform spherical planet (Radius R) has acceleration due to gravity at its surface $g$. Points P and Q located inside and outside the planet have acceleration due to gravity $\dfrac{g}{4}$. Maximum possible separation between P and Q is
A. $\dfrac{{7R}}{4}$
B. $\dfrac{{3R}}{2}$
C. $\dfrac{{9R}}{4}$
D. None

Answer 1

Use the formulae for the variation of the acceleration due to gravity with depth and height from the surface of a planet. Calculate the depth of point P and height of point Q from the surface of the planet and take addition of these distances to determine the maximum possible distance between the points P and Q.

Formulae used:
The variation of acceleration due to gravity ${g_d}$ with depth is given by
${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$ …… (1)
Here, $g$ is acceleration due to gravity on the surface of the planet, $d$ is depth from the surface of the plane and $R$ is radius of the planet.
The variation of acceleration due to gravity ${g_h}$ with height is given by
${g_h} = g\left( {1 - \dfrac{{2h}}{R}} \right)$ …… (2)
Here, $g$ is acceleration due to gravity on the surface of the planet, $h$ is height from the surface of the plane and $R$ is the radius of the planet.

Complete step by step answer:
We have given that the radius of the spherical planet is $R$ and acceleration due to gravity at its surface is $g$.We have also given that at the points P and Q which are inside and outside the planet, the value of acceleration due to gravity is $\dfrac{g}{4}$.We have asked to determine the maximum possible distance between the points P and Q for having the same value of acceleration due to gravity.

Let point P is inside the planet at depth $d$ and point Q is outside the planet at height $h$. Hence, the maximum possible distance between these two points is
$r = d + h$ …… (3)
Let us first determine the depth of point P from the surface of the planet.Substitute $\dfrac{g}{4}$ for ${g_d}$ in equation (1).
$\dfrac{g}{4} = g\left( {1 - \dfrac{d}{R}} \right)$
$\Rightarrow 1 - \dfrac{d}{R} = \dfrac{1}{4}$
$\Rightarrow \dfrac{d}{R} = 1 - \dfrac{1}{4}$
$\Rightarrow d = \dfrac{{3R}}{4}$
This is the depth of point P from the surface of the planet.

Let us first determine the height of point Q from the surface of the planet.Substitute $\dfrac{g}{4}$ for ${g_h}$ in equation (1).
$\dfrac{g}{4} = g\left( {1 - \dfrac{{2h}}{R}} \right)$
$\Rightarrow 1 - \dfrac{{2h}}{R} = \dfrac{1}{4}$
$\Rightarrow \dfrac{{2h}}{R} = 1 - \dfrac{1}{4}$
$\Rightarrow h = \dfrac{{3R}}{8}$
This is the height of point Q from the surface of the planet.
Substitute $\dfrac{{3R}}{4}$ for $d$ and $\dfrac{{3R}}{8}$ for $h$ in equation (3).
$r = \dfrac{{3R}}{4} + \dfrac{{3R}}{8}$
$\therefore r = \dfrac{{9R}}{8}$
Therefore, the maximum possible distance between the points P and Q is $\dfrac{{9R}}{8}$.

Hence, the correct option is D.

Note: The students should correctly use the formulae for the variation of acceleration due to gravity from the surface of a planet with the depth and height. If these formulae are not used correctly then the final answer for the maximum possible distance between the points P and Q will be incorrect. Also the students should not forget to take addition of these two distances from the surface of the planet.