Question

A unform spherical plant (Radius R) has acceleration due to gravity at its surface g.Points P and Q location inside and outside the planet have acceleration due to gravity g/4.Maximum possible separation between and Q is

• A. $\frac{7R}{4}$
• B. $\frac{3R}{2}$
• C. $\frac{9R}{4}$
• D. none

Answer 1

The correct option is C $\frac{9R}{4}$

Let us assume point P is outside the planet and Q inside the planet.

At a distance $r$ from the centre of the planet, the gravitational acceleration is given by

1. $r$ is outside the planet:

$=\frac{GM}{r^2}$

2. $r$ is inside the planet:

$=\frac{GMr}{R^3}$

where,

$G$ is the universal gravitational constant,

$M$ is the mass of the planet and

$R$ is the radius of the planet

If $g$ is the gravitational acceleration on the surface then,

$g=\frac{GM}{R^2}$

Therefore,

1. the position of point P is given by

$\frac{g}{4}=\frac{GM}{r^2}$

$\frac{GM}{4R^2}=\frac{GM}{r^2}$

solviing we get

$r=2R$

2. the position of point Q is given by

$\frac{g}{4}=\frac{GMr}{R^3}$

$\frac{GM}{4R^2}=\frac{GMr}{R^3}$

solviing we get

$r=\frac{R}{4}$

Thus the maximum possible distance between points P and Q will be

$2R+\frac{R}{4}=\frac{9R}{4}$

Correct option is C