Question 9: Gravitation

Question

A man of mass $m$ starts falling towards a planet of mass $$M$$ and radius $$R$$. As he reaches near to the surface, he realizes that will pass through a small role in the planet. As he enters the role, we sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass $$3 M / 4$$ and a point mass $$M / 4$$ at the centre. Change in the force of gravity experienced by the man is

Accepted Answer

When the man is outside the planet and close to its surface, the gravitational force is determined by the entire mass of the planet, $$M$$. The force of gravity at the surface can be given by Newton's law of gravitation:
$$ F = \frac{G M m}{R^2} $$
However, as the man passes through the hole in the planet, the gravitational force he experiences changes due to the unique mass distribution of the planet. The planet consists of a spherical shell of mass $$\frac{3M}{4}$$ and negligible thickness, and a point mass $$\frac{M}{4}$$ at the center.
Let's analyze the gravitational force when the man is inside the planet:
1.
Spherical Shell Contribution
:
A spherical shell of uniform density exerts no net gravitational force on a particle located inside it. Thus, the man inside the shell experiences no force from the spherical shell of mass $$\frac{3M}{4}$$.
2.
Point Mass Contribution
:
The point mass at the center, which is $$\frac{M}{4}$$, exerts a gravitational force on the man. According to Newton's law of gravitation, this force is given by:
$$ F' = \frac{G \left(\frac{M}{4}\right) m}{R^2} $$
To determine the change in the force of gravity experienced by the man, we subtract the force due to the point mass from the original force at the planet's surface:
Original force at the surface:
$$ F = \frac{G M m}{R^2} $$
Force inside due to point mass:
$$ F' = \frac{1}{4} \frac{G M m}{R^2} $$
Change in the force:
$$ \Delta F = F - F'  $$
$$ \Delta F = \frac{G M m}{R^2} - \frac{1}{4} \frac{G M m}{R^2} $$
Simplifying the expression:
$$ \Delta F = \frac{G M m}{R^2} \left(1 - \frac{1}{4}\right) $$
$$ \Delta F = \frac{G M m}{R^2} \cdot \frac{3}{4} $$
$$ \Delta F = \frac{3}{4} \frac{G M m}{R^2} $$
Hence, the change in the force of gravity experienced by the man is
$$ \frac{3}{4} \frac{G M m}{R^2} $$
The correct answer is Option A: $$\frac{3}{4} \frac{G M m}{R^2}$$.

Answer

$$\frac{3}{4} \frac{G M m}{R^2}$$

Answer

0

Answer

$$\frac{1}{3} \frac{G M m}{R^2}$$

Answer

$$\frac{4}{3} \frac{G M m}{R^2}$$

Gravitation

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