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 he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces: a spherical shell of negligible thickness of mass \(3M/4\) and a point mass \(M/4\) at the centre. Change in the force of gravity experienced by the man is

• A. \(\dfrac{3}{4} \dfrac{GMm}{R^2}\)
• B. 0
• C. \(\dfrac{1}{3} \dfrac{GMm}{R^2}\)
• D. \(\dfrac{4}{3} \dfrac{GMm}{R^2}\)

Hint:

The "sudden jump" in force when crossing a shell boundary is always equal to the force that the shell itself would have exerted if you were just outside it.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Shell Theorem, the gravitational force exerted by a uniform spherical shell on an object outside it is as if the entire mass were at the center. However, the force exerted by the shell on an object inside it is zero.

Step 2: Key Formula or Approach:
1. Force outside a sphere: \(F_{out} = \frac{GMm}{r^2}\)
2. Force inside a shell: \(F_{shell\_in} = 0\)
3. Change in force \(\Delta F = F_{just\_outside} - F_{just\_inside}\)

Step 3: Detailed Explanation:
1. Just outside the surface (\(r \approx R\)): The man experiences gravity from the entire mass \(M\) (shell + point mass). \[ F_1 = \frac{G(M)m}{R^2} \] 2. Just inside the surface (\(r<R\)): The man is now inside the spherical shell of mass \(3M/4\). The shell exerts zero force on him. He only experiences the gravitational pull of the point mass \(M/4\) located at the center. \[ F_2 = \frac{G(M/4)m}{R^2} = \frac{1}{4} \frac{GMm}{R^2} \] 3. Change in force: \[ \Delta F = F_1 - F_2 = \frac{GMm}{R^2} - \frac{1}{4} \frac{GMm}{R^2} = \frac{3}{4} \frac{GMm}{R^2} \]

Step 4: Final Answer
The change in the force of gravity experienced by the man is \(\dfrac{3}{4} \dfrac{GMm}{R^2}\).