Step 1: Understanding the Concept:
The gravitational force outside a spherically symmetric mass distribution behaves as if the entire mass is concentrated at the center. However, inside a uniform spherical shell, the net gravitational force exerted by the shell is zero (Shell Theorem).
Step 2: Key Formula or Approach:
Newton's law of universal gravitation: \[ F = \frac{GMm}{r^2} \] Step 3: Detailed Explanation:
Just outside the surface of the planet:
The planet's total mass is $M = \frac{3M}{4} (\text{shell}) + \frac{M}{4} (\text{point mass})$. At a distance $R$ (just outside), the entire mass behaves as a point mass at the center. The force experienced by the man is: \[ F_{\text{out}} = \frac{G M m}{R^2} \] Just inside the hole (inside the spherical shell):
Once the man crosses the spherical shell of mass $\frac{3M}{4}$, this shell no longer exerts any net gravitational force on him (by Gauss's Law for gravity or Shell Theorem). The only force acting on him is due to the point mass $\frac{M}{4}$ situated at the center, which is still at a distance $R$ away. The force experienced by the man inside is: \[ F_{\text{in}} = \frac{G (\frac{M}{4}) m}{R^2} = \frac{1}{4} \frac{GMm}{R^2} \] Change in force:
The sudden change in the force as he crosses the shell is: \[ \Delta F = F_{\text{out}} - F_{\text{in}} = \frac{GMm}{R^2} - \frac{1}{4} \frac{GMm}{R^2} = \frac{3}{4} \frac{GMm}{R^2} \] Step 4: Final Answer:
The change in the force of gravity experienced is $\frac{3}{4} \frac{GMm}{R^2}$.