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Question #81722

Difficulty: 5/10 | 🟢 Free

The Question

A spherically symmetric planet of total mass $M$ and radius $R$ is composed of a fluid in hydrostatic equilibrium. The gravitational potential $V(r)$ in its interior ( $r \le R$ ) is given by: $V(r) = -\frac{GM}{3R} \left[ 5 - 3\left(\frac{r}{R}\right)^2 + \left(\frac{r}{R}\right)^3 \right]$ where $G$ is the universal gravitational constant. Assuming the pressure at the surface of the planet is zero, what is the pressure at the center of the planet?

Options

A.
$\frac{7 G M^2}{12 \pi R^4}$
B.
$\frac{3 G M^2}{8 \pi R^4}$
C.
$\frac{5 G M^2}{12 \pi R^4}$
D.
$\frac{G M^2}{2 \pi R^4}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Potential vs Field Relationship|Electric Field Line Geometry|Coulomb's Law Vector Form

Why This Question Is Tricky

Common Mistakes

This question tests the inverse application of gravitational field theory and hydrostatic equilibrium.

A common trap is assuming a uniform mass density when given a macroscopic parameter like total mass $M$ , leading students to blindly apply the standard central pressure formula $P_c = 3GM^2 / 8\pi R^4$ .

To succeed, students must recognize that a non-standard potential implies a non-uniform density.

They must correctly apply the potential gradient to find the local field, use the gravitational analogue of Gauss's Law (or the differential form of the shell theorem) to extract the radial density profile $\rho(r)$ , and finally integrate the hydrostatic equilibrium equation $\frac{dP}{dr} = -\rho g$ over the varying density and field.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Gravitation - Class 11 Part 1
Concept	Potential vs Field Relationship
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free