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Question #42275

Difficulty: 8/10 | 🟢 Free

The Question

A spherically symmetric planet of radius $R$ has an internal density profile $\rho(r) = \rho_0 \left( 1 - \frac{r}{R} \right)^2$ for $r \le R$ . A satellite orbits the planet in a circular trajectory just above its surface with an orbital period $T_0$ . Due to internal thermal processes, the planet undergoes a structural phase transition, expanding to a new radius $R'$ while strictly conserving its total mass. Its new spherically symmetric density profile becomes $\rho'(r) = \frac{\rho_0}{32} \left( 1 - \frac{r^2}{R'^2} \right)$ for $r \le R'$ . If the satellite's orbit is adjusted so that it again orbits just above the new surface $R'$ , what is the new orbital period $T'$ ?

Options

A.
$2\sqrt{2} T_0$
B.
$4\sqrt{2} T_0$
C.
$2\sqrt{5} T_0$
D.
$2 T_0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Kepler’s Third Law (Time Period)|Variation of Density with Temperature|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple non-uniform mass integration with orbital dynamics (Kepler's Third Law).

The primary trap is assuming the orbital period scales inversely with the square root of the central density ( $T \propto 1/\sqrt{\rho_c}$ ), which is only true for uniform planets and would incorrectly yield $4\sqrt{2} T_0$ .

Students must realize that the Shell Theorem dictates the orbit depends only on the total enclosed mass, forcing them to correctly integrate the changing density profiles to find the new planetary radius before applying $T^2 \propto R^3$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Gravitation - Class 11 Part 1
Concept	Kepler’s Third Law (Time Period)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free