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Question #42261

Difficulty: 8/10 | 🟢 Free

The Question

A rocket of mass $m$ is propelled radially outward from the center of a uniform solid spherical planet of mass $M$ and radius $R$ . The rocket maintains a constant speed $v$ relative to the planet throughout its journey to infinity. The rocket's engine provides thrust solely to overcome the planet's gravitational pull. Let $P(t)$ be the instantaneous mechanical power delivered by the engine. Determine the discontinuity in the rate of change of power, defined as $\Delta \dot{P} = \lim_{\epsilon \to 0^+} \left( \left. \frac{dP}{dt} \right|_{r=R+\epsilon} - \left. \frac{dP}{dt} \right|_{r=R-\epsilon} \right)$ .

Options

A.
$-\frac{3GMmv^2}{R^3}$
B.
$-\frac{2GMmv^2}{R^3}$
C.
$\frac{GMmv^2}{R^3}$
D.
$0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Gravitational Field Variation (Depth vs Height)|Average vs. Instantaneous Calculus|Power-Velocity Relationship

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the piecewise nature of the gravitational field of a solid sphere and its implications on kinematics and power.

A common trap is assuming that because the gravitational field $g(r)$ (and thus the power $P$ ) is continuous at the surface $r=R$ , its derivative is also continuous, leading to an incorrect answer of 0.

Another trap is calculating only the external rate of change and forgetting to subtract the internal rate, leading to the $-2GMmv^2/R^3$ option.

It requires correctly applying the chain rule to a piecewise vector field boundary, a classic JEE Advanced technique to weed out students who rely on rote memorization of $F=mg$ without analyzing the field's gradient.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Gravitation - Class 11 Part 1
Concept	Gravitational Field Variation (Depth vs Height)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free