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Question #43502

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The Question

A satellite of instantaneous mass $m$ is in a circular orbit of radius $r$ around a planet of mass $M$ . At time $t=0$ , it fires its thruster, ejecting propellant tangentially backward (opposite to its orbital velocity $v$ ) at a constant mass rate $\mu = -\frac{dm}{dt}$ and with a constant relative exhaust speed $u$ . Which of the following correctly gives the ordered pair $\left(\frac{dE_R}{dt}, P_{sys}\right)$ at $t=0$ , where $E_R$ is the satellite's total orbital mechanical energy and $P_{sys}$ is the total mechanical power generated by the thruster (rate of conversion of internal chemical energy into mechanical energy of the system)?

Options

A.
$\left( \mu u v + \frac{1}{2} \mu v^2, \; \frac{1}{2} \mu u^2 \right)$
B.
$\left( \mu u v, \; \mu u v \right)$
C.
$\left( \mu u v - \frac{1}{2} \mu v^2, \; \frac{1}{2} \mu u^2 \right)$
D.
$\left( \mu u v, \; \frac{1}{2} \mu u^2 + \mu u v \right)$

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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

Power-Velocity Relationship|Orbital Velocity Energy Balance

Why This Question Is Tricky

Common Mistakes

This question targets a lethal misconception in variable-mass systems: the misapplication of $P = Fv$ to calculate the rate of change of a rocket's mechanical energy.

Students typically assume that the power delivered to the satellite is simply the thrust multiplied by velocity ( $\mu u v$ ), ignoring that the satellite is an open system.

Because mass is being ejected, one must account for the specific kinetic and potential energy carried away by the propellant.

Differentiating the total orbital energy $E_R = \frac{1}{2}mv^2 - \frac{GMm}{r}$ correctly yields an extra $\frac{1}{2}\mu v^2$ term.

Furthermore, students often confuse the rate of work done on the rocket with the total mechanical power generated by the engine ( $P_{sys}$ ), which is strictly the kinetic energy of the exhaust in the center-of-mass frame ( $\frac{1}{2}\mu u^2$ ).

This forces a rigorous application of the Work-Energy Theorem for a system of particles.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Power-Velocity Relationship
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free