Start Learning Free

Before You Begin

Question #42273

Difficulty: 8/10 | 🟢 Free

The Question

A comet of mass mm is in a parabolic orbit around a massive star of mass MM (M≫mM \gg m). At its perihelion distance r0r_0, it passes through a localized, stationary cloud of cosmic dust. The comet sweeps up a mass Δm=m3\Delta m = \frac{m}{3} from the cloud in a completely inelastic collision. The energy dissipated during this process is entirely absorbed by the comet, exciting its internal vibrational modes. Which of the following gives the energy EvibE_{vib} absorbed by the vibrational modes and the new semi-major axis a′a' of the comet's orbit?

Options

  • A.

    Evib=GMm4r0,a′=8r07\displaystyle E_{vib} = \frac{GMm}{4r_0}, \quad a' = \frac{8r_0}{7}

  • B.

    Evib=GMm4r0,a′=6r07\displaystyle E_{vib} = \frac{GMm}{4r_0}, \quad a' = \frac{6r_0}{7}

  • C.

    Evib=GMm3r0,a′=8r07\displaystyle E_{vib} = \frac{GMm}{3r_0}, \quad a' = \frac{8r_0}{7}

  • D.

    Evib=GMm3r0,a′=6r07\displaystyle E_{vib} = \frac{GMm}{3r_0}, \quad a' = \frac{6r_0}{7}

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 Second Law (Areal Velocity)|Internal Energy Dependency|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of conservation laws in variable-mass orbital mechanics.

A common trap is assuming the dissipated energy is simply the kinetic energy of the accreted mass evaluated in the comet's initial frame (12Δmv02=GMm3r0\frac{1}{2}\Delta m v_0^2 = \frac{GMm}{3r_0}), which ignores the reduced mass of the system.

Furthermore, students often forget to use the updated mass m′m' in the semi-major axis formula E=−GMm′/2aE = -GMm'/2a, incorrectly yielding a′=6r0/7a' = 6r_0/7.

Finally, it highlights that while angular momentum is conserved during the collision, the areal velocity dA/dt=L/2mdA/dt = L/2m strictly decreases due to the mass increase, breaking standard Keplerian assumptions.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterGravitation - Class 11 Part 1
ConceptKepler’s Second Law (Areal Velocity)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree