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Question #42324

Difficulty: 8/10 | 🟢 Free

The Question

A satellite of mass $m$ is in a bound planar orbit around a planet of mass $M$ ( $M \gg m$ ). The satellite is connected to the center of the planet by a massless elastic tether of natural length zero and unknown spring constant $k$ . The maximum and minimum distances of the satellite from the planet's center are $3R$ and $R$ , respectively. If the speed of the satellite at its closest approach is $\sqrt{\frac{2GM}{R}}$ , what is the radius of curvature of the satellite's trajectory at its farthest point?

Options

A.
$\frac{4R}{5}$
B.
$\frac{3R}{2}$
C.
$2R$
D.
$\frac{5R}{4}$

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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

Spring Potential Energy|Orbital Velocity Energy Balance|Gravitational Field Variation

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of angular momentum conservation with the energy balance in a modified effective potential.

The primary trap is geometric: students often assume any bound orbit with defined apsides is a standard Keplerian ellipse and blindly apply the geometric formula $\rho = b^2/a = 3R/2$ .

However, the addition of the spring force alters the central force law, meaning the orbit is not a simple ellipse.

A secondary trap involves forgetting to include the spring's restoring force in the local centripetal requirement at the apogee, which would incorrectly yield $\rho = 2R$ .

The correct approach requires deriving the unknown spring constant from energy conservation and applying the exact radial equation of motion $F_{net} = m v^2 / \rho$ at the apsidal point.

Quick Facts About This Question

Question Information

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