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Question #43555

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The Question

A satellite is placed in a constrained, non-Keplerian circular equatorial orbit around a planet of mass MM (in kg\text{kg}). The planet's rotation period is TT (in s\text{s}). By means of a radial propulsion system, the satellite's orbital period is forced to be T/2T/2. Despite being in this constrained orbit, the satellite's areal velocity (the rate at which its radius vector sweeps out area) is maintained to be exactly equal to the areal velocity of a standard, unconstrained Keplerian satellite in a synchronous orbit (orbital period TT) around the same planet. An onboard telescope operating at a wavelength λ\lambda (in m\text{m}) is used to observe the planet. Assume the planet's radius is negligible compared to the satellite's orbital radius. What is the minimum aperture diameter DD required for the telescope to resolve two point-like surface features on the planet separated by a distance dd (in m\text{m})? (Let GG be the universal gravitational constant in N⋅m2/kg2\text{N}\cdot\text{m}^2/\text{kg}^2).

Options

  • A.

    1.22λd(GMT282π2)13\frac{1.22 \lambda}{d} \left( \frac{G M T^2}{8 \sqrt{2} \pi^2} \right)^{\frac{1}{3}}

  • B.

    1.22λd(GMT216π2)13\frac{1.22 \lambda}{d} \left( \frac{G M T^2}{16 \pi^2} \right)^{\frac{1}{3}}

  • C.

    1.22λd(GMT24π2)13\frac{1.22 \lambda}{d} \left( \frac{G M T^2}{4 \pi^2} \right)^{\frac{1}{3}}

  • D.

    1.22λd(GMT232π2)13\frac{1.22 \lambda}{d} \left( \frac{G M T^2}{32 \pi^2} \right)^{\frac{1}{3}}

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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

Resolving Power Limits|Kepler’s Second Law (Areal Velocity)|Geostationary Satellite Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to decouple Kepler's Third Law from the fundamental kinematic definition of areal velocity in a non-Keplerian context.

The primary trap lies in assuming that a satellite with a forced period of T/2T/2 must naturally follow the Keplerian radius-period relationship (r3∝T2r^3 \propto T^2), which leads to an incorrect orbital radius (Option B).

By constraining the areal velocity instead, the problem forces students to use the fundamental definition vA=12r2ωv_A = \frac{1}{2} r^2 \omega.

Equating this to the areal velocity of a standard synchronous orbit requires careful algebraic substitution.

Finally, bridging this orbital mechanics result with the Rayleigh criterion for angular resolution (θ=1.22λ/D\theta = 1.22 \lambda / D) tests cross-domain synthesis, filtering out students who treat optics and gravitation as siloed topics.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptResolving Power Limits
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree