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PhyMedAdv 2018 • Paper 2

Match Ratios of Satellite Orbital Parameters

A planet of mass MM has two natural satellites with masses m1m_1 and m2m_2. The radii of their circular orbits are R1R_1 and R2R_2 respectively. Ignore any gravitational interaction between the satellites. Define v1,L1,K1v_1, L_1, K_1 and T1T_1 to be, respectively, the orbital speed, angular momentum, kinetic energy and period of revolution of satellite 1; and v2,L2,K2v_2, L_2, K_2 and T2T_2 to be the corresponding quantities for satellite 2. Given m1/m2=2m_1/m_2 = 2 and R1/R2=1/4R_1/R_2 = 1/4, match the ratios given in List-I with the corresponding numbers in List-II.

LIST–ILIST–IIP. v1v21. 18Q. L1L22. 1R. K1K23. 2S. T1T24. 8\begin{array}{ll} \text{LIST--I} & \text{LIST--II} \\ \text{P. } \frac{v_1}{v_2} & \text{1. } \frac{1}{8} \\ \text{Q. } \frac{L_1}{L_2} & \text{2. } 1 \\ \text{R. } \frac{K_1}{K_2} & \text{3. } 2 \\ \text{S. } \frac{T_1}{T_2} & \text{4. } 8 \end{array}

Options

A

P→4;Q→2;R→1;S→3\text{P} \rightarrow 4; \quad \text{Q} \rightarrow 2; \quad \text{R} \rightarrow 1; \quad \text{S} \rightarrow 3

B

P→3;Q→2;R→4;S→1\text{P} \rightarrow 3; \quad \text{Q} \rightarrow 2; \quad \text{R} \rightarrow 4; \quad \text{S} \rightarrow 1

Correct
C

P→2;Q→3;R→1;S→4\text{P} \rightarrow 2; \quad \text{Q} \rightarrow 3; \quad \text{R} \rightarrow 1; \quad \text{S} \rightarrow 4

D

P→2;Q→3;R→4;S→1\text{P} \rightarrow 2; \quad \text{Q} \rightarrow 3; \quad \text{R} \rightarrow 4; \quad \text{S} \rightarrow 1

Topics & Concepts

GravitationGravitationKepler's LawsAngular Momentum

Step-by-Step Solution

To determine the correct matching between List-I and List-II, we analyze the functional dependence of each physical quantity on the satellite's mass mm and orbital radius RR.

Given data: m1m2=2,R1R2=14\frac{m_1}{m_2} = 2, \quad \frac{R_1}{R_2} = \frac{1}{4}

P. Ratio of Orbital Speeds (v1v2\frac{v_1}{v_2})

For a satellite revolving in a circular orbit around a planet of mass MM, the gravitational force provides the necessary centripetal force: GMmR2=mv2R  ⟹  v=GMR\frac{GMm}{R^2} = \frac{mv^2}{R} \implies v = \sqrt{\frac{GM}{R}}

Thus, the orbital speed v∝1Rv \propto \frac{1}{\sqrt{R}}, which is independent of the mass of the satellite.

Calculating the ratio: v1v2=R2R1=4=2\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{4} = 2

Thus, P→3\text{P} \rightarrow 3.

Q. Ratio of Angular Momenta (L1L2\frac{L_1}{L_2})

The magnitude of the angular momentum of a satellite in a circular orbit about the central planet is given by: L=mvR=m(GMR)R=mGMRL = m v R = m \left(\sqrt{\frac{GM}{R}}\right) R = m \sqrt{G M R}

Thus, the angular momentum L∝mRL \propto m \sqrt{R}.

Calculating the ratio: L1L2=(m1m2)R1R2=(2)×14=2×12=1\frac{L_1}{L_2} = \left(\frac{m_1}{m_2}\right) \sqrt{\frac{R_1}{R_2}} = (2) \times \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1

Thus, Q→2\text{Q} \rightarrow 2.

R. Ratio of Kinetic Energies (K1K2\frac{K_1}{K_2})

The kinetic energy of the satellite is given by: K=12mv2=GMm2RK = \frac{1}{2} m v^2 = \frac{GM m}{2R}

Thus, the kinetic energy K∝mRK \propto \frac{m}{R}.

Calculating the ratio: K1K2=(m1m2)(R2R1)=(2)×(4)=8\frac{K_1}{K_2} = \left(\frac{m_1}{m_2}\right) \left(\frac{R_2}{R_1}\right) = (2) \times (4) = 8

Thus, R→4\text{R} \rightarrow 4.

S. Ratio of Time Periods (T1T2\frac{T_1}{T_2})

The period of revolution for a circular orbit is given by Kepler's Third Law: T=2πRv=2πR3/2GMT = \frac{2\pi R}{v} = \frac{2\pi R^{3/2}}{\sqrt{GM}}

Thus, the time period T∝R3/2T \propto R^{3/2}.

Calculating the ratio: T1T2=(R1R2)3/2=(14)3/2=18\frac{T_1}{T_2} = \left(\frac{R_1}{R_2}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}

Thus, S→1\text{S} \rightarrow 1.

Conclusion

Matching the results: P→3;Q→2;R→4;S→1\text{P} \rightarrow 3; \quad \text{Q} \rightarrow 2; \quad \text{R} \rightarrow 4; \quad \text{S} \rightarrow 1

This corresponds to Option B.

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