SATHEE: Gravitation 5 Question 11

Gravitation 5 Question 11

12. A planet of mass $M$ , has two natural satellites with masses $m_{1}$ and $m_{2}$ . The radii of their circular orbits are $R_{1}$ and $R_{2}$ , respectively. Ignore the gravitational force between the satellites. Define $v_{1}, L_{1}, K_{1}$ and $T_{1}$ to be respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and $v_{2}, L_{2}, K_{2}$ and $T_{2}$ to be the corresponding quantities of satellite 2. Given, $m_{1} / m_{2} = 2$ and $R_{1} / R_{2} = 1 / 4$ , match the ratios in List-I to the numbers in List-II.

(2018 Adv.)

	List-I		List-II
P.	$v_{1} / v_{2}$	1.	$1 / 8$
Q.	$L_{1} / L_{2}$	2.	1
R.	$K_{1} / K_{2}$	3.	2
S.	$T_{1} / T_{2}$	4.	8

(a) $P \to 4; Q \to 2; R \to 1; S \to 3$

(b) $P \to 3; Q \to 2; R \to 4; S \to 1$

(c) $P \to 2; Q \to 3; R \to 1; S \to 4$

(d) $P \to 2; Q \to 3; R \to 4; S \to 1$

Show Answer

Answer:

Correct Answer: 12. (b)

Solution:

As, $v = \sqrt{\frac{G M}{R}}$

Let $R_{1} = R$ , then $R_{2} = 4 R$

If $m_{2} = m$ , then $m_{1} = 2 m$

List-I

(P) $\frac{v_{1}}{v_{2}} = \sqrt{\frac{R_{2}}{R_{1}}} = \sqrt{\frac{4 R}{R}} = 2 : 1$

(Q) $L = m v R$

$\frac{L_{1}}{L_{2}} = \frac{R (2 m) v_{1}}{4 R (m) v_{2}} = \frac{1}{2} (2) = 1 : 1$

(R) $\frac{K_{1}}{K_{2}} = \frac{\frac{1}{2} (2 m) v_{1}^{2}}{\frac{1}{2} (m) v_{2}^{2}} = 2 (4) = 8 : 1$

(S) $\frac{T_{1}}{T_{2}} = {\frac{R_{1}}{R_{2}}}^{3 / 2} = {\frac{1}{4}}^{3 / 2} = 1 : 8$

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