JEE Advanced 2018 · Paper 2 · Physics Q16

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PhysicsList match+3 / -1Question 16 of 54

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.

LIST-I

P
$\dfrac{v_1}{v_2}$
Q
$\dfrac{L_1}{L_2}$
R
$\dfrac{K_1}{K_2}$
S
$\dfrac{T_1}{T_2}$

LIST-II

1
$\dfrac{1}{8}$
2
1
3
2
4
8

P → 4; Q → 2; R → 1; S → 3

P → 3; Q → 2; R → 4; S → 1

P → 2; Q → 3; R → 1; S → 4

P → 2; Q → 3; R → 4; S → 1

Answer and worked solution

Official answer: B

For circular orbits, $v=\sqrt{GM/R}$ and $T\propto R^{3/2}$ . With $R_1/R_2=\tfrac14$ and $m_1/m_2=2$ :

P. $\dfrac{v_1}{v_2}=\sqrt{\dfrac{R_2}{R_1}}=2$ , which is 3.
Q. $\dfrac{L_1}{L_2}=\dfrac{m_1v_1R_1}{m_2v_2R_2}=2\times2\times\tfrac14=1$ , which is 2.
R. $\dfrac{K_1}{K_2}=\dfrac{m_1v_1^2}{m_2v_2^2}=2\times4=8$ , which is 4.
S. $\dfrac{T_1}{T_2}=\left(\tfrac14\right)^{3/2}=\tfrac18$ , which is 1.

P → 3; Q → 2; R → 4; S → 1.

(B)

Topics:Satellites and orbital motion Kepler's laws

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