Question

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.

• A. P-4, Q-2, R- 1, S- 3
• B. P-3, Q-2, R- 4, S- 1
• C. P-2, Q-3, R- 1, S- 4
• D. P-3, Q-2, R- 1, S- 4

Answer 1

The correct option is B P-3, Q-2, R- 4, S- 1

As the satellites are orbiting around the plane, we can say
$\frac{m_1{v}_1^2}{r_1}=\frac{GM{m}_1}{r_1^2}$....(i)
and
$\frac{m_2{v}_2^2}{r_2}=\frac{GM{m}_2}{r_2^2}$....(ii)
Dividing (i) and (ii) we will get
$\frac{v_1}{v_2}=\sqrt{\frac{r_2}{r_1}}$
$⟹\frac{v_1}{v_2}=2$
We know that $L$ = $mvr$
$L_1=m_1{v}_1{r}_1$....(iii)
and $L_2=m_2{v}_2{r}_2$.....(iv)
Dividing (iii) and (iv) we get
$\frac{L_1}{L_2}=\frac{m_1{v}_1{r}_1}{m_2{v}_2{r}_2}$
$\frac{L_1}{L_2}=1$ ($as\frac{m_1}{m_2}=2and\frac{r_1}{r_2}=\frac{1}{4}$)
Kinetic energy = $\frac{1}{2}m{v}^2$
$⟹K_1=\frac{1}{2}{m}_1{v}_1^2$.......(v)
and $K_2=\frac{1}{2}{m}_2{v}_2^2$.........(vi)
Dividing (v) and (vi) we get
$\frac{K_1}{K_2}=\frac{m_1{v}_1^2}{m_2{v}_2^2}$
$⟹\frac{K_1}{K_2}=8$
We knoe that $T^2\propto R^3$
thus, $\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}$
$⟹\frac{T_1}{T_2}=\frac{1}{8}$