Question

A satellite of mass m is revolving around earth $(M, R)$ in elliptical orbit with minimum and maximum distance $2R$ and $4R$ respectively from centre of earth

List-I	List-II
P. Maximum speed of satellite	1. $\frac{-GMm}{6R}$
Q. Minimum speed of satellite	2. $\sqrt{\frac{GM}{3R}}$
R. Total energy of satellite	3. $\sqrt{\frac{2GM}{3R}}$
S. Time period of satellite	4. $6\pi \sqrt{\frac{3R^3}{GM}}$

                        | 5. $\sqrt{\frac{GM}{6R}}$   

• A. P→ 3; Q→ 5; R → 1; S→ 4
• B. P→ 3; Q→ 2; R→ 4; S→ 1
• C. P→ 4; Q→ 5; R → 1; S→ 4
• D. P→ 4; Q→ 2; R→ 4; S→ 5

Answer 1

Answer: (A)

P→ 3; Q→ 5; R → 1; S→ 4

Answer 2

1. The semi-major axis $a = \frac{r_{min} + r_{max}}{2} = \frac{2R + 4R}{2} = 3R$.

2. Total energy $E = -\frac{GMm}{2a} = -\frac{GMm}{6R}$. (R→1)

3. Maximum speed $v_{max}$ at $r_{min}=2R$: $v_{max}^2 = GM\left(\frac{2}{r_{min}} - \frac{1}{a}\right) = GM\left(\frac{2}{2R} - \frac{1}{3R}\right) = \frac{2GM}{3R}$. $v_{max} = \sqrt{\frac{2GM}{3R}}$. (P→3)

4. Minimum speed $v_{min}$ at $r_{max}=4R$: $v_{min}^2 = GM\left(\frac{2}{r_{max}} - \frac{1}{a}\right) = GM\left(\frac{2}{4R} - \frac{1}{3R}\right) = \frac{GM}{6R}$. $v_{min} = \sqrt{\frac{GM}{6R}}$. (Q→5)

5. Time period $T = 2\pi\sqrt{\frac{a^3}{GM}} = 2\pi\sqrt{\frac{(3R)^3}{GM}} = 2\pi\sqrt{\frac{27R^3}{GM}} = 6\pi\sqrt{\frac{3R^3}{GM}}$. (S→4)

Therefore, the correct matching is P→3, Q→5, R→1, S→4.