Consider a satellite of mass 'm' orbiting a planet of mass 'M' and radius 'R' in a circular orbit. Match the following quantities. Assume R E is Earth's radius and

Question

Consider a satellite of mass 'm' orbiting a planet of mass 'M' and radius 'R' in a circular orbit. Match the following quantities. Assume $R_{E}$ is Earth's radius and $M_{E}$ is Earth's mass.

List - I	List - II
(P) Ratio of the orbital speed of a satellite in a circular orbit of radius $2 R$ to its orbital speed in a circular orbit of radius $R$ .	(1) $2 π \frac{( 6 R _{E} ) ^{3}}{G M _{E}}$
(Q) Minimum additional energy required to shift a satellite of mass 'm' from a circular orbit of radius $R$ to a circular orbit of radius $2 R$ .	(2) $\frac{1}{2}$
(R) Ratio of the total mechanical energy of a satellite in a circular orbit of radius $R$ to its kinetic energy in the same orbit.	(3) $- 1$
(S) Orbital period of a geostationary satellite (approximately).	(4) $\frac{GM m}{4 R}$
	(5) $2 π \frac{( 42000 km ) ^{3}}{G M _{E}}$
	(6) $\frac{GM m}{2 R}$

TestHub · Accepted Answer

P: Orbital speed v = r GM ​ ​ . Thus, v R ​ v 2 R ​ ​ = GM / R ​ GM / ( 2 R ) ​ ​ = 2 ​ 1 ​ . Matches (2). Q: Total mechanical energy of a satellite in circular orbit is E = − 2 r GM m ​ . To shift from radius R to 2 R , the required energy is Δ E = E f ina l ​ − E ini t ia l ​ = − 2 ( 2 R ) GM m ​ − ( − 2 R GM m ​ ) = − 4 R GM m ​ + 2 R GM m ​ = 4 R GM m ​ . Matches (4). R: For a satellite in a circular orbit, total energy E = − 2 R GM m ​ and kinetic energy K = 2 R GM m ​ . The ratio K E ​ = GM m /2 R − GM m /2 R ​ = − 1 . Matches (3). S: For a geostationary satellite, the orbital period is 24 hours. The orbital radius is approximately 42000 km from Earth's center, which is about 6 R E ​ . Using Kepler's 3rd law T = 2 π G M E ​ r 3 ​ ​ , with r = 6 R E ​ , we get T = 2 π G M E ​ ( 6 R E ​ ) 3 ​ ​ . Matches (1).

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