250+ TOP MCQs on Gravitation – Energy of an Orbiting Satellite | Class 11 Physics

Physics Multiple Choice Questions on “Gravitation – Energy of an Orbiting Satellite”.

1. What is the magnitude of the ratio of KE and PE of a satellite revolving around a planet in a circular orbit of radius “R”?
a) 1:4
b) 1:2
c) 2:1
d) 3:2

Answer: b
Clarification:Let the mass of the satellite be “m” and mass of the planet be “M”.
Kinetic energy (KE) = 1/2 x m x v2
= (1/2) x (G x m x M)/R
Potential energy (PE) = – (G x m x M)/R
KE : PE = -1:2
Therefore, (the magnitude of KE:PE) = 1:2.

2. What is the total energy possessed by a satellite of mass “m” orbiting above the earth of mass “M” and radius “R” at a height “h”?
a) [(G*M*m)/2*(R+h)]
b) -[(G*M*m)/2*(R+h)]
c) [(G*M*m)/(R+h)]
d) -[(G*M*m)/(R+h)]

Answer: b
Clarification: Total energy (E) of a satellite is the sum of its potential energy (PE) and kinetic energy (KE).
PE = – (G x M x m)/(R+h)
KE = (G x M x m)/(2 x (R+h))
E = PE + KE
= -(G x M x m)/(R+h) + (G x M x m)/(2 x (R+h))
= -(G x M x m)/(2 x (R+h)).

3. A satellite of mass “m” is in a circular orbit of radius 1.5R around the earth of radius R. How much energy is required to move it to an orbit of radius 2R?
a) -(G*M*m)/12R
b) (G*M*m)/12R
c) -(G*M*m)/4R
d) (G*M*m)/4R

Answer: a
Clarification: Total energy at 1.5R (TE) = -[(G*M*m)/3R]
Total energy at 2R (TE’) = -[(G*M*m)/4R]
The difference in energy is the required energy.
Difference = TE’ – TE
= -[(G*M*m)/4R] – [-[(G*M*m)/3R]]
= -[(G*M*m)/4R] + (G*M*m)/3R]
= (G*M*m)/12R.

4. A satellite revolves around the earth in a circular orbit with a velocity “v”. What is the total energy of the satellite? Let “m” be the mass of the satellite.
a) – (m*v2)/2
b) (m*v2)/2
c) (3*m*v2)/2
d) – (m*v2)

Answer: a
Clarification: Total energy (E) = KE + PE
KE = (1/2) x m x v2
PE = – 2 x (KE) [For a satellite]
= – m x v2
E = (1/2) x m x v2 + (- m x v2)
= – (1/2) x m x v2.

5. For a satellite with an elliptical orbit and not a circular orbit, the kinetic energy varies with time.
a) True
b) False

Answer: a
Clarification: The kinetic energy of a satellite is;
KE = (1/2) x (G x m x M)/(R+h)
h: Height above the surface of the planet
For an elliptical orbit, “h” changes with time. Hence, the kinetic energy also varies with time.

6. For a satellite with an elliptical orbit and not a circular orbit, the potential energy varies with time.
a) True
b) False

Answer: a
Clarification: The potential energy of a satellite is;
PE = – (G x m x M)/(R+h)
h: Height above the surface of the planet
For an elliptical orbit, “h” changes with time. Hence, the potential energy also varies with time.

7. Assume that a satellite executes perfectly circular orbits around a perfectly circular sphere. The work done by the satellite in 1 complete revolution is _____
a) (1/2) x (G x m x M)/(R+h)
b) – (G x m x M)/(R+h)
c) – (1/2) x (G x m x M)/(R+h)
d) 0

Answer: d
Clarification: The net displacement of the satellite in one complete revolution is zero. Hence, the total work which is done by the satellite in one complete revolution also amounts to zero.
Work done = Force x Displacement.

8. If the kinetic energy of a satellite is halved, its orbital radius becomes _____
a) 1/2 times
b) 2 times
c) 4 times
d) infinity

Answer: b
Clarification: The kinetic energy of a satellite is;
KE = (1/2) x (G x m x M)/r
r: Orbital radius
Let new kinetic energy be KE’
KE’ = KE/2
= [(1/2) x (G x m x M)/r]/2
= (1/2) x (G x m x M)/(2r)
= (1/2) x (G x m x M)/r’
r’: New orbital radius
r’ = 2r.

9. For a satellite executing an elliptical orbit around a planet, its minimum potential energy is at the perigee.
a) True
b) False

Answer: a
Clarification: Perigee is the point in the orbit closest to the surface of the planet. Hence, it has a maximum velocity. Therefore, at perigee, the kinetic energy is maximum and potential energy is minimum.

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