Free Satellite Motion and Orbital Velocity MCQs with Answers
26 Satellite Motion and Orbital Velocity MCQs from Physics, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Satellite motion uses gravitational force as the centripetal force required for circular orbit, giving relationships among orbital radius, period and orbital velocity. Questions cover geostationary conditions and weightlessness, while distinguishing orbital velocity, escape velocity and the satellite's tangential speed.
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11. Suppose that thomass and radius of the Moon changes to 7.35 x 10^22 kg and 1.7 × 10º m respectively. The escape velocity of the Moon will then be(Note: The value of Gravitational constant is 6.63 x 10^-11 Nm²/kg².)
- A. 1.0 x 10^3 m/s
- B. 1.1 x 10^4 m/s
- C. 2.4 x 10^3 m/s
- D. 8.2 x 10^6 m/s
Explanation: The escape velocity (ve) is calculated using the formula: ve = √(2GM/R), where G is the gravitational constant, M is the mass of the celestial body, and R is its radius. Plugging the values: M = 7.35 × 1022 kg, R = 1.7 × 106 m, and G = 6.63 × 10-11 Nm²/kg², we find ve = √(2 * 6.63 × 10-11 * 7.35 × 1022 / 1.7 × 106) ≈ 2.4 x 103 m/s. Therefore, the correct escape velocity is approximately 2.4 x 103 m/s, which corresponds to Option C. The other options are incorrect because they either underestimate or overestimate the escape velocity based on the given values.
Correct answer: 2.4 x 10^3 m/s12. For the purpose of oceanography and meteorology, Pakistan has launched a Remote Sensing Satellite System (RSSS) at an altitude of 7 x 10^5 m above the ground station.At the time of launch, the orbital velocity of the artificial satellite was calculated as(Note: The value of gravitational constant, mass and radius of Earth are 6.67 x 10" Nm³/kg.5.9 x 10 kg and 6.3 × 10º m respectively.)
- A. 5.6 x 10^7 m/s.
- B. 6.2 x 10^7 m/s.
- C. 6.8 x 10^3 m/s.
- D. 7.4 x 10^3 m/s.
Explanation: To find the orbital velocity, we use the formula v = √(GM/(R+h)), where G is the gravitational constant (6.67 x 10-11 Nm²/kg²), M is the mass of Earth (5.9 x 1024 kg), R is the radius of Earth (6.3 x 106 m), and h is the altitude of the satellite (7 x 105 m). Calculating this gives an orbital velocity of approximately 7.4 x 103 m/s. The other options are either too high or incorrect based on the calculations.
Correct answer: 7.4 x 10^3 m/s.13. One complete of geo-stationary satellite orbit around the earth takes approximately
- A. 1 hour
- B. 24 hours
- C. 120 hours
- D. 365 hours
Explanation: A geo-stationary satellite completes an orbit around the Earth in 24 hours, which allows it to remain fixed relative to a point on the Earth's surface. This is because it orbits at the same rate that the Earth rotates. The other options (1 hour, 120 hours, and 365 hours) do not align with the Earth's 24-hour rotational period, making them incorrect.
Correct answer: 24 hours14. A body of mass m is projected from the Earth's surface. At the point of launch, the acceleration of free fall is g and the radius of the Earth is R. To escape from the gravitational field of the Earth, the speed of the body must be at least:
- A. √(gR)
- B. mgR
- C. √(2gR)
- D. mg/2R
Explanation: a) √(gR):This option suggests that the speed of the body must be at least √(gR) to escape from the gravitational field of the Earth. This is the correct option. The escape speed from the Earth's gravitational field can be calculated using the formula √(2gR), which takes into account the acceleration due to gravity (g) and the radius of the Earth (R). So, this option represents the correct relationship between g and R.
Correct answer: √(gR)15. Initial velocity of the object with which it goes out of the Earth's gravitational field, is called:
- A. Escape velocity
- B. Threshold velocity
- C. Maximum velocity
- D. Terminal velocity
Explanation: Initial velocity of the object, with which it goes out of the Earth's gravitational field, is called 'escape velocity'.
Correct answer: Escape velocity16. The minimum required velocity 10 put a satellite into the orbit is called
- A. Terminal velocity
- B. Escape velocity
- C. Critical velocity
- D. Average velocity
Explanation: The minimum speed required to put a satellite into a given orbit around earth is known as Critical velocity of the satellite.
Correct answer: Critical velocity17. The escape velocity corresponds to _ energy gained by the body, which carries it to an infinite distance from the surface of earth.
- A. Total
- B. Potential
- C. Initial kinetic
- D. None of these
Explanation: Escape velocity Is the velocity required by a body to get out of the earth's gravitational pull and leave the earth without further propulsion, which means no further acceleration is required by the object to leave the earth after the object has attained escape velocity. The initial kinetic energy which associates with movement allows the object to cover an infinite distance from the earth's surface.
Correct answer: Initial kinetic18. Value of escape velocity for the surface of the earth is 11 km/sec. Its value for surface of the moon is:
- A. 11 km/sec
- B. 10.4 km/sec
- C. 2.4 km/sec
- D. 4.3 km/sec
Explanation: The correct answer is Option C: 2.4 km/sec. The escape velocity for the Moon is approximately 2.374 km/sec. This is significantly lower than Earth's escape velocity due to the Moon's smaller mass and gravitational pull. Options A and B are incorrect because they refer to the escape velocities of Earth and Venus, respectively. Option D is also incorrect as it does not correspond to any known celestial body's escape velocity.
Correct answer: 2.4 km/sec19. A moon rotates about its axis. In the future, scientists may wish to put a satellite into an orbit around the moon such that the satellite remains stationary above one point on moon surface, the period of rotation of moon about its axis is 27.4 days,what is the radius of required orbit? Mass of the moon =7.35x10²²kg
- A. 3.59 x 10^7 m
- B. 4.23 x 10^7 m
- C. 8.86 x 10^7 m
- D. 6.96 x 10^6 m
Explanation: In this case, once the satellite has been launched, the gravitational force between the satellite and the moon would be providing the centripetal force hence the centripetal force, mrw2 is equal to the gravitational force, GMm/r2 where r is the radius of the orbit, w being the angular velocity that can be simplified to 2π/T(time period) and M being the mass of the moon. Equating the two expressions would give us r= (GMT2/4π2)1/3 Plugging in all the values, performing mathematical operations and then finally taking cube root of the expression gives you the answer i.e 8.8x10⁷m
Correct answer: 8.86 x 10^7 m20. The orbital speed of satellite orbiting around the Earth is:
- A. √GM/Re
- B. √GMe/Re
- C. √GMe/R²
- D. √GMe/h
Explanation: The correct formula to calculate the orbital speed (v) of a satellite orbiting the Earth is: v = √(GMe/r) where G is the gravitational constant, Me is the mass of the Earth, and r is the distance between the center of the Earth and the satellite. To derive this formula, we can start with the centripetal force equation F = m v² / r G (Me * m) / r² = m v² / r Solving for v, we get: v = √(GMe/r)
Correct answer: √GMe/Re