Free Satellite Motion and Orbital Velocity MCQs with Answers

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.

12. 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.

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.

14. 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.

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'.

16. 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.