Free Gravitation MCQs with Answers

211. A satellite appears to be at rest when seen from the equator. Its height from the Earth's surface is nearly

• A. 35600 km
• B. 356000 km
• C. Such a satellite cannot exist
• D. 6400 km

Explanation: The satellite appears to be at rest when seen from the equator because it orbits the Earth at the same angular velocity as the Earth's rotation. This type of satellite is called a geostationary satellite. The height of a geostationary satellite from the Earth's surface is approximately 35600 km.

212. Height of closest orbit of earth is

• A. 400m
• B. 40 km
• C. 400 km
• D. 36000 km

Explanation: This is the closest to the average height for Low Earth Orbit (LEO).LEO satellites typically range from an altitude of around 160 km to 2000 km.An average value of 400 km falls within this range and represents the most common altitude for LEO satellites.

213. An artificial satellite orbiting around the earth does not fall down because the attraction of the earth

• A. vanishes at such distances
• B. is balanced by the attraction of the moon
• C. is balanced by the centrifugal force
• D. provides the necessary acceleration for its motion in a curved path

Explanation: An artificial satellite orbiting around the Earth does not fall down because the gravitational force of the Earth provides the necessary centripetal acceleration for the satellite's motion in a curved path. The satellite moves at a constant speed while continuously falling toward the Earth, but its forward motion matches the curvature of the Earth, leading to a stable orbit.

214. Which is constant for a satellite in orbit?

• A. Velocity
• B. K.E
• C. Angular momentum
• D. P.E

Explanation: For a satellite in orbit, the angular momentum (L) is constant.

215. The geostationary satellites are

• A. Stationary W.R.T Earth
• B. Rotating W.R.T Earth
• C. Rotating very fast
• D. Rotating very slow

Explanation: Geostationary satellites appear stationary relative to a fixed point on Earth's surface because they orbit the Earth at the same rate as the Earth's rotation.

216. An artificial satellite of earth releases a packet. If air resistance is neglected, the point where the packet will hit will be?

• A. Behind
• B. Ahead
• C. Exactly below
• D. it will never reach earth

Explanation: Here is the explanation of the correct optionFc = mv2/rThe satellite will move in a circular orbit. When the packet is released it retains the tangential velocity of satellite.The packet will know act as a free falling body it will continue to follow a trajectory determined by initial tangential velocity and gravitational acceleration.The packet retains the horizontal velocity of the satellite upon the release and because no horizontal forces act on it will continue to move horizontally at the same speed as satellite.So it will revolve the earth like the satellite

217. 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.

218. 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.

219. 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.

220. A man of weight \( w \) is standing on an elevator which is ascending with an acceleration \( a \). The apparent weight of the man is

• A. \( mg \)
• B. \( ma - mg \)
• C. \( mg + ma \)
• D. \( mg - ma \)

Explanation: The apparent weight of a person in an elevator is the sum of the gravitational force and the force due to the elevator's acceleration