Free Gravitation MCQs with Answers

231. The ratio: gravitational force/electrostatic force between two electrons is always:

• A. Greater than unity
• B. Less than unity
• C. Equal to unity
• D. Zero

Explanation: Option A : Gravitational force / electrostatic force is not greater than unity. Option: B Electrostatic forces are much stronger than gravitational forces. This is because gravity depends on mass, atoms have tiny masses so the gravitational force between them is close to zero. Whereas, the electrostatic force related to charges is bigger. The ratio of gravitational force to electrostatic force is always less than 1.

232. What force provides the centripetal force for planets moving around the sun?

• A. Coulombs force
• B. Gravitational force
• C. Magnetic force
• D. None of these options are correct

Explanation: In the case of an orbiting planet, the force is gravity. The gravitational attraction of the Sun is an inward (centripetal) force acting on Earth. This force produces the centripetal acceleration of the orbital motion.

233. The value of acceleration due to gravity on moon is _ of earth.

• A. 1/4th
• B. 1/10th
• C. 2/3rd
• D. 1/6th

Explanation: This is a fact to be memorized. Gravitational acceleration on Moon is 1/6th the gravitational acceleration on Earth. Hence Option D is correct.

234. Acceleration due to gravity near earth is:

• A. Non uniform
• B. Uniform
• C. Decreasing with distance
• D. Increasing with time

Explanation: The acceleration due to gravity near Earth is approximately uniform. On Earth's surface, the standard acceleration due to gravity, denoted as "g," is approximately 9.81 meters per second squared (m/s²) and is relatively constant at a given location. This means that the acceleration due to gravity remains nearly the same for objects in the vicinity of Earth's surface, regardless of their mass. However, it's worth noting that the value of "g" can vary slightly depending on your location on Earth and other factors like altitude and local geology. For example, it is slightly higher at the poles and decreases slightly as you move toward the equator due to Earth's rotation and shape. Nevertheless, these variations are relatively small, and for most practical purposes, we consider the acceleration due to gravity as uniform near Earth's surface.

235. 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.

236. 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'.

237. 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.

238. On the ground, the gravitational force on a satellite is W. What is the gravitational force on the satellite when at a height R/50, where R is the radius of the Earth?

• A. 1.04 W
• B. 1.02 W
• C. 0.98 W
• D. 0.96 W
• E. 2.13 W

Explanation: Because of equality F = - m. (G.M/r ²) = - m.g one draws G = G.M/r ² = [G.M/R ²] * [1 (1+z/R)]² =g○/(1+z/R) ² ► g (z) = g○/(1+z/R) ² ▬▬▬▬▬ > W/W○ = g/g○ = (1+z/R)⁻ ² = (1 + 1/50)^-2 = 0.961

239. The dimensions of gravitational constant "G" are:

• A. [ML-2T-1]
• B. [ML-2T-2]
• C. [M2L-2T-1]
• D. [M-1L3T-2]

Explanation: Explanation: F = G (m1) (m2) / r² G = F (r²) / (m1) (m2) G = [MLT-2] [L²] / [M] [M] = [M-1L³T-2] These dimensions correspond to surface gravity or acceleration due to gravity. These dimensions correspond to force per unit mass or pressure. These dimensions correspond to the inverse of the square of a length or area.

240. 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.