Free Gravitation MCQs with Answers

21. A satellite is moving in a circular orbit at a constant speed due to gravity.Which of the following statements is correct about the above figure?

• A. The earth does zero work on the satellite.
• B. The earth does positive work on the satellite.
• C. The earth does negative work on the satellite.
• D. The concept of work is not applicable in circular motion.

Explanation: The explanation for this question has been added!

22. Which of the following statements is absolutely correct about mass?

• A. More the mass of a body connected with spring balance, more will be the elongation in the spring balance.
• B. More the mass of body kept in one pan of beam balance, more the mass has to be kept on the other pan to keep beam horizontal.
• C. More the mass of a body, lesser will be its acceleration for a given force.
• D. All of the above

Explanation: This is the following solution: Consider the situation in gravity-free space. From Newton's second law of motion, F = ma Here, F = Force on mass m, m = Mass of the object, a = Acceleration of the object Now, for a constant force, m ∝ 1/a From the above relation, we can say that, if the mass of a body increases, its acceleration decreases.Out of the given statements, option 'C' is the correct answer as it is absolutely correct. The other statements are either partially correct or correct but not absolute. Mass plays a significant role in physics and is measured using various instruments such as spring balance, beam balance, etc.

23. If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will:

• A. Continue to move in its orbit with the same velocity
• B. Move tangentially to the original orbit in the same velocity.
• C. Become stationary in its orbit.
• D. Move towards the earth.

Explanation: This is the following solution: For a body moving around an object, it is bound by the centripetal force of the object on the body. This force is along the line connecting the body and the object. But the velocity of the body will be perpendicular to this force, which is tangent. When the force of gravity is essentially zero, there is no binding for the satellite to revolve around the Earth. Since it moves in the tangent direction with velocity 'v', it continues in the same direction with the same velocity 'v'. Hence, it moves tangentially to orbit due to the inertia of direction.

24. The kinetic energy needed to project a body of mass m from the earth's surface to infinity is :

• A. 1/4 mgR
• B. 1/2 mgR
• C. mgR
• D. 2mgR

Explanation: This is the following solution: The correct option is mgR: The kinetic energy needed to project a body of mass m from the Earth's surface to infinity is equal to the initial potential energy of the body, which is mgR.When an object is on the surface of the Earth, it has gravitational potential energy due to its position relative to the Earth's center. The gravitational potential energy is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height or distance from the reference point. In this case, the reference point is taken to be at infinity.At the Earth's surface, the height or distance from the reference point (infinity) is R, which is the radius of the Earth. Therefore, the initial potential energy of the body is given by PE = mgh = mgR.As the body is projected from the Earth's surface to infinity, the potential energy is converted into kinetic energy. At infinity, the gravitational potential energy becomes zero, and the entire energy of the body is in the form of kinetic energy.

25. The velocity with which a projectile must be fired so that it escapes Earth's gravitation does not depend on:

• A. Mass of the earth
• B. Mass of the projectile
• C. Radius of the projectile's orbit
• D. Gravitational constant

Explanation: This is the following solution: The correct option is the mass of the projectile: The escape velocity is solely determined by the mass of the celestial body from which the projectile is being launched (in this case, the Earth) and the radius from the center of the body to the projectile's starting point.The escape velocity, denoted as ve, can be calculated using the formula:ve = √((2GM)/r),where G is the gravitational constant, M is the mass of the celestial body (Earth), and r is the distance from the center of the celestial body to the projectile's starting point.

26. When a body is taken from the equator to the poles, its weight:

• A. Remains same
• B. Increases
• C. Decreases
• D. Increase at N-pole & decreases at S-pole

Explanation: The following is the solution: g′=g−Rω2cos(2λ) g′=g−Rω2cos2⁡λ, where ω is the angular velocity of rotation of Earth about its polar axis, R is the radius of the Earth and λ is the latitude of a place, At poles, λ=90, ∴g p ole=g−Rω2cos(2×90) at equator λ=0, g e quator=g−R(ω)2cos(2×0)=g−R(ω)2 Thus, the acceleration due to gravity decreases from the poles to the equator. Hence, when a body is taken from the poles to the equator on the Earth, its weight decreases.

27. If the radius of the Earth becomes half of its present value (mass remaining the same), the new length of the day would be:

• A. 6 hours
• B. 12 hours
• C. 48 hours
• D. 96 hours

Explanation: This is the following solution: T ∝ (1/2)2 T= 24 hours T= 24/ 4= 6 hours

28. If the distance between two masses is doubled, the gravitational attraction between them is:

• A. Doubled: This option is not correct. When the distance between two masses is doubled, the gravitational attraction between them does not double. In fact, it decreases.
• B. Increases four times
• C. Reduced to half
• D. Reduced to a quarter

Explanation: This is the following solution: Reduced to a quarter: This is the correct answer. According to the law of universal gravitation, the force of gravity is inversely proportional to the square of the distance between two masses. When the distance is doubled, the gravitational attraction is reduced to one-fourth of its original strength.

29. The value of 'g' on poles is:

• A. Maximum
• B. Minimum
• C. Zero
• D. None of the above

Explanation: This is the following solution: The earth's radius at poles is the least, so acceleration due to gravity will be highest at poles because g is inversely proportional to the square of the radius. Gravitational force is always inversely proportional to the distance between the earth's center and the object under study.

30. If the speed of rotation of Earth about its axis increases, then the weight of the body at the equator will:

• A. Increase
• B. Decrease
• C. Remain unchanged
• D. Sometimes decrease and sometimes increase

Explanation: The following is the solution: Decrease: This is the correct answer. As mentioned above, an increase in the speed of Earth's rotation would lead to a decrease in the effective gravitational force experienced by an object at the equator. This is because the centrifugal force (outward force due to rotation) counteracts some of the gravitational force, resulting in a reduced weight for objects at the equator. Due to rotation, the effective value of g is g'= g - ω2Rcos2λ. Now, ω increases so g′ will decrease & so weight = mg′ will reduce.