# Engineering Physics Questions and Answers – Gravitational Potential Field

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Gravitational Potential Field”.

1. If the diameter of the earth becomes half its present value but its average density remains unchanged then how would be the weight of an object on the surface of the earth affected?
a) The weight of the object remains unchanged
b) The weight is doubled
c) The weight will become one-fourth of the present value
d) The weight is halved

Explanation: Acceleration due to gravity on the surface of the earth
g=GM/R2 = G/R2 ×4/3 πR3 ρ=4/3 πGRρ
When the diameter or radius becomes half its present value,
g=4/3 πG R/2 ρ=g/2
Hence the weight of the object will be halved.

2. The centre of mass of a body is often called ___________
a) Centre of attraction
b) Centre of repulsion
c) Centre of gravity
d) Centre of acceleration

Explanation: The torque due to gravity on a body acts as its entire mass were concentrated at its centre of mass. That is why the centre of mass of a body is often called its centre of gravity.

3. The orbital speed of Jupiter is ___________
a) Greater than the orbital speed of earth
b) Less than the orbital speed of earth
c) Equal to the orbital speed of the earth
d) Proportional to the distance from the earth

Explanation: v0=√((GMsun)/r)
As the Jupiter is at a larger distance from the sun than the earth, so the orbital speed of Jupiter is less than that of the earth.

4. Statement 1- The comets do not obey the Kepler’s law of planetary motion.
Statement 2- The comets do not have elliptical orbits.
a) Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
b) Statement 1 and statement 2 are correct but statement 2 is not the correct explanation of statement 1
c) Statement 1 is true but statement 2 is false
d) Both statement 1 and 2 are false

Explanation: Both the statements are true but statement 2 is not the correct explanation of statement 1. Some of the comets are non- periodic and move along a hyperbolic or parabolic path. They do not obey Kepler’s laws of planetary motion.

5. The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B?
a) 2/9 m
b) 18 m
c) 6 m
d) 2/ m

Explanation: For the same gain of potential energy in both cases
MgA hA=mgB hB
9×gB×2=gB×hB
hB=18m

6. A ball is dropped from a spacecraft revolving around the earth at a height of 120 km. What will happen to the ball?
a) It will fall down to the earth gradually
b) It will go very far in space
c) It will continue to move with the same speed along the original orbit of spacecraft
d) It will move with the same speed, tangentially to the spacecraft

Explanation: As no external torque is being applied, by the law of conservation of angular momentum, the ball will continue to move with the same angular velocity along the original orbit of the spacecraft.

7. If the radius of earth were to shrink by one percent (its mass remains the same), then the acceleration due to gravity on the earth’s surface ___________
a) Would decrease
b) Would remain unchanged
c) Would increase
d) Cannot be predicted

Explanation: Acceleration due to gravity on the earth’s surface
g=GM/R2
If R decreases, then g increases.
Hence the acceleration due to gravity on the earth’s surface would increase if the radius of earth were to shrink by one percent.

8. If the distance between the earth and the sun were half its present value, the number of days in year would have been ___________
a) 64.5
b) 129
c) 182.5
d) 730

Explanation: According to Kepler’s law of periods
T2 αR3
(T2/T1)2=(R2/R1)3=((R1/2)/R1)3=1/8
T1=1/(2√2)×T1=0.353×365days=129days.

9. The earth revolves around the sun with an angular velocity of 2×10-7rad/s in a circular orbit of radius 1.5×108km. The force exerted by the sun on the earth in newton is?
a) 30×1021
b) 27×1039
c) Zero
d) 18×1025

Explanation: Force exerted by the sun = Centripetal force
F = Mrω2
F = 6×1024×1.5×1011×(2×10-7)2 = 36×1021 N

10. What is the weight of the body at the centre of the earth?
a) Infinity
b) Same as the weight of the body on the earth
c) Zero
d) Cannot be determined

Explanation: The weight of the body at the centre of the earth is zero. This is because the value of g is zero at the centre of the earth.

11. A clock fitted with a pendulum and another with a spring indicate correct time on earth. Which shows correct time on the moon?
a) Clock fitted with pendulum
b) Clock fitted with spring
c) Both the clocks
d) Neither of the clocks

Explanation: A clock fitted with spring will show the correct time on the moon because its time period is not affected by the variation in g.

12. What is the effect of rotation of earth on the acceleration due to gravity?
a) The acceleration due to gravity decreases
b) The acceleration due to gravity increases
c) There is no effect
d) The acceleration due to gravity first increases and then decreases

Explanation: The acceleration due to gravity decreases due to the rotation of the earth. This effect is zero at the poles and maximum at the equators.

13. What is the mass of a body that weighs 1N at a [place where g = 9.80m/s]?
a) 0kg
b) 1.02kg
c) 0.102kg
d) 102kg

Explanation: m = W/g
m=1/9.80=0.102kg.

14. Which of the following is an evidence to show that here must be force acting on earth and directed toward the sun?
a) Deviation of the falling bodies towards east
b) Revolution of the earth round the sun
c) Phenomenon of day and night
d) Apparent motion of sun around the earth

Explanation: The revolution of the earth around the sun is due to the centripetal force provided but the gravitational attraction of the sun on the earth.

Sanfoundry Global Education & Learning Series – Engineering Physics.

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