Physics Questions and Answers – Acceleration due to Gravity below and above the Surface of Earth

«
»

This set of Physics Problems focuses on “Acceleration due to Gravity below and above the Surface of Earth”.

1. Assuming that the earth is a perfect sphere of uniform density, the variation of acceleration due to gravity within the earth is related to “r” as _____ (“r” equals Radius of the earth when; r=R).

a) inversely proportional to r1
b)inversely proportional to r2
c) directly proportional to r1
d) directly proportional to r2
View Answer

Answer: c
Explanation: g = (G*M1)/r2;
M1 = Mass contained in the enclosed volume
r = Distance from the centre “O”
M1 = (density) x (volume)
= (density) x [(4/3) x (pi) x (r3)]
= (density x 4 x pi x r3) / 3
Therefore;
g = (G x density x 4 x pi x r3) / (3 x r2)
= (G x density x 4 x pi x r1) / 3
Hence, ‘g’ is directly proportional to “r”.
advertisement

2. Acceleration due to gravity increases as we go towards the centre of the earth.
a) True
b) False
View Answer

Answer: b
Explanation: Acceleration due to gravity decreases as we go towards the centre of the earth because the volume enclosed by the radius vector from the centre of the earth decreases as depth increases. This reduction in volume leads to a reduction in enclosed mass and hence, the acceleration due to gravity is reduced.

3. Acceleration due to gravity increases as we move away from the surface of the earth radially towards the sky.
a) True
b) False
View Answer

Answer: b
Explanation: The volume or mass enclosed by the radius vector from the centre of the earth remains the same as we move higher from the earth’s surface. However, the increase in radius decreases the acceleration due to gravity. The mathematical relationship is as follows;
g = (G*M1)/R2;
M1 = Mass of the earth
R = Radius vector from the centre of the earth.
advertisement
advertisement

4. Assume that the earth is a perfect sphere but of non-uniform interior density. Then, acceleration due to gravity on the surface of the earth _____
a) will be towards the geometric centre
b) will be different at different points on the surface
c) will be equal at all points on the surface and directed towards the geometric centre
d) cannot be zero at any point
View Answer

Answer: d
Explanation: Since we assumed the earth to have a non-uniform density, the acceleration due to gravity will not be directed towards the geometric centre. Furthermore, for a perfect sphere, the acceleration due to gravity will be equal at all points on the surface and will be non-zero.

5. Consider an object of mass “M” moving through a tunnel dug in the earth passing through the centre of the earth as shown. Which of the following is true?

a) The object always has both inertia and weight
b) The object has zero inertia throughout the tunnel
c) The object’s inertia is zero only at the centre
d) The object’s inertia is non-zero at the centre but weight is zero at the centre
View Answer

Answer: d
Explanation: Since the object always has mass, it will always have inertia. However, since acceleration due to gravity is zero at the centre of the earth, the weight of the object becomes zero at that point.
advertisement

6. Which of the following is the variation of acceleration due to gravity at a height “h” above the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth.
a) g = (G*M1)/R2
b) g = (G*M1)/h2
c) g = (G*M1)/(R + h)2
d) g = (G*M1)/(h/R)2
View Answer

Answer: c
Explanation: If “r” is the distance between the centre of the earth and the object at a height “h” above the earth’s surface, then;
R = R + h
And; g = (G*M1)/r2
= (G*M1)/(R + h)2.

7. Which of the following is the variation of acceleration due to gravity at a height “h” above the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume h < < R)
a) g = (G*M1)/R2
b) g = (G*M1)/h2
c) g = (G*M1)/(R/h)2
d) g = [(G*M1)/R2] x [1 – (2h)/R)]
View Answer

Answer: d
Explanation: If “r” is the distance between the centre of the earth and the object at a height “h” above the earth’s surface, then;
R = R + h
And; g = (G*M1)/r2
= (G*M1)/(R + h)2
= (G*M1)/R2 x (1 + h/R)-2
Since h < < R;
Using binomial expansion and neglecting higher order terms, we can write the above expression as;
g = [(G*M1)/R2] x [1 – (2h)/R)].
advertisement

8. Which of the following is the variation of acceleration due to gravity at a depth “d” below the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume the density of the earth to be constant)
a) g = (G*M1)/(R – d)
b) g = [(G*M1) x density]/d
c) g = (G x M1/R3) / (R – d)
d) g = (G x M1/R3) x (R – d)
View Answer

Answer: d
Explanation: g = (G*M)/r2;
M = Mass contained in an enclosed volume
r = distance from centre of the earth to the depth “d”
Therefore; r = R – d; where, R = Radius of the earth
M = (density) x (volume)
= (density) x [(4/3) x (pi) x (r3)]
= (density x 4 x pi x r3) / 3
= (M1 x r3) / R3
Therefore;
g = (G x density x 4 x pi x r3) / (3 x r2)
= (G x density x 4 x pi x r1) / 3
Therefore;
g = [(G x density x 4 x pi) / 3] x (R – d)
= (G x M1/R3) x (R – d).

9. What is the relationship between height “h” above the earth’s surface and a depth “d” below the earth’s surface when the magnitude of the acceleration due to gravity is equal? (Assume h < < R; where, R = Radius of the earth)
a) h = d
b) h = 2d
c) 2h = d
d) 3h = 2d
View Answer

Answer: c
Explanation: Acceleration due to gravity above the surface of the earth;
g’ = [(G*M1)/R2] x [1 – (2h)/R)]
Acceleration due to gravity below the surface of the earth;
g’’ = (G x M1/R3) x (R – d)
g’ = g’’
Therefore;
[(G*M1)/R2] x [1 – (2h)/R)] = (G x M1/R3) x (R – d)
2h = d.
advertisement

10. At what height above the surface of the earth, the acceleration due to the gravity of the earth becomes 5% of that of the surface?
a) h = 0.5 R
b) h = 1.5 R
c) h = 2.5 R
d) h = 3.5 R
View Answer

Answer: d
Explanation:Acceleration due to gravity on the surface of the earth;
g = (G*M1)/R2
Acceleration due to gravity above the surface of the earth;
g’ = (G*M1)/(R + h)2
g’ = (5/100) x g
(G*M1)/(R + h)2 = (5/100) x (G*M1)/R2
R2 x 100 = (R + h)2 x 5
Taking square root on both sides;
R x 10 = (R + h) x 2.24
7.76 x R = 2.24 x h
h = 3.5 x R.

11. An object of mass “m” is displaced from the surface of the earth of radius “R” as shown below. The object’s weight in its new position is 50% of that when it was on the surface of the earth. What is the value of “h” if R = 6400km?

a) h = 1600km
b) h = 2600km
c) h = 3600km
d) h = 4600km
View Answer

Answer: b
Explanation: Since mass remains unchanged, the only variable is the acceleration due to gravity.
Acceleration due to gravity on the surface of the earth;
g = (G*M1)/R2
Acceleration due to gravity above the surface of the earth;
g’ = (G*M1)/(R + h)2
Since the weight is halved (50%), the acceleration due to gravity is also halved;
g’ = (1/2) x g
(G*M1)/(R + h)2 = (1/2) x (G*M1)/R2
R2 x 2 = (R + h)2
Taking square root on both sides;
R x 1.41 = (R + h)
0.41 R = h
0.41 x 6400 = h
h = 2624km; [approximately 2600km].

12. The time period of a simple pendulum on the surface of the earth is “T”. What will be the time period of the same pendulum at a height of 2 times the radius of the earth?
a) T
b) 2T
c) 3T
d) 4T
View Answer

Answer: c
Explanation: Time period of the simple pendulum on the surface of the earth;
T = (2 x pi) x (l / g)1/2
l = Length of the simple pendulum
The time period of the simple pendulum at a certain height above the earth’s surface;
T’ = (2 x pi) x (l / g’)1/2
g’ = Acceleration due to gravity at a certain height above the surface of the earth
g’ for a height of 2R (R = Radius of the earth);
g’ = (G*M1)/(R + 2R)2
= (G*M1)/(3R)2
= (1/9) x g
T’ = (2 x pi) x (l / (1/9)g) ½
= 3 x [(2 x pi) x (l / g)½] = 3T.

13. The net acceleration due to gravity is zero at all points inside a uniform spherical shell.
a) True
b) False
View Answer

Answer: a
Explanation: The point inside the spherical shell experiences gravitational pull by all points of point of the shell. However, the net gravitational force is zero due to vector addition. Hence, the net acceleration due to gravity is also zero.

14. Let the radius of the earth be R. Now, assume that the earth shrunk by 20% but the mass is the same. What would be the new value of acceleration due to gravity at a distance R from the centre of the earth if the value at the same distance in the previous case was “g’”?
a) g’
b) 2g’
c) 3g’
d) 4g’
View Answer

Answer: a
Explanation: The value of acceleration due to gravity before shrinking at a distance R from the centre of the earth, i.e., on the surface of the earth is;
g’ = (G*M)/R2
Where; M = Mass of the earth
Now, the earth shrunk by 20%, however, the mass remains the same. This implies an increase in density.
The value of acceleration due to gravity from an object at a point outside the object is dependent only on the distance between the centre of gravity of the object and the distance between the point and the object. It is independent of density.
Hence, the new value of acceleration due to gravity at the same distance will remain unchanged, i.e., g’.

15. The acceleration of the moon towards the earth is approximately 0.0027 m/s2. The moon revolves around the earth once approximately every 24 hours. What would be the acceleration due to gravity of the earth of the moon towards the earth if it were to revolve once every 12 hours?
a) Become half in magnitude
b) double in magnitude
c) Change direction but remain the same in magnitude
d) Remains unchanged
View Answer

Answer: d
Explanation: The value of acceleration due to gravity does not depend on the speed of revolution but only the distance between both the centres of gravity. Hence, the magnitude and direction would remain unchanged.

Sanfoundry Global Education & Learning Series – Physics – Class 11.

To practice all areas of Physics Problems, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter