Physics Problems on “Acceleration due to Gravity below and above the Surface of Earth”.
1. Acceleration due to gravity increases as we go towards the centre of the earth.
a) True
b) False
Answer: b
Clarification: 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.
2. Acceleration due to gravity increases as we move away from the surface of the earth radially towards the sky.
a) True
b) False
Answer: b
Clarification: 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.
3. 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
Answer: d
Clarification: 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.
4. 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
Answer: c
Clarification: 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.
5. 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)]
Answer: d
Clarification: 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)].
6. 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)
Answer: d
Clarification: 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).
7. 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
Answer: c
Clarification: 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.
8. 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
Answer: d
Clarification: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.
9. 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
Answer: c
Clarification: 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.
10. The net acceleration due to gravity is zero at all points inside a uniform spherical shell.
a) True
b) False
Answer: a
Clarification: 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.
11. 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’
Answer: a
Clarification: 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’.
12. 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
Answer: d
Clarification: 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.