Physics

1st Paper
Chapter-06
Gravitation and Gravity
➢ Important Topics of this Chapter for Creative Question of (c) & (d):
Times Questions Board & Year the Questions Have
Importance Topic Appeared Appeared
c d CQ
T-01: Use of
 Gravitational 1 1 JB’17
Force Formula
RB’23, 22; DB’18; JB’23, 18; CB’23;
T-02: Acceleration due
 10 5 MB’23; SB’22, 19, 18; Din.B’22, 19, 18;
to Gravity
Ctg.B’19; BB’19,
T-03: Gravitational
 Intensity & 2 2 Ctg.B’23; BB’22
Potential
BB’23; CB’23; DB’22, 18; Din.B’19, 18;
 T-04: Escape Velocity 1 5
SB’18; JB’18
SB’23, 19; BB’23, 19; JB’23, 22, 19;
 T-05: Satellite Velocity 9 8 CB’23; Din.B’23; RB’22; Ctg.B’22, 19;
MB’22
 T-06: Miscellaneous 2 1 RB’23; JB’23; Din.B’23

T-01: Use of Gravitational Force Formula

01. The graph shows the distance from center of moon r, the change of gravitational force F of a body of mass
1000kg at different distances on the moon surface. [JB’17]

Given, radius of earth 6.4 × 106 m. Gravitational acceleration of earth, g = 9.8ms −2 ,

G = 6.67 × 10−11 Nm2 kg −2 .
(c) Determine the mass of moon from the given data. 3
6
(d) Compare the gravitational force of that body at a height 2.55 × 10 m from earth surface and moon
using the stem’s data. 4

1
 Solution
(c) From graph, radius of moon, R m = 1.75 × 106 m
Weight of a body of mass, m = 1000 kg on moon surface,
Wm = mg m ⇒ 1.6 × 103 = 1000 × g m ⇒ g m = 1.6 ms−2
2
GMm gm R2m 1.6×(1.75×106 )
Say, mass of moon = Mm ∴ g m = R2m
⇒ Mm = G
= 6.67×10−11
kg = 7.35 × 1022 kg (Ans.)

(d) Radius of earth, R = 6.4 × 106 m

Gravitational acceleration on earth’s surface, g = 9.8 ms−2
Height, h = 2.55 × 106 m
let us assume, gravitational acceleration at a height h from earth surface is = g′
2
g′ R2 6.4×106
∴ g
= (R+h)2 ⇒ g ′ = g (6.4×106 +2.55×106) = 5.011 ms −2
∴ At height h, weight of body of mass m = 103 kg ; We = mg ′ = 5011 N
2
Rm 2 1.75×106
From graph, at a height 2.55 × 106 m from the moon’s surface, g ′m = (R ) + g m = ((1.75+2.55)×106 ) × 1.6
m +h
ge
= 0.256 ms −1 [g m = = 1.6 ms−2 ]
6
Weight of the object at height h from the moon’s surface, Wm = mg ′m = 265 N
5011
∴ Ratio of gravitational force = = 18.91
265

T-02: Acceleration due to Gravity

01. An object of mass 10 kg is moved from the equator to the pole. Earth's annual motion and diurnal motion
are 365 days and 24 hours respectively. Distance between the Sun and the Earth is 1.5 × 1011 m, radius of
the Earth is 6.4 × 106 m, and acceleration due to gravity is 9.81 ms−2 . [RB’23]
(d) If there was no diurnal motion, would there be change in weight due to the displacement of the object
from the equatorial region to the polar region? Analyze mathematically. 4
Solution
(d) If there was no diurnal motion, ω = 0 πad s−1
⸫ Acceleration due to gravity at any point between the equator and the polar regions,
g′ = g − ω2 Rcos2 λ ⇒ g′ = g − 02 × R cos2 λ ⇒ g′ = g
⸫ There would be no change of weight.

An object of mass m is released through a tunnel that goes through the center of the
Earth. The acceleration due to gravity at the surface of the Earth is g = 9.8 ms−2 ,
02. and the radius of the Earth is R = 6.4 × 106 m [OP = 5 × 105 m].
[CB’23]

(c) Calculate the value of acceleration due to gravity at point P. 3

2
 Solution

(c) We know, g ′ = (1 − R) g
d Given, Radius of Earth, R = 6.4 × 106 m
5.9×106
d = R − OP = 6.4 × 106 − 5 × 105 =
= (1 − )× 9.8 = 0.7656 ms−2 (Ans.)
6.4×106 5.9 × 106 m

03. It is not possible for the weight of an object to be the same everywhere on this Earth, which rotates on its
axis in 24 hours.
Radius of Earth, 6.4 × 103 km and gravitational acceleration, 9.8 ms−2 . [JB’23]
(c) Determine the acceleration due to gravity at the pole of the Earth. 3
(d) In order for the weight of an object in the equatorial region of the Earth to be zero, the angular
velocity must be increased. - Analyze mathematically. 4
Solution
(c) We know,
gλ = g − ω2 R cos2 λ gλ = gravitational acceleration at the pole
2
= 9.8 − ω R cos 90° 2 g = 9.8 m−2
= 9.8 ms−2 (Ans.) λ = Latitude of pole = 90°
Radius of Earth = R
(d) Weight = mgλ = 0 ⇒ gλ = 0
Now, gλ = g − ω22 Rcos 2 λ ⇒ 0 = 9.8 − 6.4 × 106 × cos 2 0 × ω22
9.8 ω2 = final angular velocity,
⇒ ω22 = 6.4×106
Radius of Earth, R = 6.4 × 106 m
⇒ ω2 = 1.23 × 10−3 rads−1
At equator, λ = 0°
2πN 2π×1
Initial angular velocity, ω = T
= 24×60×60
= 7.27 × 10−5 rads−1
ω2 > ω ∴ Angular velocity has to be increased.

04. Limon obtained the following information from Google. Using these data, he draws two graphs shown
below. [MB’23]
Given that, mass of earth,M = 6.0 × 1024 kg; Radius of earth, R = 6.4 × 106 m
Gravitational constant,G = 6.67 × 10−11 Nm2 kg −2

Graph-1: Distance from surface to the center of the earth is h and gravitational acceleration for h distance
is g h .
Graph-2: Distance from surface towards up is h and gravitational acceleration for h distance is g h .
(c) Using the data, find the value of g at earth’s surface. 3
(d) In the stem, why are the two graphs different? Explain using mathematics. 4

3
 Solution
(c) We know,
GM Given that,
g=
R2
Gravitational constant,
6⋅67×10−11 ×6×1024 G = 6 ⋅ 67 × 10−11 Nm2 kg−2
= (6⋅4×106 )2 Mass of earth, M = 6 × 1024 kg
Radius of earth, R = 6 ⋅ 4 × 100 m
= 9 ⋅ 77ms−2 (Ans.)
(d) First graph shows the change in g from the surface to the center of the Earth.
Second graph shows the change in g from the surface to the upward direction.
As we travel from the surface to the bottom, the value of g decreases proportionally with distance.
h
Here, with increase in h the value of h/R increases. As a result, the value of 1 − g decreases too.
R
R
And, at the center of earth, g h = (1 − R) g = 0 ms −2
So, from the surface to the bottom, the value of g decreases proportionally.
R 2 R2
While going from the surface to the top, g h = (R+h) g = g [x = R + h]
x2
1
∴ gh ∝ x2
As we go up, the value of g h is inversely proportional to the square of the distance from the center of the
Earth. That’s why this graph is different from the 1st graph.

05. An artificial satellite was placed at a height of 500 km from the surface of the earth. The radius of the earth
is 6400 km and the acceleration due to gravity at the surface of the earth is 9.81 ms−2 . [RB’22]
(c) Determine the magnitude of acceleration due to gravity at the height of the stimulus. 3
Solution
GM GM
(c) We know, acceleration due to gravity in case of the Earth, g = R2
and for height h, g ′ = (R+h)2
g′ GM R2 R2 6400 2
Now, = (R+h)2 × ⇒ g ′ = (R+h)2 × g = ( ) × 9.81 ⇒ g ′ = 8.44 ms −2 (Ans.)
g GM 6400+500
∴ Acceleration due to gravity at the height of the stem = 8.44 ms −2

06. The graph for the rate of change of gravitational acceleration of earth with respect to distance is given
below: [SB’22]

Radius of earth, R = 6.4 × 106 m, Mass of earth, M = 6 × 1024 kg and gravitational constant, G =
6.673 × 10−11 Nm2 kg −2
(c) Determine the gravitational acceleration when an object stays at point A. 3
(d) Between the points A & C mentioned in the stem where will an object feel more weight? Give
mathematical analysis. 4
4
 Solution
GM 6.673×10−11 ×6×1024
(c) Gravitational acceleration at point B, g = R2
= (6.4×106 )2
= 9.775 ms−2
R
A Point is h = 2
m below the surface of earth. Then,
R
h g
Gravitational acceleration at point A, g A = (1 − R) × g = (1 − R ) × g = 2 = 4.8875 ms−2
2

Gravitational acceleration at point A is 4.8875 ms −2 (Ans.)

3R R
(d) At C point height from the surface, hc = − R=
2 2
GM
On the surface of earth g = = 9.775 ms −2
R2
GM
Again, at point C, g c = (R+hc )2
2 R −2
gc R2 1 h −2
Now, = (R+hc )2
⇒ gc = ( h ) × g ⇒ gc = g × (1 + c ) = 9.775 × (1 + ) 2
g 1+ c R R
R

3 −2
∴ g c = 9.775 (2) = 4.4 ms−2
Therefore, at point C gravitational acceleration, g c = 4.4 ms−2 which is less than the gravitational
acceleration on earth surface g = 9.8 ms−2. And from ‘c’ we get, g A = 4.8875 ms−2. Here we can see,
g A > g C . So one will feel more weight at point A.

07. Radius of the earth R = 6.4 × 106 m and period of revolution is 24 hours. The height of an artificial
satellite from the surface of earth is h = 3.6 × 107 m. Mass of earth is M = 6 × 1024 kg. [Din.B’22]

(c) What is the effective gravitational force on a mass m on the point P? 3

Solution
(c) We know, λ = 60°
m = 2 kg
R = 6.4 × 106 m
2π 2π
ω= T
= 24×60×60 = 7.27 × 10−5 rads−1
At point P gravitational acceleration = g ′
We know, g ′ = g − ω2 R cos 2 λ = 9.8 − (7.27 × 10−5 )2 × 6.4 × 106 × (cos 60°)2 = 9.79154 ms−2
∴ Effective gravitational acceleration acting on the object, FG = mg ′ = (2 × 9.79154)N
= 19.58308 N (Ans.)

5
08. The mass and radius of earth are as follows 6 × 1024 kg and 6400 km respectively. Acceleration due to
gravity at the surface is 9.8 ms−2. Gravitational constant 6.673 × 10−11 Nm2 kg −2 .A satellite is lifted at
700 Km height from its surface. [Ctg.B’19]
(c) What would be the height of the satellite from the earth’s surface, so that the weight of the satellite
becomes 80% of the weight at the surface? Calculate. 3
Solution
g′ R 2
(c) We know, g
= (R+h) . Now, g ′ = 0.8g
R 2 h 2 5 √5 √5
∴ 0.8 = (R+h) ⇒ (1 + R) = 4 ⇒ h = ( 2 − 1) R = ( 2 − 1) × 6.4 × 106 = 7.554 × 105 m
From the earth’s surface the height is, 7.554 × 105 m. (Ans.)

09. A planet of 6.4 × 106 m radius revolves around its own orbit once in 24 hours. For establishing a
relation between that planet & acceleration due to gravity g, a scientist placed an object of 80 kg mass
in a place at 58° north latitude. Acceleration due to gravity,
g = 9.80 ms −2. [SB’19]
(d) Will the weight of the object at that place be greater or less than its weight at the surface of that planet?
Give your opinion by showing mathematical analysis. 4
Solution
2π 2
(d) g = g − ω2 R cos 2 λ = 9.8 − (24×3600) × 6.4 × 106 × cos 2 58° = 9.79
∴ Weight at that condition, W2 = 9.79 × 80 = 783.2 N
Usual weight, W1 = 9.8 × 80 N = 784 N ∴ Difference = W1 − W2 = 0.8 N
∴ Weight of the object at that place will be less.

10. An artificial satellite is established at a height of 600 km from the surface of the earth. The radius of the
earth is 6400 km and the acceleration due to gravity is 9.8 ms −2 . [BB’19]
(c) Calculate the acceleration due to gravity at the height in the stem. 3
Solution
2 2
R 6.4×106
(c) We know, g = (R +h) g = (6.4×106 +6×105 ) × 9.8 = 8.192 ms−2 (Ans.)

11. [Din.B’19]
Planet Mass Radius Distance from sun
23
Mars 6.39 × 10 kg 3390 km 227.9 × 106 km
24
Earth 5.97 × 10 kg 6378 km 149.6 × 106 km
−11 2 −2
And Gravitational constant G = 6.673 × 10 Nm kg
(c) Find the acceleration due to gravity on Mars. 3
Solution
(c) The acceleration due to gravity on Mars, g = GM Mass, M = 6.39 × 1023 kg
R2
6.673×10−11 ×6.39×1023
∴g= (3.39×106 )2
= 3.71ms−2 (Ans.) Radius, R = 3390 × 103 m

6
12. An object of mass 5kg due to throwing with escape velocity from earth surface reaches another planet
whose mass is sixteen times of that of earth and diameter is eight times of that of earth. (mass of earth =
6 × 1024 kg, radius of earth = 6.4 × 103 km) [DB, SB, JB, Din.B’18]
(c) Find the acceleration due to gravity on the surface of another planet. 3
Solution
(c) If the masses of new planet & earth are respectively MN & ME and radii are R N & R E then,
MN = 16ME ; DN = 8R E ⇒ R N = 4R E
GMN G16ME 6.67×10−11 ×6×1024
 Acceleration due to gravity, g N = R2N
= 16R2E
= (6.4×106 )2
= 9.77ms −2 (Ans.)

T-03: Gravitational Intensity & Potential

01. The masses of the Sun and the Earth are 2 × 1030 kg and 6 × 1024 kg respectively. The distance between
the Earth and the Sun is 1.5 × 1011 m. [Ctg.B’23]
(c) Determine the gravitational potential energy at the midpoint of the line connecting the Sun and the
Earth. 3
(d) Examine mathematically whether the intensity at multiple points on the straight line connecting the
Earth and the Sun can be zero. 4
Solution
(c) Given that, Mass of the Sun, M = 2 × 1030 kg
Mass of the Earth, m = 6 × 1024 kg
Distance between the Sun and the Earth, R = 1.5 × 1011 m
Value of gravitational potential at the midpoint of the connecting line
GM Gm 2G
− R − R =− R
(M + m) = −1.78 × 109 J kg −1 (Ans.)
2 2

(d) Let, the gravitational potential is zero at a distance of x m from the Earth.
GM Gm M m
Now, − (R−x)2 = − x2
⇒ (R−x)2 = x2
⇒ 2 × 1030 x 2 = 6 × 1024 (1.5 × 1011 − x)2
⇒ x = 2.59 × 108 m, −2.6 × 108 m
If we consider points A and B, at these two points, the values of the gravitational intensities of the Earth and the
Sun are equal. However, only at point A, where the directions of both intensities are opposite, the resultant
intensity becomes zero. Since the directions are not the same at point B, it is not possible for the gravitational
intensity to be zero at multiple points.

02. A and B are two planets (imaginary) made of the same component and with the same average density.
The mass of A is 5.93 × 1024 kg, radius is 6.93 × 106 m and radius of B is 3 × 106 m . [BB’22]
(c) Determine the gravitational potential at any point on the surface of the planet A. 3
(d) Will the value of gravitational intensity on the surface of A & B planet be same? Give your opinion
with mathematical analysis. 4

7
 Solution
24
(c) Mass of A planet, M = 5.93 × 10 kg
Radius of A planet, R = 6.93 × 106 m
gravitational potential on the surface of planet A,
Gm 6.673×10−11 ×5.93×1024
V=− R
=− 6.93×106
Jkg −1 = −5.71 × 107 Jkg −1 (Ans.)
(d) A & B are two planets of the same average density.
We know, M ∝ V and V ∝ R3 (in case of two spherical objects)
M R 3
∴ MB = (RB ) [in case of both planets’ density is constant]
A A
3
3×106
⇒ MB = (6.93×106 ) × (5.93 × 1024 ) ∴ MB = 4.81 × 1023 kg
1
Again, in case of gravitational intensity, E ∝ M and E ∝ R2
2 2
E M R E 5.93×1024 3×106 E
∴ EA = MA × (RB ) ⇒ EA = 4.81×1023 × (6.93×106 ) ∴ EA = 0.231
B B A B B
∴ So, the value of gravitational intensity on the surface of A & B planet will not be same.

T-04: Escape Velocity

01. The moon is considered to be an uniform sphere. Let us assume an artificial satellite is orbiting at an
altitude of 2 × 106 m from the center of the moon. Moon's circumference = 10.048 × 106 m and density
= 1742.3 kg m−3. [CB’23]
(d) State the final result with mathematical analysis if the artificial satellite is launched from the surface
of the Moon with a velocity of 1.7538 kms−1 . 4
Solution
(d) Given that, circumference, 2πR = 10.048 × 106 ⇒ R = 1.6 × 106 m
4 4
M = Vρ = π R3 × ρ = π × (1.6 × 106 )3 × 1742.3 = 2.98 × 1022 kg
3 3
Escape velocity on moon surface,
2GM 2×6.673×10−11 ×2.98×1022
ve = √ R
=√ 1.6×106
= 1576.6 ms−1 = 1.5766 kms −1
−1
Given, Escape Velocity, v = 1.7538 kms . ∵ Escape velocity, v > ve
So, the artificial satellite will escape the gravitational pull of the lunar surface and leave the lunar
gravitational field.

02. The mass and radius of the Earth are M = 6 × 1024 kg and R = 6.4 × 106 m, respectively. An artificial
satellite is orbiting the Earth at a height of 700 km from the surface of the Earth.
[G = 6.673 × 10−11 Nm2 kg −2 ] [BB’23]
(d) Is there a possibility of the artificial satellite in the stimulus to drift away into space? Give a
mathematical analysis to make a correct decision. 4
Solution
(d) Potential energy in space = Ep = 0
GMm 1 2
According to the Law of Conservation of Energy, Epi + Eki = Efinal ⇒ − R+h
+2 mvm =0
1
2 GMm 2 GM 2×6.673×10−11 ×6×1024
⇒ 2 mvm = R+h
⇒ vm = √ R+h = √ 6.4×106 +700×103
= 10619.92ms −1

8
 GM 6.67 3×10−11 × 6×1024
Velocity at that height, v = √R+h = √ 6.4×10+700×103
= 7509.43 ms−1

Therefore, the satellite has to get an additional velocity of (10619. 92 − 7509.43) = 3110.49 ms−1
to drift away into space.

03. The mass of an imaginary planet is equal to the mass of the Earth, but its radius is twice that of the Earth.
An object of 3 kg mass is projected vertically upwards from the surface of both planets with a velocity of
9 kms −1. Mass of the Earth is 5.97 × 1024 kg and radius is 6.4 × 106 m. [DB’22]
(c) Determine the escape velocity of the earth. 3
(d) Will the projected objects of the stem return to both the planets – Analyze mathematically. 4
Solution
2GM (2×6.673×10−11 ×5.97×1024 )
(c) Escape velocity of the Earth, ve = √ R
=√ 6.4×106
= 11.16 kms −1 (Ans.)

(d) Radius of the new planet, R′ = 2R c = (2 × 6.4 × 106 ) m

2GM 2×6.673×10−11 ×5.97×1024
Escape velocity of the new planet, ve′ = √ R′
; ve′ = √ 2×6.4×106
= 7.89 kms−1

According to ‘c’,ve = 11.16 kms −1

It can be seen that, ve > 9 and 9 > ve′ . So, the object projected from the earth will return but the object
projected from the imaginary planet will not return.

04. [Din.B’19]
Planet Mass Radius Distance from sun
23
Mars 6.39 × 10 kg 3390 km 227.9 × 106 km
Earth 5.97 × 1024 kg 6378 km 149.6 × 106 km
And Gravitational constant G = 6.673 × 10−11 Nm2 kg −2
(d) Will the escape velocities of both the planets be equal or not? Give your opinion based on
mathematical analysis. 4
Solution
2GM1 2×6.673×10−11 ×6.39×1023
(d) The escape velocity on Mars, Ve1 = √ R1
=√ 3.39×106
= 5015.63ms −1

2GM2 2×6.673×10−11 ×5.97×1024

The escape velocity on Earth, Ve 2 = √ R2
=√ 6.378×106
= 11176.8769 ms −1

∴ The escape velocities are not same.

05. An object of mass 5kg due to throwing with escape velocity from earth surface reaches another planet
whose mass is sixteen times of that of earth and diameter is eight times of that of earth. (mass of earth =
6 × 1024 kg, radius of earth = 6.4 × 103 km) [DB, SB, JB, Din.B’18]
(d) If the mass of the mentioned object is half, then will the escape velocity required to throw the object
from another planet to space, be equal to escape velocity of earth surface? Give your opinion with
mathematical explanation. 4

9
 Solution

2GMN 2G16ME 2GME

(d) Escape velocity on new planet, VN = √ RN
=√ 4RE
= 2√ RE
= 2 × VEE

= 2 × escape velocity on earth.

So if mass of the object is half then escape velocity required to throw the object from another planet to
space, will not be equal to escape velocity of earth surface.

T-05: Satellite Velocity

01. The moon is considered to be a uniform sphere. Suppose an artificial satellite is rotating at a height of
2 × 106 m above the center of the moon. The circumference of the moon is 10.048 × 106 m and density
is 1742.3 kg m−3. [CB’23]
(c) Determine the value of the linear velocity of the artificial satellite. 3
Solution

GM Given, Circumference, 2πR = 10.048 × 106 ⇒ R = 1.6 × 106 m

(c) We know, v = √R+h
4 4
M = Vρ = π R3 × ρ = π × (1.6 × 106 )3 × 1742.3
6.673×10−11 ×2.98×1022 3 3
=√ 2×106 = 2.98 × 1022 kg; R + h = 2 × 106 m
= 977.13 ms−1 (Ans.)

02. The mass and radius of the Earth are M = 6 × 1024 kg and R = 6.4 × 106 m, respectively. An artificial
satellite is orbiting the Earth at a height of 700 km from the surface of the Earth.
[G = 6.673 × 10−11 Nm2 kg −2 ] [BB’23]
(c) Calculate the centripetal acceleration of the artificial satellite. 3
Solution
(c) We know,
GM
2
Mass of earth, M = 6 × 1024 kg
(√R+h)
v2
a= R+h
= R+h
Radius of earth, R = 6.4 × 106 m
GM 6.673×10−11 ×6×1024 Height of the satellite, h = 700 km = 7 × 105 m
= (R+h)2 = (6.4×106 +7×105 )2
Centripetal acceleration, a =?
= 7.94 ms −2 (Ans.)

03. An artificial satellite is orbiting the Earth at an altitude of 3 × 104 km kilometers at a velocity of
3 kms −1 kilometers per second. Mass of Earth, 6 × 1024 kg; Radius of Earth, 6.4 × 103 km; Mass of the
satellite, 1 × 103 kg and G = 6.673 × 10−11 Nm2 kg −2 [JB’23]
(d) Will the satellite be static in the mentioned height in the stem? Analyze. 4

10
 Solution

GM 6.673×10−11 ×6×1024
(d) Velocity at the mentioned height, v = √R+h = √ 6.4×106 +3×107
= 3316.54 ms −1 = 3.316 kms −1

According to the question, velocity of the satellite, v′ = 3 kms −1

v > v ′ , So, the satellite will not be static.

04. Mass of earth, M = 6 × 1024 kg ,Radius of earth, R = 6.4 × 106 m [SB’23]

−11 2 −2
Gravitational constant, G = 6.673 × 10 Nm kg
6
Height of artificial satellite,h = 3.46 × 10 m

(c) Find the velocity of the artificial satellite. 3

(d) Will the artificial satellite in the stimulus be a geostationary satellite? Analyze your opinion with
mathematical analysis. 4
Solution
(c) GM Mass of earth, M = 6 × 1024 kg
We know, v = √R+h
Radius of earth, R = 6.4 × 106 m
=√
6.673×10−11 ×6×1024
= 6372.32 ms−1 (Ans.) Height of satellite, h = 3.46 × 106 m
6.4×106+3.46×106

4π2 (R+h)3 4π2 (6.4×106 +3.46×106 )3

(d) We know, T ′ = √ GM
=√ 6.673×10−11 ×6×1024
= 9722.07s = 2.7 hr
But time period of geostationary satellite, T = 24 hr
T ′ < 24 hr ∴ The artificial satellite in the stimulus is not a geostationary satellite.

05. The mass of earth is 6 × 1024 kg,, radius is 6.4 × 106 m. An artificial satellite of 80 kg is taken 200 km
above the ground and inside it 4.5 × 109 Joule kinetic energy is transferre [G = 6.67 ×
10−11 Nm2 kg −2 , g = 9.8 ms−2 ] [Din.B’23]
(c) Determine the time period of the artificial satellite. 3
Solution
(c) We know, time period,
Given that, Radius of earth, R = 6.4 × 106 m
4π2 (R+h)3 4π2 (6.4×106 +200×103 )
T=√ =√ Height, h = 200 km = 200 × 103 m
GM 6.67×10−11 ×6×1024

= 5325.46 s (Ans.) Mass of earth, M = 6 × 1024 kg

06. An artificial satellite was placed at a height of 500 km from the surface of the earth. The radius of the earth
is 6400 km and the acceleration due to gravity at the surface of the earth is 9.81 ms−2 . [RB’22]
(d) Is it possible to transform the satellite in the stem into a geostationary satellite? Give your opinion
with mathematical analysis. 4

11
 Solution

gR2 9.8×(6400×103 )2
(d) Velocity of the artificial satellite, vi = √R+h = √ (6400+500)×103 = 7627.26 ms−1
2π(𝑅+h) 2×3.1416×(6400+500)×103
Time period of the artificial satellite, Ti = vi
= 7627.26
= 5684.1 s = 1.58 h
For transforming into a geostationary satellite, time period of artificial satellite must be, Tf = 24 h.
In that case, the height of the artificial satellite will be,
1 1
2
′ gR2 T2 3 9.8×(6400×103 ) ×(24×3600)2 3
h = ( 4π2 ) −R={ 4×(3.1416)2
} − (6400 × 103 ) = 3.6 × 107 m = 36000 km
2π(R+h′ ) 2×3.1416×(6400+36000)×103
∴ Velocity of the artificial satellite will be, vf = Tf
= 24×3600
= 3083.42 ms−1
∴ For transforming the artificial satellite into a geostationary satellite, the artificial satellite must be raised
higher by (36000 − 500) km = 35500 km and its velocity should be reduced by
(7627.26 − 3083.42) km = 4543.84 ms−1.

07. An artificial satellite was placed 26000 km above the surface of Dhaka. The radius and mass of the Earth
is 6400 km and 6 × 1024 kg respectively. [Ctg.B’22]
(c) Determine the orbital velocity of the artificial satellite. 3
(d) Will the satellite be seen over Dhaka all the time? Show mathematical analysis. 4
Solution
(c) Mass of the Earth, Me = 6 × 1024 kg
Radius, R e = 6400 km = 6400 × 103 m
Height, h = 26000 km = 2.6 × 107 m
GM 6.67×10−11 ×6×1024
Orbital velocity of the artificial satellite, v = √ =√
R+h 6.4×106 +2.6×107

∴ v = 3514.52 ms −1 = 3.51 kms−1 (Ans.)

(d) If the satellite is geostationary, its time period will be 24 hours and the satellite will be visible above Dhaka
at all times.
Orbital velocity of the artificial satellite, v = 3514.52 ms−1
2π(R+h) 2×3.1416×(6.4×106 +2.6×107
We know, T = v
= 3514.52
s = 57924.17 s = 16.09 hour
Since T ≠ 24 hour, so it is not a geostationary satellite. That is, the satellite will not be seen over Dhaka
at all times.

08. At a height of 3.6 × 104 km from the launching point of an artificial satellite, an artificial satellite is
launched, where gravitational acceleration is g = 9.8 ms −2 . Radius of earth and gravitational constant are
6.4 × 106 m and 6.7 × 10−11 Nm2 kg −2 respectively. [JB’22]
(c) Determine the velocity of the artificial satellite. 3
(d) Will the artificial satellite be geostationary? Analyze mathematically. 4

12
 Solution
(c) Now, mass of the earth,
gR2 9.8×(6.4×106)
2
g = 9.8 ms −2
Me = = 6.67×10−11
= 6.02 × 1024 kg
G R = 6.4 × 106 m
GM 6.7×10−11 ×6.02×1024 h = 3.6 × 104 km = 3.6 × 107 m
Velocity of artificial satellite, v = √ =√
R+h 6.4×106 +3.6×107
G = 6.67 × 10−11 Nm2 kg −2
= 3084.27 ms−1 (Ans.)
V =?
(d) Time period of the artificial satellite,
R+h 6.4×106 +3.6×107
T = 2π(R + h)√ GM = 2 × 3.1416 × (6.4 × 106 + 3.6 × 107 ) × √6.02×1024 ×6.67×10−11
∴ T = 86376.25 s = 23.99 hour
Since the time period of the artificial satellite is ≈ 24 hour, so it will be geostationary.

09. At a height of 3.6 × 104 km from the surface of Earth the Bangabandhu Satellite is placed. It is
determined to broadcast the U-19 cricket world cup. Radius of Earth, R = 6.4 × 106 m. [MB’22]
(c) Determine the linear velocity of the satellite. 3
(d) Will the satellite mentioned in the stem be able to broadcast the U-19 world cup matches-explain. 4
Solution
(c) Height of the satellite, h = 3.6 × 104 km = 3.6 × 107 m
Mass of the earth, M = 6 × 1024 kg
Radius of the earth, R = 6.4 × 106 m
GM 6.673×10−11 ×6×1024
Then, Linear velocity, v = √ =√ = 3072.93 ms−1 (Ans.)
R+h 6.4×106 +3.6×107
(d) The satellite mentioned in the stem will be capable of broadcasting the matches of U-19 world cup if it is a
geo-stationary satellite.
Now, Time period of a geo-stationary satellite is, T = 24 h = 86400 s
Then, height of the geo-stationary satellite is,
1 1
GMT2 3 6.673×10−11 ×6×1024 ×(86400)2 3
h= ( 2 ) −R= { } − 6.4 × 106 = 3.6 × 107 m
4π 4π2
The Bangabandhu Satellite mentioned in the stem is also at this height.
Therefore, it is capable of broadcasting U-19 matches of World Cup.

10. The mass and radius of earth are as follows 6 × 1024 kg and 6400 km respectively. Acceleration due to
gravity at the surface is 9.8 ms−2. Gravitational constant
6.673 × 10−11 Nm2 kg −2. A satellite is lifted at 700 km height from its surface. [Ctg.B’19]
(d) Is there any possibility of the projected satellite in the stem to be just like the moon? Explain
mathematically. 4
Solution
(R+h)3 (6.4×106 +7.554×105 )3
(d) Time period of rotation, T = 2π √ = 2π √ = 6010.292 s
GM 6.673×10−11 ×6×1024
So, time period of rotation would be 6009.92 s .This time period is quite less than that of the moon.

13
11. A planet of 6.4 × 106 m radius revolves around its own orbit once in 24 hours. For establishing a
relation between that planet & acceleration due to gravity g, a scientist placed an object of 80 kg mass
in a place at 58° north latitude. Acceleration due to gravity, g = 9.80 ms−2 [SB’19]
(c) What will be the linear velocity of the object for revolution of that planet in that place? 3
Solution
2π
(c) We know, ω = 24×3600 = 7.272 × 10−5 rads−1
∴ Linear velocity of the object, v = ωr = 7.272 × 10−5 × 6.4 × 106 = 465.422 ms−1 (Ans.)

12. An artificial satellite is established at a height of 600 km from the surface of the earth. The radius of the
earth is 6400 km and the acceleration due to gravity is 9.8 ms −2 . [BB’19]
(d) Is it possible to convert the satellite in the stem into a geo-stationary satellite? Explain
mathematically. 4
Solution
1 1
GMT2 3 gR2 T2 3
(d) For a geo-stationary satellite, h = ( ) −R=( ) −R
4π2 4π2
1
2
9.8×(6.4×106 ) ×(24×3600)2 3
={ 4π2
} − 6.4 × 106 = 3.6 × 107 m
∴ The artificially satellite can be converted into a Geo-stationary satellite if it is raised by
(3.6 × 107 − 6 × 105 ) or 3.534 × 107 m.

13. An artificial satellite after launching from Kennedy space center is rotating 3.58 × 107 m above the ground
along the equator. The mass of earth is 5.972 × 1024 kg, radius is 6.4 × 106 m and gravitational constant,
G = 6.67 × 10−11 Nm2 kg −2 . [JB’19]
(c) Find the time period of the satellite. 3
Solution
(c) Here, (R + h) = (6.4 + 35.8) × 106 m ; M = 5.972 × 1024 kg ; G = 6.67 × 10−11 Nm2 kg −2
GMT2 (R+h)3 .4π2 (42.2×106 )3 ×4×(3.1416)2
We know, (R + h)3 = 4π2
⇒T=√ GM
= √6.673×10−11 × 5.972 × 1024 = 86283.7672 s = 23.96 hrs

T-06: Miscellaneous

01. An object of mass 10 kg is moved from the equator to the pole. Earth's annual motion and diurnal motion
are 365 days and 24 hours respectively. Distance between the Sun and the Earth is 1.5 × 1011 m, radius of
the Earth is 6.4 × 106 m, and acceleration due to gravity is 9.81 ms−2 . [RB’23]
(c) Determine the linear velocity of the Earth around the Sun. 3
Solution
(c) We know,
Linear velocity, v =
2πR
=
2π×1.5×1011
ms −1 R = distance between the Sun and the Earth = 1. 5 × 1011 m
T 31.536×106
= 29.885 × 103 ms −1
(Ans.) Time, T = 365 days = 31.536 × 106 s

14
02. An artificial satellite is orbiting the Earth at an altitude of 3 × 104 km kilometers at a velocity of
3 kms −1 kilometers per second. Mass of Earth, 6 × 1024 kg, Radius of Earth, 6.4 × 103 km, Mass of the
satellite, 1 × 103 kg and G = 6.673 × 10−11 Nm2 kg −2 [JB’23]
(c) How much work was done to take the satellite to that height? 3
Solution

(c) We know, Ei + W = Ef
⇒−
GMm
+ W=−
GMm Mass of Earth, M = 6 × 1024 kg
R R+h
GMm GMm Mass of the satellite, m = 103 kg
⇒W= R
− R+h
Height, h = 3 × 104 m = 3 × 107 m
6.673×10−11 ×6×1024 ×103 6.673×10−11 ×6×1024 ×103
= − Initial energy, = Ei
6.4×106 6.4×106 +3×107

= 5.16 × 1010 J Changed energy, = Ef

03. The mass of earth is 6 × 1024 kg,, radius is 6.4 × 106 m. An artificial satellite of 80 kg is taken 200 km
above the ground and inside it 4.5 × 109 Joule kinetic energy is transferred.
[G = 6.67 × 10−11 Nm2 kg −2 , g = 9.8 ms −2 ] [Din.B’23]
(d) In the stem, do we need to change the kinetic energy in order to send the artificial satellite to space-
Explain mathematically. 4
Solution
−GMm 1 GMm 1 GM
(d) Total energy at 200 km, Etotal = EP + Ek = R+h
+ 2 mv 2 =− R+h
+2m × R+h
−GMm 6.67×10−11 ×6×1024 ×80
= (R+h)×2 = − (6.4×106 +200×103 )×2
= −2.43 × 109

Let, in order to send in space, we need to apply additional W work, −2.43 × 109 + W = 0
⇒ W = 2.43 × 109 J ∴ W > 4.5 × 103 J So, we need to change the kinetic energy.

CQ Knowledge-Based Questions & Sample Answers from Previous Board Exams

01. What is gravitational constant? [RB’23, SB, Ctg.B’17]

Ans: The magnitude of the force of attraction between two bodies of unit masses placed at unit distance is
called gravitational constant.
02. What is escape velocity. [Ctg.B’23, 22; MB’23, Din.B’22; DB’19; SB’19; CB’19; BB’17]
Ans: The minimum velocity at which an object thrown from the surface of the earth does not return to the
earth is known as the escape velocity of the object from the surface of the earth.
03. What is parking orbit? [CB’23, JB, MB’22; SB’19]
Ans: Orbit of geostationary satellite is called parking orbit.

15
04. What is geostationary satellite? [SB’22, DB’22; Ctg.B’19]
Ans: If the period of rotation of an artificial satellite is equal to the period of rotation of the Earth around
its own axis, the satellite will appear stationary with respect to the Earth. Such satellites are called
geostationary satellites.
05. What is gravitational potential? [RB, CB’22; JB’19; DB’17]
Ans: The amount of work done to bring an object of unit mass from an infinite distance to a point in the
gravitational field is known as the gravitational potential of that point.
06. What is gravitational field? [BB’22]
Ans: The region around an object in which its gravitational effect is present, that is if another object is
placed there that gains attractive force, then it is called gravitational field.
07. What is gravitational intensity? [JB’17]
Ans: The gravitational force experienced by an object of unit mass placed at a point in the gravitational
field is called gravitational field intensity of the point.
08. Write the 2nd law of Kepler on planetary motion. [JB’17]
Ans: The straight line connecting the planet and the sun covers equal area in equal time.
09. What is center of gravity? [CB’17]
Ans: The point through which the total weight of a body acts, no matter how it is kept, is called center of
gravity of that body.

CQ Comprehensive Questions & Sample Answers from Previous Board Exams

01. Are the escape velocities equal in all the planets of the Sun? Explain. [RB’23]
2GM
Ans: The value of escape velocity is not equal in all the planets of the Sun. Escape velocity, ve = √ R

As it can be seen, the escape velocity depends on the mass and radius of the planet. Because the mass and
radius of each of the planets of the Sun are different, they all have different escape velocities.
02. Explain the formula that follows the change in the value of acceleration due to gravity from the
Earth's surface upwards. [CB’23]
GM
gh (R+h)2 R 2
Ans: g
= GM ⇒ g h = (R+h) × g
R2

here R is a constant and suppose, R + h = x

R 2 1
⇒ gh = ( ) × g ⇒ gh ∝
x x2
As R and g are constants.

When, h = 0; that is, in the ground g h = g

16
 Therefore, the value of acceleration due to gravity from the Earth's surface upwards is inversely
proportional to the square of the distance from the center of the Earth. In other words, the value of
acceleration due to gravity decreases with increasing altitude.
03. Can humans live on the moon? Explain. [BB’23]
Ans: The moon does not provide the oxygen, water, and other necessities for humans to survive. Again, the
value of the gravitational acceleration on the moon is one-sixth of that on Earth. Therefore, it will be
difficult for humans to move on the moon in the way they are accustomed to. Therefore, living on the
moon is difficult for humans.
04. The more we go down from the Earth’s surface, the less the value of the gravitational acceleration
gets – explain. [JB’23]
x
Ans: Below the Earth’s surface g ′ = (1 − R ) g
In this context, x refers to the distance from that point to the Earth's surface. Here, the value of x as well as
x x
will increase as it goes down from the surface. As a result, the value of (1 − ) g will decrease.
R R
Therefore, the value of the acceleration due to gravity decreases downwards.
05. The escape velocity from the surface of Jupiter is greater than the escape velocity from the surface
of Mars- Explain. [SB’23]
2GM
Ans: We know, escape velocity, ve = √ R
M
The ratio of mass to radius ( R ) of Jupiter is greater than that of Mars. As a result, the escape velocity of
Jupiter is greater. In other words, to free an object from the gravity of the planet, the object must be
launched at a higher speed from Jupiter than to Mars.
06. What does it mean by a force of 12 N acting on an object? [Din.B’23]
−2 −2
Ans: The amount of force by applying which on an object of 1 kg mass 12 ms or 4 ms acceleration is
created on 3 kg mass is called 12 N.
07. “At definite place, weight is proportional to mass” – Explain. [MB’23]
Ans: We know, weight, W = mg. At any definite place, as the value of g is constant, W ∝ m. So, at any
definite place, the weight increases as mass increases, and the weight decreases as mass decreases. So,
weight is proportional to mass at any given place.
08. Will there be any change in acceleration due to gravity because of change in the density of the
Earth? [RB’22; Ctg.B’19]
GM 4
Ans: For the Earth having mass ‘M’ and radius ‘R’, g = R2
= 3 πGRρ
∴ g ∝ ρ i.e. acceleration due to gravity is proportional to density. If the density increases, the acceleration
due to gravity will increase; if the density decreases, the acceleration due to gravity will decrease.
Acceleration due to gravity will change at a rate proportional to the change in density.
09. Will the velocity of two satellites of different masses placed in the same orbit be the same or
different? Clarify. [Ctg.B’22]
Ans: The time period of an artificial satellite orbiting the Earth at a height above the surface of the earth
does not depend on the mass of the artificial satellite. The equation for the velocity of artificial satellite is
GM
v=√ .
R+h
It can be seen that the equation does not depend on the mass of the artificial satellite. Therefore, two
satellites of different masses placed in the same orbit will have the same velocity.

17
10. What do you mean by gravitational potential is 12 𝐉𝐤𝐠 −𝟏 ? [SB’22]
Ans: The work done to bring an object of one unit mass from infinite distance to any point in a gravitational
field is called gravitational potential.
W
Gravitational potential, V = m
.
By 12 Jkg −1 gravitational potential it means that the work done to bring an object of 1 kg mass from
infinite distance to any point in a gravitational field is 12 J.
11. When is the center of mass and center of gravity of an object situated at the same point? Explain.
[BB’22]
Ans: Even though the center of mass of an object depends only on the structure of the object, but the center of
gravity of an object depends on the gravitational acceleration effective on each of the points of the object along
with the structure of the object. When the magnitude of gravitational acceleration will be same on every point of
the object, only then center of mass and center of gravity will be at the same point. Usually, it is more probable
for smaller objects to have the center of mass and center of gravity at the same point.
12. Escape velocity does not depend on the mass of the object-explain [JB’22]
2GM
Ans: We know, Escape velocity, ve = √ R

We can see that, ve depends on the mass and radius of the planet. But the escape velocity formula does not
contain the mass of the object. That is why, escape velocity is not dependent on the mass of the object.
13. Explain the reason for the loss of weight of an object in an equatorial region. [Din.B’22]
Ans: Earth is not completely round. The north-south sides are somewhat flat and somewhat swollen in the
equatorial region. The radius of the equatorial region of the earth is 22km more than the radius of the
GM
equatorial region. Now for gravitational acceleration, g = R2
That means, the gravitational acceleration is inversely proportional to the radius of the earth. Since the value of R
in the equatorial region is more, so the gravitational acceleration g decreases. If the gravitational acceleration of
that place decreases, then the weight in that region also decreases.
14. Gravitational potential at infinity is maximum but zero – explain. [MB’22
Ans: The work done to bring an object of unit mass from infinity to a point inside the gravitational field is
called the gravitational potential at that point. If the gravitational potential V of an object of mass M at a
GM
distance of r is, V = − r
. That is, the gravitational potential is inversely proportional to the distance.
Here, negative sign expresses that, infinite gravitational potential is maximum and it is zero. The more the
object is brought closer to the gravitational field, the lesser the value of the gravitational potential becomes.
15. Planets revolving around the sun have different time periods-Explain [SB’19]
Ans: Time period of different planets differ due to variation in the distance of orbits. According to
Kepler’s law, T 2 ∝ R3
As the value of R varies for different planets, T or time period will also be different.
16. If a planet with a fixed rest mass is expanded, will the escape velocity of an object on it change?
Explain. [BB’19]
2GM 1
Ans: ve = √ R
⇒ ve ∝ . So, if the planet is expanded, the escape velocity will change.
√R

18
17. How will the value of g in the earth surface be changed if the rotation of the earth on its own axis.
[JB’19]
Ans: If the rotation of the earth on its own axis stops, then the value of g will be changed.
Generally, g ′ = g − ω2 R cos 2 λ . That is, if the rotation stops, the value of g will be increased by
ω2 R cos 2 λ.
18. Is the escape velocity of any planet dependent on its radius or not? Explain [Din.B’19]
Ans: The escape velocity of any planet depends on the radius. If escape velocity is ve then,
2GM
ve = √ R
… … … (i)

Here, M = Mass of the planet and R = radius of the planet. From equation (i) it is seen that if the mass
1
remains constant, then ve ∝ i.e the escape velocity is inversely proportional to the square root of the
√R
radius.
19. Acceleration due to gravity of freely falling body from different height are not uniform- Explain it.
[RB’17]
Ans: Acceleration due to gravity is different for different height. We know at height h, acceleration due to
GM
gravity is, g = (R+h)2 . That is when height increases then value of g decreases. When height is too small
h
that is R ≪ 1 then, change in acceleration due to gravity is negligible. That is at small height acceleration
due to gravity is constant.
20. If the earth comes closer to the sun, its velocity is raised- explain this in light of Kepler’s laws.
[Ctg.B’17]
Ans: When the earth comes closer to the sun, its velocity increases and this phenomenon is described by
Kepler’s 2nd law.
According to Kepler’s 2nd law, the line joining the planet and
sun intercepts equal area in equal time. Let, in figure, the sun is
at position F, the earth comes

from A to B in time t and the same time is taken by the earth to come from C to (d) Then, area AFB = area
CFD. From figure, it is seen that if the planet is closer, then more distance is to be traversed to cover equal
area. Hence velocity is more at (c)
21. Explain why apparent weight becomes less in equatorial region? [SB’17]
Ans: Difference in gravitation occurs due to the earth rotation on its own orbit.
At latitude λ, gravitational acceleration,
g ′ = g − ω2 Rcos2 λ
In case of equatorial region, λ = 0
∴ g ′ = g − ω2 R
Again, the earth is tapered like orange in the north and the south poles. So, the distance between center to
equator is more than the distance between center to pole. For this reason, also, equatorial region has less
gravitational acceleration. As such weight of the body also decreases.

19
22. Why is gravitational constant a scalar quantity? [BB’17]
M M
Ans: Gravitational force, ⃗F = G |r1⃗ |3 2 r

Here gravitational force is dependent on radius vector. The direction of the force is towards the radius
vector. Hence gravitational constant G is a scalar quantity.
23. Why does earth’s rotation have no torque? Explain. [JB’17]
Ans: The earth rotates around the sun. but the force between sun and the earth
does not create any torque. Because while rotation the position vector and force
remain parallel to each other.
We can see from the figure, angle between position vector of earth (r) and
⃗ ) is 180°.
force (F

⃗ ⇒ |τ⃗| = rFsinθ = r F sin 180° = 0

So, torque, τ⃗ = r × F
24. The acceleration due to gravity at any point inside the earth is proportional to the distance from the
center of the earth – explain. [SB’19, CB’17]
Ans: Let, R = radius of earth. Then the acceleration due to gravity on a point at a distance r(r < R) from
the centre is,
4 g 4
g = 3 πρGr ⇒ r = 3 πρG = constant

∴ g∝r
Which means, the acceleration due to gravity increases linearly from the center of the earth to the surface.
It can be seen from g vs r graph also.

20