Physics 1st Paper Gravitation and Gravity

Chapter-6

Md. Sabbir Hossain

BISC, NIRJHOR
 Definitions
Gali1eo’s laws for a falling body
First Law: In vacuum all the freely falling bodies starting from rest traverse equal distance at equal interval
of time or in vacuum all bodies starting from rest fall with equal rapidity.
Second law: Starting from rest the velocity of a freely falling body is proportional to the time taken to fall. If
a falling body gets velocity v at time t, it can be mathematically written as v  t.
Third law: Starting from rest, the distance traversed by a freely falling body is proportional to the square of
the time of fall. If a falling body traverses distance h in time t, it can be mathematically written as h  t2.
Kepler’s law about motion of the planets
Law of ellipse: The orbit of each planet is an ellipse which has the sun at one of its foci.
Law of area: Each planet moves in such a way as the imaginary line joining it to the sun sweeps out equal
areas in equal times.
Law of time: The square of the period of revolution of any planet around the sun is proportional to the cube
of its mean distance from the sun.
Gravitation: The force of attraction between two objects or particles in the universe is called gravitation.
Gravity: The force of attraction between the earth and any other object or particle is called gravity.
Newton’s law of gravitation:
Any two particles in the universe attract each other with a force which is directly proportional to the product
of their masses and inversely proportional to the square of the distance between them, this force acts along the
line joining the two particles or objects.
Weight
The amount of force by which an object is attracted towards the centre of the earth is called its weight.
Gravitational field: The space around an object in which its gravitational attraction is felt, that is, if any other
body is kept and the body gains force of attraction that is called the gravitational field.
Gravitational field intensity: The gravitational field intensity or strength of an object at any point in the
gravitational field is defined as the force experienced by a unit mass placed at that point.
Gravitational potential difference: The amount of work done in moving a unit mass from one point to
another point in the gravitational field is called gravitational potential difference.
Centre of mass: Centre of mass of a body is such a point at which if force is applied only linear motion of
the body is generated, but no rotational motion is produced.
Centre of gravity: Centre of gravity of a body is such a point through which the total weight of the body will
act vertically downward.
Escape velocity: The minimum velocity with which a body is thrown upward so that it does not return to the
earth is called the escape velocity.
Information : If an object is thrown upward from the surface of the earth with velocity different phenomena
of the object may happen by the attractive force of the earth. viz :
𝑣2
(1) If v2 < 2𝑒 i.e., if the velocity of ejection is less than 7.88 kms–1, then it will revolve round the earth in
elliptical orbit and finally will come back to the earth
 𝑣2
(2) If v2 > 2𝑒 i.e., velocity of ejection is equal to T88 kms–1, then the object will rotate round the earth and will
be a satellite like the moon
𝑣2
(3) If v2 > 2𝑒 but < ve2, i.e., velocity of ejection is between 7.88 kms–1 and 11.2 kms–1, then the object will
continue to rotate round the earth in elliptical orbit keeping the earth at one focus
(4) If v = ve, i.e., velocity of ejection is 11.2 kms–1, i.e., equal to the escape velocity, the object will leave the
surface of the earth in a parabolic path and will go outside the influence of gravitation force of attraction.
(5) If v > ve, i.e., velocity of ejection is greater than the escape velocity, then in· parabolic path the object will
leave the earth and will not come back to the earth.
Artificial satellites: The satellites which revolve round the earth-keeping themselves at a particular height
from earth’s surface are called artificial satellites.
Geostationary satellite: If the period of revolution of a satellite is equal to the period of revolution of the
earth about its own axis the satellite will remain stationary with respect to the earth. This type of satellite is
called geostationary satellite.
 Mathematical Problem
Type 1: Gravitational Acceleration
1. If radius of the earth, R = 6.4  103 km, acceleration due to gravity, g = 9.8 ms–2 and gravitational
constant, G = 6.67  10–11 Nm2kg–2, then calculate the mass M and density of the material of the earth.
2. If the weight of a person on earth’s surface is 648 N, what amount of weight will the person lose if he
goes to the moon? Mass and radius of the earth are respectively 81 and 4 times that of the moon.
3. Radius of the earth is 6.4  106 m and acceleration due to gravity or earth’s surface is 9.8 ms–2. Find
the magnitude of acceleration due to gravity at a height of 6.4  105m from earth’s surface.
4. What is the acceleration due to gravity at a height of 2 km above earth’s surface? (Radius of the earth
is 6400 km, value of g at earth’s surface is 9.8 ms–2)
5. If the earth is considered as a sphere of radius of 6400 km, at what height the value of the acceleration
1
due to gravity will be 64 of the acceleration due to gravity on the earth’s surface?
6. Angular speed of diurnal rotation of the earth is 7.3  10–5 rad s–l. If the radius of the earth is 6400 km
and acceleration due to gravity at the pole is 9·8 ms–2, then calculate the magnitude of acceleration due
to gravity at the equator?
7. Latitude of Dhaka is 230 north. It there would have no diurnal rotation then how much increase of g
would have been in Dhaka? (Radius of the earth is 6400 km).
8. Prithila and Mithila the two sisters were discussing about the universe. They were also discussing the
rotational mechanism of the earth.
(a) If the distance of the earth from the sun is half the present distance, then what is the number of days
in a year?
(b) If the rotation of the earth is stopped what change of weight of a body in equator will take place?
Give opinion with analysis.
9.

In the figure an imaginary satellite has been shown whose mass

is 12  1024 kg and radius 8  106 m. O is its centre, b is a point
on its surface. aO = ab = bc, G = 6.67  10–11 Nm2kg–2.

(a) Calculate the escape velocity of the surface of the said satellite.
(b) Between points a and b which one has higher magnitude of acceleration due to gravity? Give
justification in favour of your answer.

10. Radius and mass of a terrestrial body are respectively 3.2  106 m and 4  1024 kg. Gravitational
constant G = 6.675  10–11 Nm2kg–2. The terrestrial body was divided into eight equal parts due to the
strike of a comet.
(a) Calculate the gravitational acceleration on the surface of the terrestrial body.
(b) Whether the escape velocity of each part will be one-eighth of the escape velocity of the main
body-verify.
11. We know that weight of a body of mass of 1 kg is 9·8 N on earth, 1·6 N on moon and no weight in
the upper space but mass of the body remains 1 kg on earth, moon and in the space. Answer the
following questions:
(c) Radius of the earth is 6·4  106 m and acceleration due to gravity on earth’s surface is 9·8 ms–2.
Calculate the value of acceleration due to gravity at a height of 6·4  105 m from earth’s surface.
 (d) Are the data given in the stimulus correct or confusing? Explain with justification the reasons of
variation of this weight.
12. In the figure a body of mass m in polar region is attracted by the earth to its centre by a force, F. The
mass of the earth is M and radius is R.
(c) Explain what change of the weight of the body will take place
if it is taken from polar region to equatorial region.

(d) What is the velocity of the satellite? If the height is doubled then in case of same velocity what will be the
period of rotation of the satellite?
13. Weight of a body of mass 20 kg is 196 N on earth, Radius of the earth 6.37  106 m and gravitational
constant is 6.67  10–11 Nm2kg–2.
(c) Calculate the mass of the earth according to the stimulus,
(d) At what height the magnitude of g will be 25% than the magnitude on earth?
Type 2: Escape Velocity
1. Diameter of mars is 6000 km and acceleration due to gravity on its surface is 3.8 ms –2. Calculate the
escape velocity of a body from the surface of mars.
2. If the mass and radius of Neptune are 1.9  1027 kg and 7  107 m respectively, find the escape velocity.
3. An artificial satellite is rotating around the earth concentrically with the earth. Prove that the escape
velocity of the satellite is 1·414 times its velocity.
4. How much additional velocity is to be added to a satellite orbiting in a circular path near earth’s surface
so that it will go beyond the attraction of the earth?
5. If a stone is thrown upward it comes back to the earth. But if a body is thrown at a particular or more
velocity, then the body does not come back to the earth. The magnitude of this velocity is different for
moon, mars, Neptune. Answer the following questions:
(c) If the radius of the earth R = 6·4  106 m and acceleration due to gravity, g = 9·8 rns–2, then find
the escape velocity of a body from the earth.
(d) Why escape velocity from the earth, moon, mars, Neptune and other planets are different? Give
justification in favour of your answer.
6. An artificial satellite is revolving concentrically round the earth. Give answer of the following
questions.
(c) Prove that the escape velocity of the satellite is 1·414 times its velocity of motion.
(d) The height of the satellite is dependent on its time period of rotation – prove it by appropriate
reasons.
Type 3: Kepler’s law and Artificial Satellite
1. Radius of the planet mars rotating around the sun is 1.53 times the radius of the earth. If 365 days in
earth makes one year then in how many days a year will be in mars?
2. The radius of the earth moving around the sun is 1.5  1011 m and its time period is 3.156  107 sec.
Calculate the mass of the sun. (G = 6.673  10–11 Nm2 kg–2)
3. At a particular height where the value of the acceleration due to gravity, g = 8 ms–2, the velocity of a
satellite there is 8 kms–1. Determine at what height, from the surface of the earth, the satellite is
revolving, (Radius of the earth, R = 6.4  106 m)
4. An artificial satellite is revolving round the earth at a height 700 km from the surface of the earth. Find
the horizontal velocity of the satellite. [Radius of the earth, R = 6400 km and on the surface of earth,
g = 9.8 ms–2]
 5. Observe the following stimulus :

M = mass of the earth = 6  1024 kg

R = radius of the earth = 6.4  106 m
(a) What is the horizontal velocity of the satellite?
(b) What change of time period of revolution is to be made in order to keep the satellite 1000 km above
the earth’s surface?
6. 3.6  109 Joule of kinetic energy was generated to an artificial satellite of mass 120 kg by raising it to
a fixed height. Mass and radius of the earth are respectively 6  1024 kg and 6.4  106 m. (G = 6.673 
10–11 Nm2kg–2, g = 9.8 ms–2.)
7.

M  Mass of the earth

R  Radius of the earth
[M = 6  1024 kg and R = 6.4  106 m]
a) Calculate the horizontal velocity of the satellite.
b) What change of time period will appear if the satellite is placed 1000 km above the earth’s
surface? Analyse it.
8. An artificial satellite of mass 120 kg is raised to a fixed height and generated kinetic energy of 3.6 
109 J in it, Mass and radius of the earth are respectively 6  1024 kg and 6.4  106 m, G = 6.673  10–
11
Nm2 kg–2. g = 9.8 ms–2.
(c) At what height the satellite is from the earth?
(d) Prove mathematically that the generated kinetic energy is not sufficient to send it in the outer space.
9.
E = earth
S = geostationary satellite
R = 6.4  106m
M = 6  1024 kg
G = 67  10–11 Nm2kg–2
Singapore took decision to launch a geostationary satellite of mass 4000 kg.
(c) At what height the geostationary satellite is to be ejected?
(d) If the value of h is increased three times then how much the velocity of the satellite is to be increased?
Analyse mathematically
 Explanatory Questions
Q.1. When an iron ball and a piece of paper are allowed to fall freely from the roof, why they do not fall
simultaneously?
Ans. While falling the two bodies buoyancy or upward force of air acts opposite to their weight. Since the
buoyancy of air or upward force for lighter body is more than the heavier one, so lighter body reaches the
ground late.
Q.2. Acceleration due to gravity does not remain uniform in case of a falling body from any height–explain.
Ans. Acceleration due to gravity depends on height. So, acceleration due to gravity is different at different
heights. At small height acceleration due to gravity is more than that at large height.
Q.3. Is it profitable to buy a thing in Darjeeling by measuring in a spring balance or by measuring in ordinary
balance?
Ans. Since Darjeeling is located much above the sea level, so the value of g is slightly smaller. In a spring
balance we measure the weight of an object and in an ordinary balance we measure the mass of an object.
Since the value of g is smaller in Darjeeling and we measure weight of an object using a spring balance, so it
is profitable to buy a thing using spring balance.
Q.4. Why passengers in a boat filled with passengers are asked not to keep on standing?
Ans. The centre of gravity of the boat rises slightly when the passengers remain standing then when they are
seated, so the boat filled with passengers reaches to unbalanced condition. As a result, the boat can capsize;
so passengers are asked not to keep on standing.
Q.5. What change of weight of body at equator will take place if the rotation of the earthy is stopped for
any reason? Explain.
Ans. Magnitude of acceleration due to gravity at equator, g = g – 2R. here g = acceleration due to gravity
on earth’s surface,  = angular velocity of the earth. So, due to diurnal rotation of the earth weight of a
body at equator.
W = W – m2R
Now, if rotation of the earth is stopped, then  = 0, So, W = W.
In this case increase of weight of the body, W = m2R.
Q.6. How will be the graph of the acceleration due to gravity versus distance from the centre of the earth?
Ans.

We know that outside the earth acceleration due to gravity is inversely proportional to the square of distance
from the centre of the earth. Inside the earth acceleration due to gravity is proportional to distance from the
centre of the earth. At the centre of the earth acceleration due to gravity is zero.
Q.7. Explain the variation of gravitational potential with respect to distance in gravitational field.
Ans.
 −GM GM
We know, gravitational potential, V = . So with this increase of distance value of decreases inversely
𝑟 𝑟
with distance, but as potential is negative so magnitude increases. For infinite distance potential is zero.
Q.8. What is the difference between the gravitational potential with other potentials that you know?
Ans. As gravitation is always attractive, so gravitational potential is always negative. But as electric and
magnetic forces are both positive and negative so potential can be both positive and negative. Further,
gravitational potential does not depend on the medium, but other potentials depend on the nature of medium.
Q.9. If a piece of matter is released through the tunnel extended between one end to the opposite end along
the diameter of the earth, will it reach the other end of the tunnel? What will be the nature of motion?
Ans.

If the earth is considered a homogeneous sphere of density , then applied gravitational force on a body of
mass m at a distance x is,
4 4
F = 3  Gmx, that means F  x [  G𝑚 = Constant]
3

Again, this force is always centripetal, So, the body will continue to oscillate from one end of earth to
the other end in simple harmonic motion keeping the centre of the earth at the mid position.
Q.10. Why work done in bringing any object of unit mass from infinity to any point becomes negative? When
a rocket starts ascending from the surface of the earth, does the potential for the earth according to the
position of the rocket continue to increase or decrease? Does the gravitational potential at a point inside
the earth depend on the position of that point? Explain.
Ans. First part : In this case work is done on the body by the gravitational attractive force and by no other
outside force. So, work done becomes negative.
Second part : As the rocket ascends high and high, potential increases accordingly. Because this potential V
GM
= . i.e., as r increases V also increases. If r is infinity V will be maximum or zero.
𝑟

Third part : By considering the earth as a solid sphere, potential at a point inside the sphere of distance r
3R2 − 𝑟 2
measured from the centre is, V = – GM  ,where M and R are respectively mass and radius of the earth.
2R3
So, due to change of the value of r potential will also change.

Q.11. A body is thrown vertically upward from the surface of the earth with a velocity of v. At what height
will it rise from the centre of the earth? Consider this body is rising upward in a frictionless path.
 1
Ans. Kinetic energy on earth of a body of mass m thrown with velocity v = mv2. Let us consider that the
2
body can rise to a height r from the center. There the velocity of the body is zero. As a result kinetic energy =
GM𝑚
0. If mass of the earth is M and radius is R, then potential energy of the body = – and at height r
R
GM𝑚
potential energy = – .
𝑟

Now, according to the conservation principle of energy, total energy of the body on earth = total energy of the
body at a height r.
1 GM𝑚 GM𝑚
So, mv2 – =0–
2 R 𝑟
GM𝑚 GM𝑚 1
or, = – 2 mv2
𝑟 R

1 1 𝑚𝑣 2
or, 𝑟 = R – 2 GM𝑚
2 GMR
or, r = 2 GM𝑚−𝑣2 R

2 gR2 .R
= 2 gR2 −𝑣2 R [ GM = gR2]

2 gR2
maximum height = 2 gR2−𝑣2

Q.12. Can the centre of gravity stay outside the material of the body?
Ans. Centre of gravity of a metallic ring, or a circular ring or bangle is in the centre of the ring, but there does
not exist material of the body. So, in special cases, centre of gravity of a body can be outside the material of
the body.
Q.13. Can the kinetic energy of a body be negative? –explain.
Ans. We know the kinetic energy of any moving body,
1
𝐸= 𝑚𝑣 2
2
Here, m is mass of the body and v is the velocity. Mass cannot be negative. In this equation we have square
of velocity, which is why kinetic energy of anybody cannot be negative.
Q.14. For what reasons the value of acceleration due to gravity changes?
Ans. Gravitational acceleration of any body involves mass of earth and distance of the body from the center
of the earth. Mass of earth is constant. So, gravitational acceleration changes with the change of distance from
the center of the earth. Furthermore, we know earth is circulating around its own axis. And due this rotation
gravitational acceleration also varies for different places on earth.
Q.15. If a body is thrown upward why it returns to the earth?
Ans. If a body thrown upward the body returns to the earth due to the gravitational attraction force of earth.
Q.16. Why gravitational constant is called universal constant?
Ans. The value of gravitational constant is same for any two object in this universe placed at any distance
from each other. It does not depends on distance, mass or place. That is why gravitational constant is called
universal constant.
Q.17. What is the difference between artificial satellite and geostationary satellite?
 Ans. Any object that is circulating around a planet is called a satellite of that planet. Now some satellite are
natural and others are not. For example, moon is the natural satellite of earth. Some satellites are sent to the
space by us. Those satellites are known as artificial satellite. If any satellite moves around the earth with same
time period as earth then the satellite is called geostationary satellite. Not all the artificial satellite moves with
same time period as earth does. So we can say that, all geostationary satellites are artificial but not all the
artificial satellites are geostationary.
Q.18. If the radius of the earth is double but mass remains unchanged then what will be the change of weight
of a body on the earth?
Ans. We know, weight of a body depends on the gravitational acceleration of earth. The equation of
gravitational acceleration is,
𝐺𝑀
𝑔=
𝑅2
Now if the mass of earth remain unchanged and radius becomes double then gravitational acceleration will
become one fourth of present gravitational acceleration. That is why the weight of any body on earth will also
become one fourth of present weight.
Q.19. Why the weight of a body at the centre of the earth becomes zero?
Ans. The value of gravitational acceleration at h depth from the surface of the earth,
ℎ
𝑔′ = 𝑔(1 − )
𝑅
Now, at the center of the earth h=R. Now if we put this value in the above equation,
𝑔′ = 0
Since weight of a body depends on gravitational acceleration that is why weight of a body at the center of the
earth becomes zero.
Q.20. Why the values of acceleration due to gravity are different in polar, equatorial and tropical regions?
Ans. Earth is circulating around its own axis. Due to this rotation the value of g changes according to the
following equation,
𝑔′ = 𝑔 − 𝜔2 𝑅𝑐𝑜𝑠 2 𝜆
For equator the value of 𝜆 = 00
So the value of gravitational acceleration is, 𝑔′ = 𝑔 − 𝜔2 𝑅
For pole, 𝜆 = 900
So the value of gravitational acceleration is, 𝑔′ = 𝑔
Now from pole to equator the value of g changes due to the rotation of the earth. If there is no rotation then
the value of g will be unchanged then.
Q.21. A mango drops on earth but why an artificial satellite does not fall on earth? Explain.
Ans. Artificial satellites are moving around the earth due to the balance of earths gravitational force and
centrifugal force. That is why artificial satellite does not fall on earth. But in case of a mango there is no force
to balance earths gravitational force. That is why a mango fall on earth.

Q.22. Why astronauts feel weightless while travelling in space ship?

 Ans. Astronauts orbiting the earth in a spacecraft feel weightless because two equal and opposite force acts
on a spacecraft. Since there is no gravity on space craft to pull the spacecraft towards earth, that is why
astronauts feel weightless.
Q.23. Why the velocity of a rocket is not escape velocity?
Ans. If a rocket wants to move into the space then it needs to acquire minimum velocity of 7.9 Kms-1. And it
is much lesser than earths escape velocity. If the rocket wants to enter into the orbit of the earth then it does
not need the escape velocity. That is why velocity of rocket is not escape velocity.