Chapter 7 gravitation

CHAPTER 7:
GRAVITATION
WONG LIHERNG
TAN SI YOUNG
TAN SZE MING
YEE SIEW QI

7.1
NEWTON’S
LAW
OF
GRAVITA
TION

Newton ‘s law of gravitation states that the
force of attraction between particles is
directly proportional to their mass and
inversely proportional to the square of
distance apart.

where
The negative sign shows that the force is an
attractive force.

Case 1: Determine the force of gravitational
attraction between the earth 5.98 x 1024 kg and a 70
kg boy who is standing at sea level, a distance of
6.38 x 106 m from earth's center.
m1 = 5.98 x 1024 kg, m2 = 70 kg, r = 6.38 x 106 m, G
= 6.6726 x 10-11N-m2/kg2

7.2
GRAVITATIONAL
FIELD
STRENGTH

• A gravitational field is a region where gravitational force acts
on massive bodies. Eg . gravitaional field of the Earth.
• The gravitational field strength tells us how strong a
gravitational field is. The gravitational field strength of the
Earth near its surface is 9.81m / s2. This means an object that
is near the surface of the earth will accelerate towards it
at 9.81ms2. We could then define the gravitational field
strength as the acceleration an object will experience within
that gravitational field.
• A better definition, however, can be derived from the
equation, F = ma.
• The gravitational field strength , E at a point is the force of
gravity per unit mass exerted on a mass placed

• Gravitational field strength is a vector quantity
which measured in ， though it is
perfectly acceptable to use ms − 2 for situations
where it is treated as an acceleration (such as the
acceleration of an object in free fall).
• （r > R）

• The –ve sign shows that the direction of E is in
the direction of decreasing r, that is, towards
the centre of the Earth.
• (by substituting F for mg)

• (by cancelling the lower case 'm's)
• On the surface of the Earth，
r=R and

Questions:

• 1. A point mass produces a gravitational field
strength of 3.4 x 10-2 N/kg at a distance of 4.9
x 102 m away. What is the mass of the point?

Ans:

g = Gm/r2
Thus, m = gr2/G
= 1.2 x 1014 kg

• 2. Estimate the gravitational field strength at
the surface of an interstellar body whose
density is 5.5 x 103kg/m3 and radius is 6.4 x
106 m.

Ans :
g = Gm/r2
and m = density x volume = D x V
= D x 4/3 x p x r3
Substitute for m in the field equation above.

Thus, g = 4pGDr/3
= 4 x p x 6.67 x 10-11 x 5.5 x 103 x 6.4 x
106/3 = 9.8 N/kg

Variation of Acceleration Due to Gravity g
with Distance r from the centre of
the Earth
• Weight = mg = gravitational force
• Acceleration due to gravity,
g=gravitational force/m
=gravitational field strength,E
# The variation of the acceleration due to gravity g with distance r
from the centre of the Earth equals the variation of the
gravitational field strength.

• For a mass m on the surface of the Earth,
weight =mg =

Acceleration due to gravity on the surface of
the Earth ,

Where g = accleration due to gravity
G = universal gravitational constant
M = mass of the earth
R = radius of the earth

• The variation of the acceleration due to
gravity g’ with distance r from the centre of
the Earth is illustrated by the graph above.

• Figure above shows a mass m at a distance of
r < R from the centre of the Earth.

Variation of g with latitude
• The acceleration due to gravity, and the gravitational field strength on the
surface of the Earth is not the same at different points on the Earth’s
surface because :-
• The Earth is not a perfect sphere, but it is an
ellipsoid as shown in below.
• The rotation of the Earth about its axis.

• The Earth’s rotation on its axis causes it to bulge at the equator and
flatten at the poles, forming an ellipsoidal shape.

• The figure below shows that the radius of the
Earth at the equator R1 is greater than the
distance of the poles from the centre of the
Earth R2.

• g α ⅟r²
• The value of g at the equator is less than the value of
g at the poles.

-The centripetal force
m(R cos Ѳ)ω² is provided
by the component of mo
In the direction to the axis
of the rotation.
-At the N-pole, Ѳ=90°
g’=g
-At the equator, Ѳ=0°
g’=g-Rω²

Gravitational Potential

• The strength of the gravitational force at a
point in a gravitational field is described by
the gravitational field strength E or g is a
vector quantity.

• Another quantity associated with the point in
the gravitational field is the gravitational
potential. It is a scalar quantity.

• The gravitational potential V at a point P in a
gravitational field is defined as the work done per
unit mass to bring a body from infinity to P. The
unit for gravitational potential is Jkgˉ¹.
• The gravitational potential energy U of a body at
a point P in a gravitational field is defined as the
work done to bring the body from infinity to P.
The unit for gravitational potential energy is J.
• Hence the gravitational potential energy U of a
body of mass m at a point where the gravitational
potential, V is given by
U = mV
Conversely,
U
V= m

• Figure below shows a body of mass m at a distance of x from
Mm
The gravitational force on the body is F = - G x²

P -dx m
M
R Mm
F=-G
x²
r
x

• The work done to move the body a small distance dx towards
the Earth, that is, a displacement ( dx ) is
dU = F (-dx)

= (-G Mm )( -dx)
x²
Mm
= G x² dx

• The work done to bring the body from infinity to the point P
which is at a distance r from the centre of the Earth is

r

∞

Gravitational potential energy, U = - GMm
r
• Gravitational potential at a distance r from the centre of the
Earth is
U
V= m
GM
Gravitational potential, V = - r
• The negative sign in the expression for U and V shows that the
work is done by the Earth to bring the mass from infinity.

• On the surface on the Earth, r = R
GM
– Gravitational potential, V = - R

GMm
– Gravitational potential energy, U = - R

• The graph illustrates the variation of the
gravitational potential V with distance r from
V
1
R r²
0 Eα r

1
- GM r
R Vα

Changes in Gravitational Potential Energy
• A body of mass m on the surface of the Earth has
Mm
gravitational potential energy, U = - G R
• When the body is raised through a small height
dR, the change in its gravitational potential energy
dU is obtained by differentiating U with respect to R.
Mm
dU = G R² dR
= mg (dR)

• If dR is written as h, then the change in gravitational
potential energy of a body of mass m when it raised
through a height h is mgh.
∆ U = mgh

 G is the Universal Gravitational Constant.
It is a scalar quantity with dimension

g is the acceleration due to gravity .
It is a vector quantity with dimension

Where g = accerelation due to
gravity
R = constant radius of
earth
G = universal gravitational
constant
M = mass of Earth

EXAMPLE :
A planet of mass M and radius r rotates about its axis
with an angular velocity large enough to substances on
its equator just able to stay on its surface . Find in
terms of M , r and G the period of rotation of the
planet.

7.5 Satellite Motion in
Circular Orbits

Satellite is a body that revolves round a
planet. Satellites can be categorized as
natural satellites or man-made satellites.
The moon, the planets and comets are
examples of natural satellites. Examples of
man-made satellites are Sputnik I , Measat
I ,II and III which are communication
satellites.

In order to launch satellite into orbit
, rockets are used. When rocket that
carries the satellite reaches the required
height , the satellite is launched into
circular orbit with a certain velocity v that
is tangential to intended orbit.

Since the centripetal force required for
the circular motion of the satellite is
provide by the gravitational attraction
of the Earth on the satellite,
(mv2) / r = G(Mm) / r2
Velocity of satellite,
v = [ (GM) / r]1/2
since GM= gR2,
v = [ (gR2) / r]1/2

For a satellite orbiting close to the
Earth’s surface , the radius of the
satellite’s orbit is approximate to the
radius of the surface.
( r = R)
V = 7.92 x 103 ms-1
Period of satellite’s orbital , T = 85
minutes

Synchronous Satellite is the communication
satellite in which the orbit of it is synchronised
with the rotation of the Earth about its axis .
It should have the characteristics of :
Period of its orbit = 1.0 day (which is equals to
the period of the earth’s rotation about its
axis)

The satellite orbits the Earth in the same
direction as the direction of the Earth.
The satellite’s orbit is in the same plane as the
equator because the centripetal force is
provided by the gravitational attraction of the
Earth.

As Qravitational force = Centripetal force
G[ (Mm) / r2] = mrw2
r3 = GM * T / (2π) + 2

*Radius of satellite’s orbit , r = 4.24 x 10 7 m
(6.6 times the radius of the Earth)

1. Planet is a body in orbit round the Sun which
sastifies the condition of
•Has enough mass to form itself into a spherical shape

•Has cleared its immediate neighbourhood of all smaller
objects

2. Due to the above definition of ‘planet”, Pluto was
excluded from planethood and reclassified as a
‘dwarf planet”.

3.Planets orbit the Sun in elliptical orbits. As an
approximation, we assume that the orbits of the
planets are circular.

If a planet of mass m is in a circular orbit of radius r
round the Sun (mass Ms), the centripetal force is
provided by the gravitational attraction between the
planet and the Sun.

G[ (Mm) / r2] = mrw2
r3 = GM * T / (2π) + 2
T2/ r3 = (4 π2) / GMs
= constant

(which sastifies Kepler’s Third Law)

Energy of a Satellite
Total energy of the satellite ,
E = U+ K
E = -(GMm)/r + (GMm)/2r
As an object approaches the Earth ,
*Its kinectic energy K increases as the
gravitational pull of the Earth on it increase.
*Its gravitational potential energy U decrease ,
becoming more and more negative.
*Its total energy E also increase.

1.The apparent loss of weght is known
as weightlessness.
For example , an astronaut in a
spacecraft feels that he has no mass as
all objects in the spacecraft experience
an apparent loss of weght.

2. A spacecraft orbiting the Earth has
an acceleration that equals the
acceleration due to gravity g at the
position of th spacecraft.

3. If N = normal reaction of the floorboard of the
spacecraft on an object of mass m ,
Using F = ma,
Mg – N = ma
N=0
*means that the floorboard does not exert any
force on the object.

4. Condition for a person to experience
weightlessness :

The person must fall towards the Earth
with an acceleration that equals the
acceleration due to gravity.

When a rocket is launched
from the earth’s surface , it can
escape from gravitational pull
of the earth if it has sufficient
energy to travel to infinity.

The energy required come
from the kinetic energy when
the rocket is launched .

The minimum velocity is also
known as escape velocity .

Expression for escape velocity
Using the principle of conservation of energy,

Kinetic energy required by the body on the
Earth’s surface

=
Gain in gravitational potential energy of the
body when it moves from the Earth ‘s surface
to infinity.

Case 1: What is the
escape velocity on
earth with respect to
Earth's gravity?

Mass of the earth =
5.98x1024kg,

R = 6378100 m,

G = 6.6726 x 10-11N-
m2/kg2.

Chapter 7 gravitation

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