This document discusses orbital velocity of satellites. It explains that artificial satellites revolve around planets like the Earth. Communication satellites take 24 hours to complete one revolution, appearing stationary relative to Earth. This stable orbit is called a geostationary orbit. The gravitational force between a satellite and Earth provides the necessary centripetal force for the satellite to maintain its orbit. An equation shows the relationship between orbital velocity, radius of orbit, and acceleration due to gravity.

2. Class: 9th
Subject: Physics
Unit 1: Gravitation
Topic: Orbital velocity of Satellite

3. ARTIFICIAL SATELLITES
.
ARTIFICIAL SATELLITES
An object that revolves around a planet is called a satellite.
Types of Satellite
Natural Satellite:The moon revolves around the Earth so
moon is a natural satellite of the Earth.
Artificial Satellite:Scientists have sent many objects into
space. Some of these objects revolve around the Earth. These
are called artificial satellites. Most of the artificial satellites,
orbiting around the Earth are used for communication
purposes. Artificial satellites carry instruments or passengers
to perform experiments in space

4. Communication satellites
They take different time to complete their one revolution
around the Earth depending upon their distance h from
the Earth. Communication satellites take 24 hours to
complete their one revolution around the Earth. As Earth
also completes its one rotation about its axis in 24 hours,
hence, these communication satellites appear to be
stationary with respect to Earth. It is due to this reason
that the orbit of such a satellite is called geostationary
orbit.

5. Motion of Artificial Satellite
A satellite requires centripetal force that keeps it to move around the
Earth. The gravitational force of attraction between the satellite and the
Earth provides the necessary centripetal force
Consider a satellite of mass m revolving round the Earth at an altitude h
in an orbit of radius r with orbital o velocity vo . The necessary
centripetal force required is given by equation
R+h
r0= R+h
R> > >h
r0 = R

8. Numerical
5.1:Find the gravitational force of attraction between two
spheres each of mass 1000 kg. The distance between the
centres of the spheres is 0.5 m.
m1 = 1000kg m2 = 1000kg d=0.5m
F= 6.673 Nm2kg-2x 10-11x1000kgx1000kg
0.5mx0.5m
F= 26.67x 10 -5N
F =2.67x 101 x10-5 N
F =2.67x10-4 N

9. Numerical
5.2:The gravitational force between two identical lead spheres kept at 1
m apart is 0.006673 N.Find their masses. F=0.0 06673 N
F=0.0 0 6 6 7 3 N
F= 6.673x 10-3 N
6.673x 10-3 N = 6.673x 10-11 Nm2kg-2 x mass xmass
1mx1m

10. The gravitational force between two identical lead spheres kept at
1 m apart is 0.006673 N.Find their masses. F=0.0 06673 N
6.673x 10-3 N = 6.673x 10-11 Nm2kg-2 x mass x mass
1m x1m
6.673x 10-3 N x 1m2 = mass2
6.673x 10-11 Nm2kg-2
108 kg2 = mass2
Taking Under root
104 kg = mass

11. ASSIGNMENT
5.2:The gravitational force between two
identical lead spheres kept at 1 m apart is
0.006673 N.Find their masses. F=0.0 06673 N
5.1:Find the gravitational force of
attraction between two spheres each of
mass 1000 kg. The distance between the
centres of the spheres is 0.5 m.

12. Thanks

15. 14th Sep,21

16. VARIATION OF G WITH ALTITUDE
Equation
shows that the value of acceleration due to gravity
g depends on the radius of the Earth at its surface.
The value of g is inversely proportional to the
square of the radius of the Earth. But it does not
remain constant. It decreases with altitude.
Altitude is the height of an object or place above
sea level. The value of g is greater at sea level than
at the hills.

17. VARIATION OF G WITH ALTITUDE
Consider a body of mass m at an altitude h as shown in
figure 5.5. The distance of the body from the centre of
the Earth becomes R + h. Therefore, using equation (5.6),
we get

18. According to the above equation, we come to
know that at a height equal to one Earth radius
above the surface of the Earth, g becomes one
fourth of its value on the Earth. Similarly at a
distance of two Earths radius above the Earth's
surface, the value of g becomes one ninth of its
value on the Earth.

22. Mass of Mars = Mm = 6.42 × 1023 kg
Radius of Mars = Rm = 3370 km = 3370 × 1000 m = 3.37 × 108
m
Acceleration due to gravity of the surface of Mars = gm =?
gm = G Mm/R2
or gm = 6.673 × 10-11 × 6.42*1023 /(3.37*106)2
= 6.673*10-11 * 6.42*1023 / 11.357
2 3370 km. = 42.84/11.357 = 3.77 m-2
Find the acceleration due to gravity on the surface of
the Mars. The mass of Mars is 23 6.42x10 kg and its
radius is 3370km

23. Rough work

24. A satellite requires centripetal force that keeps it to move
around the Earth. The gravitational force of attraction
between the satellite and the Earth provides the necessary
centripetal force
Consider a satellite of mass m revolving round the Earth at an
altitude h in an orbit of radius r with orbital o velocity vo .
The necessary centripetal force required is given by equation
MOTION OF ARTIFICIALSATELLITES
gh r0= v0
2

26. gh= GMe
(R+2R)2
gh=GMe
(3R)2
gh= GMe
9 R2
gh= 1 g :g=GMe
9 R2
g=GMe
R2

27. gh= GMe
(R+R)2
gh=GMe
(2R)2
gh= GMe
4 R2
gh= 1 g :g=GMe
4 R2

28. √ gh r0= √ v0
2
√ gh r0 = v0
r0= R+h
R> > >h
r0 = R
√ gh R= v0 g= gh
√ g R= v0
A satellite revolving around very close to the Earth,
has speed v nearly 8 km/s or 29000 km/h