Le Fevre High School
SACE Stage 2 Physics
Gravitation and Satellites 1
1. A satellite is in orbit 480 km above the earth's surface. Its period of rotation about the earth is
100 minutes. If the orbit is taken to be circular, calculate its linear speed. The mean radius of
the earth is 6.4 x 106 m.
Period, T = 100 min. = 6000 s
radius of satellite's orbit = radius of earth + height above earth
= 6400 + 480
= 6880 km
= 6.88 x 107 m
Therefore linear speed
v=
3
2R 2(6.88 x 10 6 )

= 7205 = 7.2 x 10 m s-1
T
6000
2. (a) State Newton's Law of Gravitation.
For any two point masses in the universe there is a force of attraction acting on each
mass along the line joining their centres. Each of these forces has the same magnitude and
this magnitude is directly proportional to the product of their masses and inversely
proportional to the square of their distance apart.
F
F
m1
F
m1m2
r2
(b) Calculate the gravitational force of attraction acting on a person of mass 80 kg as a result of the
presence of another person of mass 40 kg at a distance of 2 m. What is the force on the 40 kg
person?
F=
6.67 x 1011 x 80 x 40
22
-8
= 5.3 x 10 N
The force on the 80 kg person is the same size but opposite in direction. Note that the
forces are the same size even though the masses are quite different. Obviously this is a
negligible force compared to say the force due to the interaction with the much larger
earth.
Le Fevre High School
3. Use Newton's law of gravitation and the value of ‘g’ at the earth's surface to estimate the mass
of the earth. The mean radius of the earth is 6.4 x 106 m.
Using Newton's second law the force on a mass, m at the earth's surface = ma (where a = g)
Using the law of gravitation the force is given by:
G m mE
F=
where R = radius of earth
R2
mE = mass of earth
G m mE

= mg
R2
G mE
so
=g
R2
g R2
hence mE =
G
9.8 x (6.4 x 106 ) 2
=
6.67 x 1011
24
= 6.0 x 10 kg
note that this method of "weighing" the Earth requires measurement of the two constants G and
g. Both can be measured in the laboratory.
4. (a) Calculate the radius of the orbit of a satellite, which always stays vertically above the same
place on the earth's surface. Such a satellite is said to be in synchronous orbit or geostationary
orbit.
T = 24 hours = 24 x 3600 s
F = F gravity
m S v 2 = G mS m E
where R is radius of orbit and ms is mass of satellite.
R
R2

so
2
G mS
 2 
R  
T
R2
3
G mE T2
R =
42
6.67 x1011 x 6.0 x1024 x (24 x 3600)2
=
39.5
22 3
= 7.56 x 10 m
 R = 4.2 x 107 m
height above earth's surface
= 4.2 x 107 - .64 x 107
= 3.6 x 107 m (i.e. 6 earth radii above surface)
(b) Give a reason why synchronous orbits are used for telecommunication satellites.
Note that three such satellites could be positioned as shown and
communicate directly with one another. They are far enough out
for the earth to be out of the way.
Le Fevre High School
5. (a) Use the following data for two of Jupiter's satellites to check Newton's Law of Gravitation.
Satellite
period of revolution
radius of orbit
(x 105 s)
(x 105 km)
Europa 11
3.06
6.67
Ganymede 111
6.16
10.6
If Newton's Law is valid then
F = F gravity
m v 2 = G m J mS
where ms = mass of satellite
R
R2
2
G mJ
 2 
 R  
T
R2
R 3 G mJ

so
= constant
T2
4 2
R 3 (6.67 x 108 )3

(b) For Europa II.
=
T 2 (3.06 x 105 ) 2
= 3.17 x 1015 m3 s-2
For Ganymede III.
R3
T2

(1.07 x 109 )3
(6.16 x 105 ) 2
= 3.14 x 1015 m3 s-2
Hence
R3
very nearly constant, differences are a result of non circular orbits and
T2
consequent errors in R. We conclude that Newton's Law of gravitation is
consistent with this data.
(c) How can you measure masses of distant planets? Just measure the period and radius of
rotation of one or more of its satellites of course! Use the data above to estimate the mass
of Jupiter.
R3
T2


G mJ
4 2
42 R 3
R3
mJ =
use ave value of 2
G T2
T
39.5
mJ =
x 3.2 x 1015
11
6.67 x 10
= 1.9 x 1027 kg (over 300 earth masses!)
Le Fevre High School
6.
Show that for a satellite to be in a circular orbit about the centre of the earth its speed must
be given by v = g R where R is the radius of the orbit and g is the acceleration due to gravity
at the radius R.
Fc  Fg
m = mass of satellite
R = radius of Earth
v = speed of the satellite
2
mv
 mg
R
v 2  gR
v  gR
7.
Calculate:
(i) the linear speed of a point on the Earth's equator due to the Earth's rotation.
T  24hrs  24  60  60s
r  6380km  6.38  10 6 m
2r
v
2r
v
T
2    6.38  10 6
v
24  60  60
v  463.88ms 1  4.63  10 2 ms 1
T
(ii) the magnitude of the centripetal force acting on a mass of 2.0 kg at the equator due to
the Earth's rotation.
Fc 
mv 2
r


2
2.0  4.63  10 2
Fc 
6.38  10 6
Fc  0.0672  6.72  10 2 N