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Satellite Motion
Consider a satellite with mass Msat orbiting a central body with a mass of mass MCentral. The central body could
be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive
nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting
satellite is given by the relationship
Fnet = ( Msat • v2 ) / R
This net centripetal force is the result of the gravitational force that attracts the satellite towards the central
body and can be represented as
Fgrav = ( G • Msat • MCentral ) / R2
Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force can be set equal to each
other. Thus,
(Msat • v2) / R = (G • Msat • MCentral ) / R2
Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by
dividing through by Msat. Then both sides of the equation can be multiplied by R, leaving the following
equation.
v2 = (G • MCentral ) / R
Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a
central body in circular motion
where G is 6.673 x 10-11 N•m2/kg2, Mcentral is the mass of the central body about which the satellite orbits, and
R is the radius of orbit for the satellite.
Similar reasoning can be used to determine an equation for the acceleration of our satellite that is expressed in
terms of masses and radius of orbit. The acceleration value of a satellite is equal to the acceleration of gravity
of the satellite at whatever location that it is orbiting. The equation for the acceleration of gravity is given as
g = (G • Mcentral)/R2
Thus, the acceleration of a satellite in circular motion about some central body is given by the following
equation
where G is 6.673 x 10-11 N•m2/kg2, Mcentral is the mass of the central body about which the satellite orbits, and
R is the average radius of orbit for the satellite.
The final equation that is useful in describing the motion of satellites is Newton's form of Kepler's third law.
Since the logic behind the development of the equation has been presented elsewhere, only the equation will
be presented here. The period of a satellite (T) and the mean distance from the central body (R) are related by
the following equation:
where T is the period of the satellite, R is the average radius of orbit for the satellite (distance from center of
central planet), and G is 6.673 x 10-11 N•m2/kg2.
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There is an important concept evident in all three of these equations - the period, speed and the
acceleration of an orbiting satellite are not dependent upon the mass of the satellite.
None of these three equations has the variable Msatellite in them. The period, speed and acceleration of a
satellite are only dependent upon the radius of orbit and the mass of the central body that the satellite is
orbiting. Just as in the case of the motion of projectiles on earth, the mass of the projectile has no affect upon
the acceleration towards the earth and the speed at any instant. When air resistance is negligible and only
gravity is present, the mass of the moving object becomes a non-factor. Such is the case of orbiting satellites.
Example Problems
Practice Problem #1
A satellite is to orbit the earth at a height of 100. km (approximately 60 miles) above the
surface of the earth. Determine the speed, acceleration and orbital period of the satellite.
(Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
Like most problems in physics, this problem begins by identifying
known and unknown information and selecting the appropriate
equation capable of solving for the unknown. First, the radius of a
satellite's orbit can be found from the knowledge of the earth's radius
and the height of the satellite above the earth. As shown in the
diagram at the right, the radius of orbit for a satellite is equal to the
sum of the earth's radius and the height above the earth. These two
quantities can be added to yield the orbital radius. In this problem, the
100 km must first be converted to 100 000 m before being added to
the radius of the earth. The equations needed to determine the unknown are listed below. For this problem,
the knowns and unknowns are listed below.
Given or Known
Rsat = Rearth + height = 6.37 x 106 m + 1.00 x 105 m = 6.47 x 106 m
Unknown
v = ???
Mearth = 5.98x1024 kg
a = ???
G = 6.673 x 10-11 N m2/kg2
T = ???
Begin by determining the orbital speed of the satellite using the following (Equation #1 above):
v = SQRT [ (G•MCentral ) / R ]
The substitution and solution are as follows:
v = SQRT [ (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m) ]
v = 7.85 x 103 m/s
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The acceleration can be found from either one of the following equations:
(Equation #2) a = (G • Mcentral)/R2
a = v2/R
Equation (2) was derived above. The other is a general equation for circular motion. Either equation can be
used to calculate the acceleration. The use of equation #2 will be demonstrated here.
a = (G •Mcentral)/R2
a = (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m)2
a = 9.53 m/s2
Observe that this acceleration is slightly less than the 9.8 m/s2 value expected on earth's surface. Since the
satellite is above the earth’s surface, as the distance increases from the surface, the value of g decreases.
Finally, the period can be calculated using the following (Equation #3):
The equation can be rearranged to the following form
T = SQRT [(4 • pi2 • R3) / (G*Mcentral)]
The substitution and solution are as follows:
T = SQRT [(4 • (3.1415)2 • (6.47 x 106 m)3) / (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ]
T = 5176 s = 1.44 hrs
Practice Problem #2
The period of the moon is approximately 27.2 days. Determine the radius of the moon's orbit
and the orbital speed of the moon. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
Like before, begin by identifying known and unknown values.
Given/Known:
T = 27.2 days
Unknown
Mearth = 5.98 x 1024 kg
G = 6.673 x 10-11 N m2/kg2

Question: Which equation can be used to calculate the radius of orbit?
Rearrange the equation to solve for radius3:
R3 =
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Now substitute:
R3 =
Now take the cubed root to solve for R:
R = 3.82 x 108 m

What are the two formulas for determining the orbital speed of an object:
Substitute and calculate:
v = 1.02 x 103 m/s
Practice Problem #3
A geosynchronous satellite is a satellite that orbits the earth with an orbital period of 24 hours, thus
matching the period of the earth's rotational motion. A special class of geosynchronous satellites is a
geostationary satellite. A geostationary satellite orbits the earth in 24 hours along an orbital path that is
parallel to an imaginary plane drawn through the Earth's equator. Such a satellite appears permanently
fixed above the same location on the Earth. If a geostationary satellite wishes to orbit the earth in 24
hours (86400 s), then how high above the earth's surface must it be located? (Given: M earth = 5.98x1024 kg,
Rearth = 6.37 x 106 m)
Known:
Unknown:
T=
Mearth =
Rearth
G=
The unknown in this problem is the height (h) of the satellite above
the surface of the earth. Yet there is no equation with the variable h.
The solution then involves first finding the radius of orbit and using
this R value and the R of the earth to find the height of the satellite
above the earth. As shown in the diagram at the right, the radius of
orbit for a satellite is equal to the sum of the earth's radius and the
height above the earth. The radius of orbit can be found using the
following equation:
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1. Rearrange:
R3 =
2. Substitute:
R3 =
3. Solve for R3:
4. Solve for R:
R = 4.23 x 107 m
The radius of orbit indicates the distance that the satellite is from the center of the earth. Now that the radius
of orbit has been found, the height above the earth can be calculated. Since the earth's surface is 6.37 x 106 m
from its center (that's the radius of the earth), determine the height of the satellite:
So the height of the satellite is 3.59 x 107 m.
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Check Your Understanding
1. A satellite is orbiting the earth. Which of the following variables will affect the speed of the satellite?
a. mass of the satellite
b. height above the earth's surface
c. mass of the earth
2. Use the information below and the value of G [ 6.673 x 10-11 N•m2/kg2 ] to answer the following.
Sun
M = 2.0 x 1030 kg
Earth
M = 6.0 x 1024 kg
Saturn
M = 5.7 x 1026 kg
a. One of Saturn's moons is named Mimas. The mean orbital distance of Mimas is 1.87 x 108 m. The mean
orbital period of Mimas is approximately 23 hours (8.28x104 s). Use this information to estimate a mass for the
planet Saturn.
b. Consider a satellite which is in a low orbit about the Earth at an altitude of 220 km above Earth's surface.
Determine the orbital speed of this satellite.
c. Suppose the Space Shuttle is in orbit about the earth at 400 km above its surface. Use the information from
the previous question to determine the orbital speed and the orbital period of the Space Shuttle.
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