Download Circular Motion

Planetary & Satellite Motion
Kepler’s 3 Laws
 Law of Ellipses
 The path of the planets about the sun is elliptical in
shape, with the center of the sun being located at one
focus.
 Law of Equal Areas
 An imaginary line drawn from the center of the sun to
the center of the planet will sweep out equal areas in
equal intervals of time
 Law of Harmonies
 The ratios of the squares of the periods of any two
planets is equal to the ratio of the cubes of their average
distances from the sun.
Law of Ellipses
 All planets are orbiting the sun in a path described as
an ellipse
 An ellipse is a special curve in which the sum of the
distances from every point on the curve to two other
points in constant
Law of Equal Areas
 Describes the speed at which
any given planet will move
when orbiting the sun
 A planet moves fastest when
it is closest to the sun and
slowest when it is furthest
from the sun
Law of Harmonies
 Compares the orbital period and radius of orbit of a
planet to those of other planets
 The comparison being made is the ratio of the squares
of the periods to the cubes of their average distances
from the sun
Planet
Period
(s)
Average Dist.
(m)
Earth
3.156 x 107
1.4957 x 1011
Mars
5.93 x 107
2.278 x 1011
T2/R3
(s2/m3)
What did Newton Do???
 His comparison of the moons acceleration and the
acceleration of objects on earth allowed him to
conclude the moons is held in a circular orbit by the
force of gravity.
 The force that is inversely proportional to the distance
between the centers
 He used Kepler’s 3rd law (T2/R3) to find an average ratio
of
 k = 2.97 x 10-19 s2/m3
Example 1
 Suppose a planet is discovered that is 14 times as far
from the sun as the Earth’s distance is from the sun (1.5
x 1011 m). Use Kepler’s law of harmonies to predict the
orbital period of such a planet. GIVEN: T2/R3 = 2.97 x
10-19 m2/s3.
2
2
p
e
3
3
e
p
T
T

R
R
Example 2
 The average orbital distance of Mars is 1.52 times the
average orbital distance of the Earth. Knowing that
the Earth orbits the sun in approximately 365 days, use
Kepler’s law of harmonies to predict the time for Mars
to orbit the sun.
2
mars
3
mars
T
R
2
Tearth

3
Rearth
Satellites
 Any objects that is orbiting the earth, the sun, or an
other massive body
 A satellite is a projectile
 The only force governing its motion is the force of
gravity
 How fast do you have to launch it to get it into orbit?
Therefore, must be
launched at a speed
greater than 8000 m/s
to orbit earth!
Velocity, Acceleration, Force Vectors
 Velocity
 Directed tangent to the circle at every
point along its path
 Acceleration
 Directed toward the center of the circle
– toward the central body that it is
orbiting
 Net Force
 Directed inwards in the same direction
as acceleration
Elliptical Orbit of Satellites
 Velocity is tangent to the ellipse
 Acceleration is directed toward the focus of the ellipse
 Net Force is directed in the same direction in the same
direction as the acceleration
Mathematics of Satellite Motion
 Consider a satellite with mass, Msat orbiting a central
body with mass, Mcentral. If it moves in the circular
motion, the net centripetal force is
Fnet
(M sat * v 2 )

R
 The net centripetal force is a result of the gravitation
force:

(G * Msat * Mcentral )
Fgrav 
2
R
Mathematics of Satellite Motion cont.

 Since Fgrav = Fnet
M sat * v 2 (G * M sat * M central )

2
R
R
(G * M central )
v 
R
2
v = velocity of the satellite moving
G = 6.673 x 10-11 N*kg2/m2
Mcentral = mass of the central body which the satellite is orbiting
R = 
radius of the orbit of the satellite
Mathematics of Satellite Motion cont.
 Acceleration of gravity
G * M central
g
2
R
 Motion of Satellites


T2
4 * 2

3
R
G * M central
Example Problem 1
 A satellite wishes 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)
Example Problem 2
 The period of the moon is approximately 27.2 days
(2.35 x 106 s). 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)
Example 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: Mearth = 5.98x1024 kg, Rearth =
6.37 x 106 m)