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Let’s say you were able to build a tunnel through
the center of the Earth to the opposite side. If
you were to jump in, you would accelerate
toward the center. However your acceleration
decrease as your distance to center decreases.
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Let consider a very fast runner.
Being that the shape of the Earth is
spherical, the runner should experience
an centripetal force.
RE
Centripetal Force
Ignoring air drag and the terrific sonic boom that would
accompany such a speed, how fast must the runner
achieve to balance the force of gravity?
= 8km/sec
At this speed, the runner is in a circular orbit right above the surface r=RE.
What time is required for the runner to make a single circumference or orbit? So why are the periods the same?
x
RE
θ
r
y
Magnitude of
gravitational force
does not change
Gravitational force
vector is changing
Let’s look at the force
projected onto the X axis
RE
Let say the fast runner entered an circular orbit with distance of 3.84x108m from the center of Earth. What is the period of this orbit? About a month
The average distance separating Earth and the Moon center to center.
A geosynchronous satellite is one whose orbital period is
equal to one day. If such a satellite is in a circular orbit above
the equator moving directly East or West?, it will be in a fixed
position with respect to the ground.
Find the altitude above the Earth’s surface where a satellite orbits with a period of one day R
E

 637  106 m, M E  597  1024 kg, T  1 day  864  104 s
r  T/ 2 
2/ 3
 GM E 
1/ 3
Substitute numerical values
r  422  107 m
Find the altitude above the Earth’s surface
r  RE  358 107 m
358  107 m  22,300 mi
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Kepler’s Laws of Orbital Motion
Johannes Kepler made detailed studies of the
apparent motions of the planets over many years,
and formulate three empirical laws:
The Danish astronomer Tycho
Brahe (1546–1601) recorded
the paths of the planets for
many years and was joined in
his work by Johannes Kepler
(1571–1630) in 1600, and
after Brahe’s death, Kepler
“inherited” his astronomical
observations.
Kepler made good use of
Brahe’s life work, extracting
from his carefully collected
data the three laws of orbital
motion we know today as
Kepler’s laws.
1571–1630
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#1. Planets follow elliptical orbits, with the Sun at one focus of the ellipse.
Kepler tried long and hard to find a circular orbit around the Sun that would match Brahe’s observations of Mars. Up to that time everyone from Ptolemy to Copernicus believed that celestial objects moved in circular paths of one sort or another. Though the orbit of Mars was exasperatingly close to being circular, the small differences between a circular path and the experimental observations just could not be ignored. Eventually, after a great deal of hard work and disappointment over the loss of circular orbits, Kepler discovered that Mars followed an orbit that was elliptical rather than circular. The same applied to the other planets
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y
Equation for an ellipse with the origin at center r1
r1+r2=2a
x
b
r2
straight line drawn from one of the focal points to any point on the curve and then back to the other focal point has the same length for every point on the curve. a
y
Polar equation for an ellipse with origin at one focus r
θ
x
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Kepler’s Laws of Orbital Motion
#2. As a planet moves in its orbit, it sweeps out an
equal amount of area in an equal amount of time.
When Kepler plotted the position of a planet on its elliptical orbit, indicating at each position the time the planet was there, he made an interesting observation. First, draw a line from the Sun to a planet at a given time. Then a certain time later—
perhaps a month—draw a line again from the Sun to the new position of the planet. The result is that the planet has “swept out” a wedge‐shaped area. 9
What is the torque on a planet orbiting a star?
lanet
We know r and F, but what about θ
the angle between r and F?
For the case for gravity, θ=1800
Star
Angular momentum L=constant
For a central force acting on a body in orbit, there will be no net torque, as the force will be anti‐parallel to the radius. Since the net torque is zero, the body will have a constant angular momentum. L = mvr = constant
r is the radius of the orbit, m is the mass of the body, and v is the velocity
The area A swept out in a time t will be (For small time t)
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Kepler’s Laws of Orbital Motion
#3. The period, T, of a planet increases as its
mean distance from the Sun, r, raised to the 3/2
power.
This can be shown to be a consequence of the
inverse square form of the gravitational force.
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Clicker
Two identical planets orbit a star in concentric
circular orbits in the star's equatorial plane. Of the
two, the planet that is farther from the star must
have
a.
b.
c.
d.
e.
the smaller period.
the greater period.
the smaller gravitational mass.
the larger gravitational mass.
the larger universal gravitational constant.
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General form of Kepler’s 3rd Law of Orbital Motion
Recall from Kepler’s 2nd Law
If the eccentricity of a planetary orbit is zero, then Kepler's laws state:
1.The planetary orbit is a circle
2.The Sun is in the center (or center of mass)
3.The speed of the planet in the orbit is constant
4.The square of the period is proportionate to the cube of the distance from the Sun.
If the eccentricity of a planetary orbit is not zero, then Kepler's laws state:
1.The planetary orbit is not a circle, but an ellipse
2.The Sun is not at the center but at a focal point (or center of mass)
3.Neither the linear speed nor the angular speed of the planet in the orbit is constant, but the area speed is constant.
4.The square of the period is proportionate to the cube of the mean between the maximum and minimum distances from the Sun.
Gravitational Potential Energy
Very close to the Earth’s surface, the gravitational potential increases linearly with altitude:
Gravitational potential energy, just like all other forms of energy, is a scalar. It therefore has no components; just a sign.
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