A circular orbit is the result of choosing a tangential velocity such that Earth’s surface curves away at the same rate as the object falls toward Earth.
We can summarize our discussion of orbiting satellites in the following Problem-Solving Strategy.
Problem-solving strategy: orbits and conservation of energy
Determine whether the equations for speed, energy, or period are valid for the problem at hand. If not, start with the first principles we used to derive those equations.
To start from first principles, draw a free-body diagram and apply Newton’s law of gravitation and Newton’s second law.
Along with the definitions for speed and energy, apply Newton’s second law of motion to the bodies of interest.
The international space station
Determine the orbital speed and period for the
International Space Station (ISS).
Strategy
Since the ISS orbits
above Earth’s surface, the radius at which it orbits is
. We use
[link] and
[link] to find the orbital speed and period, respectively.
which is about 17,000 mph. Using
[link] , the period is
which is just over 90 minutes.
Significance
The ISS is considered to be in low Earth orbit (LEO). Nearly all satellites are in LEO, including most weather satellites. GPS satellites, at about 20,000 km, are considered medium Earth orbit. The higher the orbit, the more energy is required to put it there and the more energy is needed to reach it for repairs. Of particular interest are the satellites in geosynchronous orbit. All fixed satellite dishes on the ground pointing toward the sky, such as TV reception dishes, are pointed toward geosynchronous satellites. These satellites are placed at the exact distance, and just above the equator, such that their period of orbit is 1 day. They remain in a fixed position relative to Earth’s surface.
Check Your Understanding By what factor must the radius change to reduce the orbital velocity of a satellite by one-half? By what factor would this change the period?
In
[link] , the radius appears in the denominator inside the square root. So the radius must increase by a factor of 4, to decrease the orbital velocity by a factor of 2. The circumference of the orbit has also increased by this factor of 4, and so with half the orbital velocity, the period must be 8 times longer. That can also be seen directly from
[link] .
Determine the mass of Earth from the orbit of the Moon.
Strategy
We use
[link] , solve for
, and substitute for the period and radius of the orbit. The radius and period of the Moon’s orbit was measured with reasonable accuracy thousands of years ago. From the astronomical data in
Appendix D , the period of the Moon is 27.3 days
, and the
average distance between the centers of Earth and the Moon is 384,000 km.
Solution
Solving for
,
Significance
Compare this to the value of
that we obtained in
[link] , using the value of
g at the surface of Earth. Although these values are very close (~0.8%), both calculations use average values. The value of
g varies from the equator to the poles by approximately 0.5%. But the Moon has an elliptical orbit in which the value of
r varies just over 10%. (The apparent size of the full Moon actually varies by about this amount, but it is difficult to notice through casual observation as the time from one extreme to the other is many months.)
