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Satellite Motion
Radial and angular motion
We will be concerned here with the orbiting motion of a satellite, such as a planet around the sun, or the moon around the earth. We neglect the effects of all other bodies.
We will assume that the satellite's mass (m) is much smaller than that (M) of the body it moves around. Both bodies actually orbit around the CM of the system, but M moves only a small amount because the CM is so close to it. We ignore this motion as an approximation, and take the center of M to be the origin of our coordinate system.
Because gravity (a central and conservative force) is the only force doing work:
The total mechanical energy E of the system is conserved.
The gravitational force acts along the line between the*…show more content…*

We will now prove the other two laws. Proof of the 1st law requires an extensive use of calculus, so we will not give it here. Shown is an elliptical orbit and the position of the satellite at two times differing by dt . The satellite moves a distance ds along the orbit, while the line connecting the bodies turns through angle d. The area swept out by the line is that of the narrow triangle Since ds r dwe have The rate at which area is swept out is Since L is a constant, we have shown that the rate at which area is swept out is constant. This is Kepler's 2nd law, which we see is a simple consequence of conservation of angular momentum. If we integrate this rate over the time for a complete revolution (the period T) we obtain the total area of the ellipse. Thus Area LT 2m The area of an ellipse is ab. Substituting from the formulas given above relating E and L to a and b, we find after some algebra: Kepler’s 3rd law a3 T2 GM 42 This shows that the ratio a3 /T2 is the same for all objects orbiting a given mass M. This is Kepler's 3rd law. r d ds By measuring a and T one can use this relation to determine the mass M. It is in

We will now prove the other two laws. Proof of the 1st law requires an extensive use of calculus, so we will not give it here. Shown is an elliptical orbit and the position of the satellite at two times differing by dt . The satellite moves a distance ds along the orbit, while the line connecting the bodies turns through angle d. The area swept out by the line is that of the narrow triangle Since ds r dwe have The rate at which area is swept out is Since L is a constant, we have shown that the rate at which area is swept out is constant. This is Kepler's 2nd law, which we see is a simple consequence of conservation of angular momentum. If we integrate this rate over the time for a complete revolution (the period T) we obtain the total area of the ellipse. Thus Area LT 2m The area of an ellipse is ab. Substituting from the formulas given above relating E and L to a and b, we find after some algebra: Kepler’s 3rd law a3 T2 GM 42 This shows that the ratio a3 /T2 is the same for all objects orbiting a given mass M. This is Kepler's 3rd law. r d ds By measuring a and T one can use this relation to determine the mass M. It is in

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