Satellite Speed at Extremes

Question:

A satellite in moving in an eliptical orbit of eccentricity 0.25. Determine its speed at aphelion (maximum distance) and perihelion (minimum distance). The perihelion is 2e6 meters above the Earth.

Answer:

A satellite 2e6m above the Earth is 8.378e6m from the Earth's center. In this problem that is given as the perihelion. The eccentricity e=.25 is the ratio of the distance from the center to a focus of the elliptical orbit, cf, to the length of the semimajor axis, a, of the orbit. The semimajor axis, a, is also the sum of the perihelion and cf. This gives us cf/a=0.25 and cf+8.378e6=a. Solving for a we get
a*0.25+8.378e6=a, or, 0.75*a=8.378e6, or a=1.117e7m
cf=2.7925e6m

The total energy of the satellite in elliptical is -G*M*m/2*a where G is the universal gravitation constant, M is the mass of the Earth, m is the mass of the satellite and a is the semimajor axis of the orbit. This energy is the sum of the kinetic and potential energy of the satellite. The potential energy is -G*M*m/r where r is the distance from the satellite to the center of the Earth. So we have
1/2*m*v²-G*M*m/r=-G*M*m/2*a, v²=G*M/r-G*M/2*a
Plugging in the known values for G, M, r(at perhelion) and a, we get v at perihelion
v²=2*6.673e-11*5.976e24*(1/8.378e6-1/2.234e7)
v²=2*6.673e-11*5.976e24*7.46e-8=5.95e7
v=7.71e3m/s at perihelion.

At aphelion
r=cf+a=0.27925e7+1.117e7=1.3962e7

The speed at aphelion=speed at perihelion times the ratio of the distances
v=7.71e3*8.378e6/1.396e7=4.63e3m/s

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