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*v^{2}-G*M*m/r=-G*M*m/2*a,
v^{2}=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

v^{2}=2*6.673e-11*5.976e24*(1/8.378e6-1/2.234e7)

v^{2}=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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