Consider a spacecraft in an elliptical orbit around the earth. At the low point,
ID: 1322543 • Letter: C
Question
Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 200km above the earth's surface; at the high point, or apogee, it is 4500km above the earth's surface.
Using conservation of energy, find the speed at perigee and the speed at apogee.
It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this?
What if the rockets were fired at apogee?
Which point in the orbit is more efficient to use?
Explanation / Answer
total energy of the satellite is constant and is just the sum of the potential and kinetic energies.
Kinetic = (1/2)m*v^2
Potential = -GMm/r
Total Energy = (1/2)m*v^2 - GMm/r
This total energy is also equal to 1/2 the potential energy at a distance equal to the semi-major axis of the orbit.
Energy = -(1/2)GMm/a
Equating these two gives:
(1/2)m*v^2 - GMm/r = -(1/2)GMm/a
Now solve this for "a" since we do not have the apogee distance.
v^2 = GM(2/r - 1/a)
v2/(GM) = 2/r - 1/a
1/a = 2/r - v2/(GM)
a = 1/[2/r - v^2/(GM)]
For the elliptical orbit the changes in velocity occur at perigee and apogee. Call these distances from the center of the earth r1 and r2.
2*a = r1 + r2
a = (r1 + r2)/2
(r1 + r2)/2 = 1/[2/r - v^2/(GM)]
r2 = 2/[2/r - v^2/(GM)] - r1
Get r2. We do this at the apogee point where r=r1, which we know, and we also know v.
G = 6.67300x10^(-11) m^3 kg^-1 s^-2
M = 5.9742x10^24 kg
h = 5.99x10^5 m
r1 = Re + h = 6378 + 599 km = 6.977x10^6 m
v = (circular orbit speed) + 2.4 km/s
v = SQRT(GM/r) = 9960 m/s
Using all this gives:
r2 = 4.58x10^7 m
We now know the altitude for the new circular orbit and can calculate the circular orbital velocity using SQRT(GM/r). We get:
V(circular new) = 2948 m/s
To get the speed change at this point we need to know the speed in the eliiptical orbit at apogee. We can use the equation for speed as a function of r from above and remembering that r = r2 at this point.
v^2 = GM(2/r2 - 1/a)
v = 1515 m/s
The speed change is then 2948 - 1515 = 1433 m/s = 1.43 km/s
So the speed change required to put the satellite into a circular orbit is 1.43 km/sec.
Check. The formulas for change in velocity are:
Dv1 = SQRT(GM/r1)[SQRT(2*r2/(r1+ r2)) - 1]
Dv2 = SQRT(GM/r2)[1 - SQRT(2*r2/(r1+ r2))]
If you use what was calculated and the given data these work out to:
Dv1 = 2.4 and Dv2 = 1.43