Q. A satelite is in a circular orbit around the earth with a period of 6.0 hours. A retrorocket thruster is fired which slows the satelite down to half its speed and places it in a new orbit.
G = 6.67 x 10 power of -11 N.m2/kg2
mass of earth = 5.98 x10 power of 24 Kg
radius of the earth = 6,38 x 10 power of 6 m
a) Find the satellites new speed after the thruster is fired.
Answer = 2400m/s
b) Find the altitude of the satelites new orbit.
Answer = 61000000m
I know the answers for both questions a) and b) and solutions to a) but i don't know the solutions to b)
How did they get 61000000m for question b)???
a) Use G Me/R^3 = 4 pi^2/P^2 for the original orbit radius, R.
P = 6 h = 21,600 s
R^3 = G Me P^2/(4 pi^2)
(This is Keper's third law)
R = 1.678*10^7 m = 2.63 Re
original orbit speed
= 2 pi R/(21,600 s) = 4.88*10^3 m/s
Final orbit speed = half of above = 2440 m/s. They must have rounded it to 2 significant figures to get 2400 m/s.
b) If the new orbit is circular also, it will require some special maneuvers. A short retro burn will yield an elliptical orbit. If we assume a new circular orbit at half the speed, use the fact that V^2 * R = constant for circular orbits. If V is reduced by half, R must increase by a factor of 4.
R = 4 * 1.68*10^7 = 6.72*10^7 m
That does not quite agree with your "book answer" of 6.1*10^7. Check my numbersposted by drwls