Consider a spacecraft in an elliptical orbit around the earth. At the low point,
ID: 1324582 • Letter: C
Question
Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 450km 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 thespeed at apogee.
It is necessary to have the spacecraft escape from the earthcompletely. If the spacecraft's rockets are fired at perigee, byhow much would the speed have to be increased to achieve this?
What if the rockets were fired at apogee?
Explanation / Answer
(a) finding the semimajor axis:
a = (1/2) (4500+6380 + 450+6380) = 8855 km
= 8.855 x106 meters
from the Keplers third law:
T2 = (4?2 / GM)a3
= (4?2 / 6.673 x 10-11 * 5.98 x1024 ) * (8.855 x 106 )3
= 459.89 x 105
T = 8288.03 sec = 2.30 hours
(b) distance from center of Earth at apogee = 4500+ 6380 = 10880 km
distance from center of Earth at perigee = 450 + 6380 = 6830 km
by conservation of angular momentum: distance * speed at perigee = distance * speed at apogee
speed at perigee / speed atapogee = distance at apogee / distance at perigee
= 10880 km / 6830 km
= 1.5929
= 1.60 (rounded)
(c) you know v = 1.5929 u
where v is speedat perigee and u is speed at apogee
diff in potential energy = diff in kinetic energy
-GMm /ra - GMm /rp = (1/2) mv2 - (1/2) m u2
eliminate m and multiply by 2,
2GM( 1 /rp - 1 /ra ) = v2 - u2
2 * 6.673 x 10-11 * 5.98 x1024 ( 1 /10880000 - 1/6830000 ) = v2 - (1.5929 v)2
-120.43 x106 = -0.7929v2
v = 8565.21 m/s ( speed at perige).
u = 8565.21 / 1.5929 = 5377.11 m/s (speed at apogee)
c)it is more efficient to fire the rocketsat apogee because a certain amount of work must be done by therockets to increase the energy of the ship. Work is force *distance, and since the ship is moving faster at apogee, it moves agreater distance while the rockets fire... and therefore therockets do greater work.