Question
Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 250 above the earth's surface; at the high point, or apogee, it is 2500 above the earth's surface.
a.)Using conservation of energy, find the speed at perigee and the speed at apogee.
b.)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?
c.)What if the rockets were fired at apogee?
Explanation / Answer
(a) finding the semimajor axis: a = (1/2) (2500+6380 + 250+6380) = 7755km = 7.75 x106 meters from the Keplers third law: T2 = (42 / GM)a3 = (42 / 6.673 x 10-11 * 5.98 x1024 ) * (7.75 x 106 )3 = 459.89 x 105 T = 6781.51 sec or 1.88 hours (b) distance from center of Earth at apogee = 2500+ 6380 = 8880 km distance from center of Earth at perigee = 250 + 6380 = 6630 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 = = 8880/6630 = 1.339 (c) you know v = 1.339u 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 /8880000 - 1/6630000 ) = v2 - (1.339 v)2 -120.43 x106 = -0.7929v2 v = 12324.27 ( speed at perige. u = 12324.27 / 1.339 = 9204.09 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. (c) you know v = 1.339u 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 /8880000 - 1/6630000 ) = v2 - (1.339 v)2 -120.43 x106 = -0.7929v2 v = 12324.27 ( speed at perige. u = 12324.27 / 1.339 = 9204.09 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. v = 12324.27 ( speed at perige. u = 12324.27 / 1.339 = 9204.09 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. 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.