Question
Consider a spacecraft in an elliptical orbit around theearth. At the low point, or perigee, of its orbit, it is 400km above the earth's surface; at the high point, or apogee, it is4000 km above the earth's surface. a) What is the period of the spacecraft's orbit? b) Using conservation of angular momentum, find the ratio ofthe spacecraft's speed at perigee to its speed at apogee. c) Using conservation of energy, find the speed at perigee andthe speed at apogee. d) It is necessary to have the spacecraft escape from theearth completely. If the spacecraft's rockets are fired atperigee, by how much would the speed have to be increased toachieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use? Consider a spacecraft in an elliptical orbit around theearth. At the low point, or perigee, of its orbit, it is 400km above the earth's surface; at the high point, or apogee, it is4000 km above the earth's surface. a) What is the period of the spacecraft's orbit? b) Using conservation of angular momentum, find the ratio ofthe spacecraft's speed at perigee to its speed at apogee. c) Using conservation of energy, find the speed at perigee andthe speed at apogee. d) It is necessary to have the spacecraft escape from theearth completely. If the spacecraft's rockets are fired atperigee, by how much would the speed have to be increased toachieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?
Explanation / Answer
(a) first find the semimajor axis: a = (1/2) (4000+6380 + 400+6380) = 8580 km = 8.58 x106 meters . Then use Keplers third law: T2 = (42 / GM)a3 = . = (42 / 6.673 x 10-11 * 5.98 x1024 ) * (8.58 x 106 )3 . = 624.887 x 105 . T = 7905 seconds is the period, whichis 2.20 hours . (b) distance from center of Earth at apogee = 4000+ 6380 = 10380 km . distance from center of Earth at perigee = 400 + 6380 = 6780 km . by cons of angular momentum: distance * speed at perigee = distance * speed at apogee . So... speed at perigee / speed atapogee = distance at apogee / distance at perigee = . = 10380 /6780 = 1.531 . (c) you know v = 1.531u where v is speedat perigee and u is speed at apogee . also... diff in potential energy = diff in kinetic energy (because energy isconserved) . -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 /10380000 - 1/6780000 ) = v2 - (1.531 v)2 . -40.8251 x106 = -1.3439v2 . v = 5512 m/s (speed at perigee) . u = 5512 / 1.531 = 3600m/s (speed at apogee) . (d) 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 = 5512 m/s (speed at perigee) . u = 5512 / 1.531 = 3600m/s (speed at apogee) . (d) 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.