(a) What is the speed of a 237 kg satellite in an approximately circular orbit 6
ID: 1969858 • Letter: #
Question
(a) What is the speed of a 237 kg satellite in an approximately circular orbit 646 km above the surface of Earth?__ m/s
(b) What is the period?
__ min
Suppose the satellite loses mechanical energy at the average rate of 1.7 105 J per orbital revolution. Adopt the reasonable approximation that the trajectory is a "circle of slowly diminishing radius".
(c) Determine the satellite's altitude at the end of its 1700th revolution.
__ m
(d) Determine its speed at this time.
___m/s
(e) Determine its period at this time.
__ min
(f) What is the magnitude of the average retarding force on the satellite?
___ N
(g) Is angular momentum around Earth's center conserved for the satellite and the satellite Earth system (assuming that system is isolated)?
(A)The angular momentum of the satellite is conserved.
(B)The angular momentum of the satellite Earth system is conserved.
(C)Both are conserved.
(D)Neither are conserved.
Explanation / Answer
a. r.earth = 6370 km g.earth = 9.8 m/s^2 find a at 635 km a = g * (6370 km/(6370km+653km))^2 a = ..8227g Set a = centripetal force a = v^2/r v = v (a r) v = v (.8227 * 9.8m/s^2 * (6370km + 653 km) * 1000m/km) v = 7.524 km/s b. period v = 2p r/P P = 2pr/v P = 5,864 sec = 97.7 min c. Use PE = -mMG/r MG = 3.986 x10^14 For circular orbits, KE = 1/2 PE E = PE - KE E = 1/2 PE E loss through 1600 revolutions ? E = -1.2x10^5J * 1600 ? E = 1/2 mMG (1/(6370km + 653km) - 1/r) r = 1/(1/(6370km + 653km) - 2? E/mMG) r= 6928 km ..... (Note, because ?E is negative, you add it) altitude = 6928 - 6370 = 458km d. speed v = v (2/m(1/2 m(7.524 km/s)^2 + 1/2?E)) v = v ((7.524 km/s)^2 + ?E/m) v = 7,577 m/s ... This is a surprizing result. You slow a spaceship down to get to a lower orbit, and surprize, the velocity is actually faster. e. period P = 2pr/v P=5744 sec = 95.74 min f. retarding force = ?E per orbit/Circumference of Orbit g. 3) The angular momentum of the satellite Earth system is conserved Note: Angular momentum is only conserved when the force is directed towards the center of the earth, ergo, the angular momentum of the satellite is not conserved. However, the Earth-satellite system is conserved because there are no outside forces. The deceleration of the satellite adds to the earth's spin.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.