A satellite travels initially in an approximately circular orbit 640 km above th

Question

A satellite travels initially in an approximately circular orbit 640 km above the surface of the Earth; its mass is 220 kg. (a) Determine its speed. (b) Determine its period of revolution. (c) For various reasons the satellite loses mechanical energy at the (average) rate of 1.40 x 105 J per orbital revolution. Adopting the reasonable approximation that the trajectory is a circle of slowly diminishing radius, determine the distance from the surface of the Earth, the speed, and the period at the end of its 1500th orbital revolution. (d) What is the magnitude of the average retarding force? (e) Is angular momentum conserved?

Explanation / Answer

(A) r = 640 km + 6371 km = 7011 km

r = 7011 x 10^3 m

for circular path,

Grvaitational force = m x centripetal force

G M m / r^2 = m v^2 / r

v = G M /r

v = 7423 m/s

(b) T = 2pi r / v = 5934.34 sec or 1.65 hr

(c) E = KE + PE

E = m v^2 /2 - G M m / r

= G M m / 2r - G M m / r

= - G M m / r

after 1500 th orbita,

Energy = - G M m / r - (1500 x 1.40x 10^5 )

- G M m / 2r' = - G M m / 2r - (1500 x 1.40x 10^5 )

- (6.67 x 10^-11)(5.972 x 10^24)(220) / 2r' =6.04 x 10^9

r' = 7036 km

distance from surafce = 7036 - 6371 = 665 km

Speed = v^2 = G M / r'

v' = 7524.2 m/s

T = 2pi r' / v' = 5875.5 sec or 1.63 hrs

(d) F d = change in E

F ( 1500 x 2 x pi x (7023.5 x 10^3)) = 1500 x 1.40x 10^5

F = 3.17 x 10^-3 N

(e) Li = m v r = 1.145 x 10^13 kg m^2 /s

Lf = m v' r' = 1.165 x 10^13 kg m^2 /s

Li is not equal to Lf.

so angular momentum is not conserved.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.