Physics Gravitation General MCQ (Single Correct)

What are:

A
the speed and
B
the period of a 220 kg satellite in an approximately circular orbit 640 km above the surface of the earth? Suppose the satellite loses mechanical energy at the average rate of 1.4 × 10 5 J per orbital revolution. Adopting the reasonable approximation that due to atmospheric resistance force, the trajectory is a “circle of slowly diminishing radius”. Determine the satellite’s
C
altitude
D
speed & (e) period at the end of its 1500 th revolution. (f) Is angular momentum around the earth’s center conserved for the satellite or the satellite-earth system.

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Text Solution

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The correct answer is:
CHECK THE SOLUTION.

km/s = 7.527 km/s

sec. ≈ 1.63 hour

km ≈ 411.92 km

km/sec. ≈ 7.67 km/s

(e) sec. ≈ 1.55 hour

(f) No

Sol.

Orbital speed

v = =

= 7.53 km/sec.

Time period

T 2 =

T 2 = (6400 + 640 km) 3

T = 1.63 hours

Initial mechanical energy =

=

Total loss in mech. energy during 1500 rev.

= (1.4 × 10 5 ) × 1500

= 21 × 10 7 J.

Final mechanical energy remaining after 1500 rev.

TE f = TE i – Δ E loss

– 21 × 10 7 J

Solving get r g = 6812 km

Height from earth's surface = 6812 – 6400 = 412 km

Final orbital velocity

v f = =

v f = 7.67 km/sec.

(e) Time period T = = = 1.55 hours

(f) Due to Air resistance, net torque about the earth is non–zero.

So, angular momentum about the earth will not remain conserved.

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