Consider a spacecraft in an elliptical orbit around the earth. At the lowest point or perigee, of its orbit it is 300 km above the earth’s surface at the highest point or apogee, it is 3000 km above the earth’s surface.
Text Solution
Verified by Experts(i) (v 1 ); (ii) Where tgk; (i) ; (ii) we get v 1
T = 
= 7.16 × × 10 3 sec.
= 
= 1.4
V p = 896×10 2 
m/sec. = 8.35 × × 10 3 m/s,
V a = 896×10 2 
m/sec = 5.95 × × 10 3 m/s
Δ V = 14× 10 2 
– V P = 3.09 × × 10 3 m/s, perigee
Sol. Total distance from apogee to perigee
300 + 2(6400) + 3000 = 2a
a = 8050 km
Time period of the spacecraft
T 2 = 
T = 
= 
T = 
= 
Apply angular momentum conservation about the center of earth, between perigee and Apogee.
mv 1 r min = m v 2 r max .............(i)
(v 1 ) (300 + 6400) = v 2 (3000 + 6400)
= 
Also apply energy conservation between perigee an Apogee
mv 1 2 + 
= 
mv 2 2 + 
.............(ii)
Where tgk¡ r min = (300 + 6400)km and r max = (3000 + 6400)km
From eqn. (i) & (ii) we get
v 1 = 8.35 × 10 3 m/sec
v 2 = 5.95 × 10 3 m/sec.
To escape, velocity at r → ∞ should be zero.
Applying energy conservation between perigee and r → ∞ .
ki + U = k f + U f
mv 1 2 + 
= 0 + 0
v 1 = 11.44 × 10 3 m/sec.
Increase in speed = 11.44 × 10 3 – 8.35 × 10 3
= 3.09 × 10 3 m/sec.
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