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A satellite is put in an orbit just above the earths atmosphere with a velocity √(1.5) times the velocity for a circular orbit at that height. The initial velocity imported is horizontal. What would be the maximum distance of the satellite from the surface of the earth when it is the orbit?

A satellite is put in an orbit just above the earths atmosphere with a velocity √(1.5) times the velocity for a circular orbit at that height. The initial velocity imported is horizontal. What would be the maximum distance of the satellite from the surface of the earth when it is the orbit?

Grade:11

1 Answers

Vikas TU
14149 Points
7 years ago
From Energy conservation we get,
1/2 mv1^2 - GMm/r1 = 1/2 mv2^2 - GMm/r2 …...........(1)
and momentum conservation as no external forces are acting, 
m r1 v1 = m r2 v2 ….............(2)
From Newton’s second law of equation we get, 
1/2 v1^2 - GM/r1 = 1/2 (r1 v1/r2)^2 - GM/r2 ….............(3)
From eqns. (1) (2) and (3) on solving we get,
(v1^2 - 2GM/r1) r2^2 = (r1 v1)^2 - 2GMr2 
Given, velocity is sqrt(1.5) times the velocity of the circular orbit at the initial radius.
Hence,
GMm/r1^2 = mv^2/r1 
or 
v1^2 = 1.5GM/r1 
Furthet solving,
(-0.5GM/r1) r2^2 = 1.5GMr1 - 2GMr2 
r2^2 - 4r1 r2 + 3r1^2 = 0 
and hence solving the quadratic,
r2 = 2r1 +- sqrt(16r1^2 - 12r1^2)/2 
r2 = 2r1 +- r1 
r2 = r1, 3r1 
 Let R = radius of the earth,
 B0 = initial altitude of the orbit,
 A = maximum altitude. 
Hence,
R+A = 3(R+B0) 
or
A = 2R + 3B0 
On putting earth’s radii and B0 value we get
A = 18742 km .

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