Arun
Last Activity: 6 Years ago
Dear Rajesh
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.
Regards
Arun (askIITians forum expert)
