Question

Orbit of the earth with a velocity is root 1.5 times the velocity for circular orbit at that height

Answer 1

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 .