BITSAT 2015 Paper

The escape velocity from earth is given by
ve = √2gRe ... (i)
The orbital velocity of a satellite revolving around earth is given by
v0 =

√GMe

(Re+h)

where, Me = mass of earth, Re = radius of earth, h = height of satellite from surface of earth.
By the relation GMe = gRe2
So, v0 =

√gRe2

(Re+h)

... (ii)
Dividing equation (i) by (ii), we get
v−

e

v0

=

√2(Re+h)

(Re)

Given, v0 =

ve

2

2ve

ve

=

√2(Re+h)

Re

Squaring on both side, we get
4 =

2(Re+h)

Re

or Re + h = 2Re , i.e. h = Re