Question

How much extra velocity is needed to escape the satellite from gravitational pull of earth, if it is orbiting with the velocity v in an orbit around earth?

Answer 1

Explanation:

On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.

Answer 2

Answer:

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Explanation:

For a satellite in orbit, the velocity can be evaluated from the following equation,

r

2

GMm

=

r

mv

o

2

v

o

=

r

GM

Here, r=R+h

For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be v

e

.

E=PE+KE=0

r

−GMm

+

2

1

mv

e

2

=0

v

e

=

r

2GM

Difference in velocities is given by

Δv=v

e

−v

o

Δv=

r

GM

(

2

−1)

Δv≈

R

GM

(

2

−1) as h<<R

But g=

R

2

GM

∴Δv≈

gR

(

2

−1)