devarshi-dt-logo

Question:

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is closed to (Neglect the effect of atmosphere).

√2gR

√gR/2

√gR(√2𕒵)

√gR

Solution:

For a satellite in orbit, the velocity can be evaluated from the following equation,
GMmr²=mv²
or vo = √GMr
Here, r = R + h
For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be ve.
E = PE + KE = 0
-GMmr + 1/2mv²e = 0
ve = √2GMr
Difference in velocities is given by
Δv = ve - vo
Δv = √GMr(√2 - 1)
Δv ≈ √GMR(√2 - 1) as h<<R
But g = GMR²
∴ Δv ≈ √gR(√2 - 1)