A satellite is revolving in a circular orbit at a height ' h' from the earth's surface (radius of earth R ; h

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 close to: (Neglect the effect of atmosphere.) (A) √2 gR (B) √g R (C) √g R / 2  (D) √gR ( √2 - 1 )

Tardigrade Admin · Accepted Answer

Orbital velocity v = √(GM/R + h) = √(GM/R) as h < < R Velocity required to escape (1/2) mv'2 = (GMm/R+h) ; v' = √(2GM/R+h) = √(2GM/R) (h < < R) ∴ Increase in velocity v' - v = √(2GM/R) - √(GM/R) = √2gR - √gR = √gR ( √2 - 1 )

$2 g R$

$g R$

$g R /2$

$g R (2 - 1)$

Orbital velocity $v = \frac{GM}{R + h} = \frac{GM}{R}$ as h < < R
Velocity required to escape
$\frac{1}{2} m v^{' 2} = \frac{GM m}{R + h}; v^{'} = \frac{2 GM}{R + h} = \frac{2 GM}{R} (h << R)$
$∴$ Increase in velocity
$v^{'} - v = \frac{2 GM}{R} - \frac{GM}{R} = 2 g R - g R = g R (2 - 1)$

Q. 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 close to : (Neglect the effect of atmosphere.)

2gR​

gR​

gR/2​

gR​(2​−1)

Orbital velocity v=R+hGM​​=RGM​​ as h < < R Velocity required to escape 21​mv′2=R+hGMm​;v′=R+h2GM​​=R2GM​​(h<<R) ∴ Increase in velocity v′−v=R2GM​​−RGM​​=2gR​−gR​=gR​(2​−1)

$2 g R$

$g R$

$g R /2$

$g R (2 - 1)$

Orbital velocity $v = \frac{GM}{R + h} = \frac{GM}{R}$ as h < < R
Velocity required to escape
$\frac{1}{2} m v^{' 2} = \frac{GM m}{R + h}; v^{'} = \frac{2 GM}{R + h} = \frac{2 GM}{R} (h << R)$
$∴$ Increase in velocity
$v^{'} - v = \frac{2 GM}{R} - \frac{GM}{R} = 2 g R - g R = g R (2 - 1)$