A spaceship is launched into a circular orbit close to earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Take, radius of earth = 6400kmand g = 9.8 ms^-2)
[AIIMS 2017]

The orbital velocity of spaceship in circular orbit, $v_o = \sqrt{\frac{GM}{r}}$

If spaceship is very close to earth surface, then r = radius of earth (R).

$\therefore v_{\mathrm{o}}=\sqrt{\frac{GM}{R}}=\sqrt{\frac{R^{2}g}{R}}=\sqrt{Rg} \left(\text{ Since, }g=\frac{GM}{R^2}\right)$

Here,  $R=6400 \mathrm{km}=6.4\times {10}^6 \mathrm{m}\text{ and }g=9.8{ \mathrm{ms}}^{-2}$

$\therefore v_o=\sqrt{6.4\times {10}^6\times 9.8}$

$= 7.92\times {10}^3 m/s = 3.28 \mathrm{km} {\mathrm{s}}^{-1}$

$\therefore$  Additional velocity required

$= 11.20 - 7.92 = 3.28 \mathrm{km} {\mathrm{s}}^{-1}$

Therefore, velocity of $3.28 \mathrm{km} {\mathrm{s}}^{-1}$ must be added to orbital velocity of spaceship to overcome the gravitational pull.