Question

A space ship 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? (Radius of earth $=6400km,g=9.8m/s^2$ )

• A. 3.28 km / s
• B. 12 km / s
• C. 10 km / s
• D. 40 km / s

Answer 1

Answer: (A)

The orbital velocity of space ship in circular orbit,
$v_0=\sqrt{\frac{GM}{r}}$
If space ship is very close to earth surface, $r=$ radius of earth $=R$
$\therefore v_0=\sqrt{\frac{GM}{R}}=\sqrt{\frac{R^{2}g}{R}}=\sqrt{Rg}$
$($ since $g=\frac{GM}{R^2})$
Here, $R=6400km=6.4\times 10^{6}m$
$g=9\cdot 8m/s^2$
$\therefore v_0=\sqrt{6\cdot 4\times 10^6\times 9\cdot 8}$
$=7\cdot 92\times 10^{3}m/s$
$=7\cdot 92km/s$
The escape velocity of space ship
$v_e=\sqrt{(2Rg)}=\sqrt{2}{v}_0=\sqrt{2}\times 7.92$
$=11.20km/s$
$\therefore$ Additional velocity required
$=11\cdot 20-7\cdot 92=3\cdot 28km/s$
Therefore, velocity $3.28km/s$ must be added to orbital velocity to spaceship to overcome the gravitational pull.