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 $=6400\, km,\, g=9.8\, m / s ^{2}$ )

• 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,
$v_{0}=\sqrt{\frac{G M}{r}}$
If space ship is very close to earth surface, $r=$ radius of earth $=R$
$\therefore v_{0}=\sqrt{\frac{G M}{R}}=\sqrt{\frac{R^{2} g}{R}}=\sqrt{R g}$
$\left(\right.$ since $\left. g=\frac{G M}{R^{2}}\right)$
Here, $R=6400\, km =6.4 \times 10^{6} m$
$g =9 \cdot 8\, m / s ^{2}$
$\therefore v_{0} =\sqrt{6 \cdot 4 \times 10^{6} \times 9 \cdot 8}$
$=7 \cdot 92 \times 10^{3} m / s$
$=7 \cdot 92\, km / s$
The escape velocity of space ship
$v_{e} =\sqrt{(2 R g)}=\sqrt{2} v_{0}=\sqrt{2} \times 7.92$
$=11.20\, km / s$
$\therefore $ Additional velocity required
$=11 \cdot 20-7 \cdot 92=3 \cdot 28\, km / s$
Therefore, velocity $3.28\, km / s$ must be added to orbital velocity to spaceship to overcome the gravitational pull.