Question

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]

Moderate

Solution

The orbital velocity of spaceship in circular orbit, $v_{o} = \sqrt{\frac{G M}{r}}$
If spaceship is very close to earth surface, then r = radius of earth (R).
$∴ v_{o} = \sqrt{\frac{G M}{R}} = \sqrt{\frac{R^{2} g}{R}} = \sqrt{R g} (Since, g = \frac{G M}{R^{2}})$
Here, $R = 6400 km = 6.4 \times 10^{6} m and g = 9.8 {ms}^{- 2}$
$∴ v_{o} = \sqrt{6.4 \times 10^{6} \times 9.8}$
$= 7.92 \times 10^{3} m / s = 3.28 km s^{- 1}$
$∴$ Additional velocity required
$= 11.20 - 7.92 = 3.28 km s^{- 1}$
Therefore, velocity of $3.28 km s^{- 1}$ must be added to orbital velocity of spaceship to overcome the gravitational pull.

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