A space-ship is put into a circular orbit close to Earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of Earth(approximately)? $(R = 6400 k m, g = 9.8 \frac{m}{s^{2}})$

11.2 km/s

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3.3 km/s

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6.8 km/s

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9.3 km/s

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Solution

The correct option is B
3.3 km/s

The orbital velocity in a circular orbit close to Earth is:

$v = \sqrt{g R}$ .

The velocity required to escape is $v_{e} = \sqrt{2 g R}$

$v_{e} - v = (\sqrt{2} - 1) \sqrt{g R}$

Hence additional velocity required is: $v_{e} - v = 0.414 \times \sqrt{9.8 \times 6400 \times 10^{3}} = 3278.71 \frac{m}{s} = 3.278 \frac{k m}{s}$ .

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A space-ship is put into a circular orbit close to Earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of Earth(approximately)? (R=6400km,g=9.8 ms2)

The correct option is B 3.3 km/s The orbital velocity in a circular orbit close to Earth is: v = √gR. The velocity required to escape is ve = √2gR ve − v = (√2 − 1) √gR Hence additional velocity required is: ve − v = 0.414 × √9.8×6400×103 = 3278.71 ms = 3.278 kms.

A space-ship is put into a circular orbit close to Earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of Earth(approximately)? $(R = 6400 k m, g = 9.8 \frac{m}{s^{2}})$