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

7

s^{- 1}

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3.278

s^{- 1}

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5

s^{- 1}

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9

s^{- 1}

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Solution

The correct option is B $3.278$ km $s^{- 1}$ .
The orbital velocity in a circular orbit close to the earth is $v = \sqrt{g R}$ .
The velocity required to escape $v_{e} = \sqrt{2 g R}$ .
Hence additional velocity required is $v_{e} - v = (\sqrt{2} - 1) \sqrt{g R}$ .
Therefore,
$v_{e} - v = 0.414 \times \sqrt{9.8 \times 6400 \times 10^{3}}$
$= 3278.71$ m/s $= 3.278$ km $s^{- 1}$ .

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A spaceship goes into a circular orbit close to the earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth? (R=6400km, g=9.8 m s−2)

The correct option is B 3.278km s−1.The orbital velocity in a circular orbit close to the earth is v=√gR.The velocity required to escape ve=√2gR.Hence additional velocity required is ve−v=(√2−1)√gR.Therefore,ve−v=0.414×√9.8×6400×103=3278.71m/s =3.278km s−1.

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