A rocket is launched straight up from the surface of earth. When its altitude is one fourth of the radius of earth, its fuel runs out and therefore it coasts. The minimum velocity which the rocket should have when it starts to coast if it is to escape from the gravitational pull of the earth is approximately:

1 k m s^{- 1}

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6 k m s^{- 1}

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10 k m s^{- 1}

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15 k m s^{- 1}

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Solution

The correct option is C $10 k m s^{- 1}$
$R' = \frac{R}{4}$
$∴ V_{s c a p e} = \sqrt{\frac{2 G M}{R'}} = \sqrt{\frac{2 G M}{R / 4}} = \sqrt{8 \frac{G M}{R}} = \sqrt{8 \times 11.2} \approx 10 \frac{k m}{s}$

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Q. A rocket is launched straight up from the surface of the earth. When its altitude is one fourth of the radius of the earth, its fuel runs out and therefore it coasts. The minimum velocity which the rocket must have when it starts to coast if it is to escape from the gravitational pull of the earth is [escape velocity on surface of earth is

11.2 k m / s

]

Q. A rocket is launched with velocity of

10 k m s^{- 1}

. If radius of earth is

R

then maximum height attained by it will be:

A rocket is launched normal to the surface of the earth, away from the sun, along the line joining the sun and the earth. The sun is $3 \times 10^{5}$ times heavier than the earth and is at a distance $2.5 \times 10^{4}$ times larger than the radius of the earth. The escape velocity from earths gravitational field is $V_{c} = 11 .2 k m s^{- 1}$ . The minimum initial $(V_{c})$ required for the rocket to be able to leave the sun-earth system is closest to (Ignore the rotation and revolution of the earth and the presence of any other planet)