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 $11.2 k m / s$ ]

1 k m / s

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5 k m / s

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10 k m / s

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15 k m / s

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Solution

The correct option is B $10 k m / s$
Let $R$ be the radius of Earth and $m$ be the mass of the rocket.
At the altitude $\frac{R}{4}$ from surface of Earth:
Let $v$ be the total speed of the rocket at that altitude of $\frac{R}{4}$ . This speed is inclusive of the angular velocity of Earth's rotation and due to the propelling engine.
$K . E = \frac{1}{2} m v^{2}$

$P . E = - \frac{G M m}{(1.25 R)}$

While the object is inside the Earth's gravitational field, the total energy is negative. For the rocket to escape the Earth's gravitational field, the total energy must be 0 or more. Then at infinite distance the energy of the rocket will be 0 or more. So it will not come back towards Earth.

$\frac{1}{2} m v^{2} = \frac{G M m}{(1.25 R)}$

$v = \sqrt{\frac{2 G M}{R}} \times \sqrt{0.8}$

$= 11.2 k m / s e c \times 0.9 \approx 10 k m / s$

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