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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 should have when it starts to coast if it is to escape from the gravitational pull of the earth is approximately ,

A 1Km s-1 B 6Km s-1 C 10Km s-1 D 15 Km s-1 answer with explanation

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Answer: 10 km/sec (c)

let R be the radius of Earth. and m be the mass of the rocket. At the altitude R/4 from surface of Earth: Let v be the total speed of the rocket at that altitude of R/4. This speed is inclusive of the angular velocity of Earth's rotation and due to the propelling engine.

KE = 1/2 m v² PE = - 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.

1/2 m v² = G M m / (1.25 R) v = √(2G M / R) * √0.8 = 11.2 km / sec * 0.9 approximately = 10.08 km / sec. approximately.

=============================== we can remember the formula for escape velocity as :

v = √ [ 2 G M / d ] d = distance of rocket from Earth's center.