JEE PHYSICS MCQ: GRAVITATION

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Application +4 | -1

Find the velocity of escape at the earth given that its radius is

6.4 \times 10^6 \mathrm{~m}

and the value of

g

at its surface is

9.8 \mathrm{~ms}^{-2}

View answer and explanation

Correct option: A

11.2 \mathrm{~kms}^{-1}

Option A (

11.2 \mathrm{~kms}^{-1}

) is correct because it is the result of applying the escape velocity formula using the given values of Earth's radius and gravitational acceleration.

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The escape velocity for a body projected vertically upwards from the surface of the earth is

11.2\mathrm{\;\;km}{\mathrm{s}^{ - 1}}

. If the body is projected in a direction making an angle of

{45^0}

with the vertical, the escape velocity will be The escape velocity of a body projected from the surface of earth is

v_e

. If the body is projected at an angle

\theta

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