The escape velocity for a body projected vert | vedclass

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(D) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{2gR_e}$,where $g$ is the acceleration due to grav

Vedclass · Accepted Answer

$11.2$

Vedclass · Answer

(D) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{2gR_e}$,where $g$ is the acceleration due to gravity and $R_e$ is the radius of the Earth.This expression shows that the escape velocity depends only on the mass and radius of the planet (or the gravitational potential at the point of projection).It is independent of the direction or the angle at which the body is projected.Therefore,if the body is projected at an angle of $45^o$ with the vertical,the escape velocity remains the same,which is $11.2 \ km/s$.

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

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