The escape velocity for a body projected vert | vedclass

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(B) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{\frac{2GM}{R_e}}$,where $G$ is the gravitational

Vedclass · Accepted Answer

$11$

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(B) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{\frac{2GM}{R_e}}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,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 body from which the object is being projected).It is independent of the angle of projection.Therefore,if the body is projected at an angle of $45^o$ with the vertical,the escape velocity remains the same,i.e.,$11 \ km/s$.

The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. 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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