A body of mass m is situated at a distance 4R | vedclass

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(C) The gravitational potential energy $U$ of a body of mass $m$ at a height $h$ above the Earth's surface is given by $U = -\frac{GMm}{R_e + h}$.Give

Vedclass · Accepted Answer

$\frac{mgR_e}{5}$

Vedclass · Answer

(C) The gravitational potential energy $U$ of a body of mass $m$ at a height $h$ above the Earth's surface is given by $U = -\frac{GMm}{R_e + h}$.Given $h = 4R_e$,the potential energy is $U = -\frac{GMm}{R_e + 4R_e} = -\frac{GMm}{5R_e}$.Since $g = \frac{GM}{R_e^2}$,we have $GM = gR_e^2$.Substituting this into the expression for $U$,we get $U = -\frac{(gR_e^2)m}{5R_e} = -\frac{mgR_e}{5}$.To escape the Earth's gravitational field,the total energy of the body must be at least zero. Therefore,the minimum energy required is $E = -U = \frac{mgR_e}{5}$.

$A$ body of mass $m$ is situated at a distance $4R_e$ above the Earth's surface,where $R_e$ is the radius of Earth. How much minimum energy must be given to the body so that it may escape?

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