What is the escape energy of an object of mas | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{mgR_e}{5}$

Vedclass · Answer

(C) The gravitational potential energy $U$ of an object of mass $m$ at a height $h$ above the Earth's surface is given by $U = -\frac{GMm}{R_e + h}$.Since $g = \frac{GM}{R_e^2}$,we have $GM = gR_e^2$.Substituting this into the potential energy formula: $U = -\frac{gR_e^2 m}{R_e + h} = -\frac{mgR_e}{1 + h/R_e}$.Given $h = 4R_e$,the potential energy is $U = -\frac{mgR_e}{1 + 4} = -\frac{mgR_e}{5}$.The escape energy is the energy required to bring the object to infinity (where potential energy is $0$),which is $E = 0 - U = \frac{mgR_e}{5}$.

What is the escape energy of an object of mass $m$ at a height of $4R_e$ from the surface of the Earth? ($R_e$ = radius of the Earth)

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