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(C) The gravitational potential energy of a body of mass $m$ at the surface of the Earth is given by $PE = -\frac{GMm}{R}$.Since $g = \frac{GM}{R^2}$,

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About $3.1 	imes 10^{10} \, J$

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(C) The gravitational potential energy of a body of mass $m$ at the surface of the Earth is given by $PE = -\frac{GMm}{R}$.Since $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.Substituting this into the potential energy formula: $PE = -\frac{(gR^2)m}{R} = -mgR$.Given $m = 500 \, kg$,$g = 9.8 \, m/s^2$,and $R = 6.4 	imes 10^6 \, m$:$PE = -(500) 	imes (9.8) 	imes (6.4 	imes 10^6) = -3.136 	imes 10^{10} \, J$.To make the body escape,we must provide energy equal to the magnitude of its potential energy,which is $3.1 	imes 10^{10} \, J$.

How much energy is necessary for a body of $500 \, kg$ to escape from the Earth? $[g = 9.8 \, m/s^2$,radius of Earth $R = 6.4 \times 10^6 \, m]$

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