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(A) The gravitational potential energy at the center of the Earth is given by $U_i = -\frac{3GMm}{2R}$.The gravitational potential energy at the surfa

Vedclass · Accepted Answer

$1.28 	imes 10^8 \, J$

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(A) The gravitational potential energy at the center of the Earth is given by $U_i = -\frac{3GMm}{2R}$.The gravitational potential energy at the surface of the Earth is given by $U_f = -\frac{GMm}{R}$.The energy required to move the mass is the change in potential energy: $\Delta U = U_f - U_i = -\frac{GMm}{R} - (-\frac{3GMm}{2R}) = \frac{GMm}{2R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the equation:$\Delta U = \frac{(gR^2)m}{2R} = \frac{mgR}{2}$.Given $m = 4 \, kg$,$g = 10 \, m/s^2$,and $R = 6400 \, km = 6.4 	imes 10^6 \, m$:$\Delta U = \frac{4 	imes 10 	imes 6.4 	imes 10^6}{2} = 2 	imes 6.4 	imes 10^7 = 1.28 	imes 10^8 \, J$.

Calculate the energy needed to move a mass of $4 \, kg$ from the center of the Earth to its surface (in joules),given that the radius of the Earth is $6400 \, km$ and the acceleration due to gravity at the surface of the Earth is $g = 10 \, m/s^2$.

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