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(C) The gravitational potential energy $U$ of a particle of mass $m$ at the surface of a sphere of mass $M$ and radius $R$ is given by $U = -\frac{GMm

Vedclass · Accepted Answer

$6.67 	imes 10^{-10}\,J$

Vedclass · Answer

(C) The gravitational potential energy $U$ of a particle of mass $m$ at the surface of a sphere of mass $M$ and radius $R$ is given by $U = -\frac{GMm}{R}$.To take the particle to infinity (where potential energy is $0$),the work done $W$ against the gravitational force is equal to the change in potential energy: $W = U_{final} - U_{initial} = 0 - (- \frac{GMm}{R}) = \frac{GMm}{R}$.Given values:$M = 100\, kg$$m = 10\, g = 0.01\, kg$$R = 10\, cm = 0.1\, m$$G = 6.67 	imes 10^{-11}\, Nm^2/kg^2$Substituting these values into the formula:$W = \frac{6.67 	imes 10^{-11} 	imes 100 	imes 0.01}{0.1}$$W = \frac{6.67 	imes 10^{-11} 	imes 1}{0.1} = 6.67 	imes 10^{-10}\, J$.

$A$ particle of mass $10\, g$ is kept on the surface of a uniform sphere of mass $100\, kg$ and radius $10\, cm$. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $G = 6.67 \times 10^{-11}\, Nm^2 / kg^2$).

Similar Questions

Why does the potential energy of an object increase when it is raised above the surface of the Earth?

The potential energy of a satellite of mass $m$ revolving at a height $R_e$ above the surface of the Earth,where $R_e$ is the radius of the Earth,is:

The mass of the earth is $6.00 \times 10^{24} \ kg$ and that of the moon is $7.40 \times 10^{22} \ kg$. The gravitational constant is $G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$. The potential energy of the system is $-7.79 \times 10^{28} \ J$. The mean distance between the earth and the moon is:

Write an equation for the potential energy of a satellite. Why is the potential energy of a satellite negative?

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