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# A body of mass is placed on the earth s surface. It is taken from the earth s surface to a height $h=3R$, R is the radius of earth. The change in gravitational potential energy of the body is

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## Step1: Given data The radius of the earth is R.The height of the body from the earth's surface is $h=3R$.Step2: Gravitational potential energyThe gravitational potential energy at any point in a gravitational field of a system of masses is the amount of work done in bringing a unit mass from infinity to that point.Gravitational potential energy due to a point mass m at a distance r is, $V=-{âˆ«}_{âˆž}^{r}GM\frac{m}{{x}^{2}}dx=-GM\frac{m}{r}$, where G is the universal gravitational constant and M is the mass of the earth.Step3: Finding the change in Gravitational potential energyWe know that the gravitational potential energy of a body of mass m placed on the earth's surface is$V=-G\frac{Mm}{R}$ â€¦â€¦â€¦â€¦â€¦(1)where M and R are the mass of the earth and the radius of the earth.Now, gravitational potential energy at a height $h=3R$ from the earth's surface is ${V}^{\text{'}}=-G\frac{mM}{R+3R}$ â€¦â€¦â€¦â€¦(2)So, the change in gravitational potential energy is $âˆ†V=V-V\text{'}=-G\frac{Mm}{R}$$-$$\left(-G\frac{mM}{R+3R}\right)$. (From equations 1 and 2 )or $âˆ†V=-G\frac{Mm}{R}+G\frac{Mm}{4R}=-G\frac{3Mm}{4R}$or $âˆ†V=-G\frac{3mM}{4R}$Therefore, the change in gravitational potential energy is $\mathbf{âˆ†}\mathbit{V}\mathbf{=}\mathbf{-}\mathbit{G}\frac{\mathbf{3}\mathbf{m}\mathbf{M}}{\mathbf{4}\mathbf{R}}$.

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