Question

# 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

Open in App
Solution

## 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, $V=-{\int }_{\infty }^{r}GM\frac{m}{{x}^{2}}dx=-GM\frac{m}{r}$, 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$V=-G\frac{Mm}{R}$ ……………(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 ${V}^{\text{'}}=-G\frac{mM}{R+3R}$ …………(2)So, the change in gravitational potential energy is $∆V=V-V\text{'}=-G\frac{Mm}{R}$$-$$\left(-G\frac{mM}{R+3R}\right)$. (From equations 1 and 2 )or $∆V=-G\frac{Mm}{R}+G\frac{Mm}{4R}=-G\frac{3Mm}{4R}$or $∆V=-G\frac{3mM}{4R}$Therefore, the change in gravitational potential energy is $\mathbf{∆}\mathbit{V}\mathbf{=}\mathbf{-}\mathbit{G}\frac{\mathbf{3}\mathbf{m}\mathbf{M}}{\mathbf{4}\mathbf{R}}$.

Suggest Corrections
0
Explore more