The magnitude of the gravitational force between a particle of mass${\mathit{m}}_{\mathbf{1}}$and one of mass${\mathit{m}}_{\mathbf{2}}$is given by$\mathit{f}\left(x\right)\mathbf{=}\frac{\mathbf{G}{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}$where Gis a constant and xis the distance between the particles. (a) What is the corresponding potential energy function U(x)? Assume that$\mathit{U}\left(x\right)\mathbf{\to }\mathbf{0}$as$\mathit{x}\mathbf{\to }\mathbf{\infty }$and that xis positive. (b) How much work is required to increase the separation of the particles from$\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{1}}$to$\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathit{d}$?

a) Force is given as:$F\left(x\right)=\frac{G{m}_1{m}_2}{x^2}$

b) Assumption, $U\left(x\right)\to 0$for$x\to \infty$

We use the concept of gravitational force. To find gravitational potential energy, we integrate the equation of gravitational force over infinity to reference position. Then we can find the work required to increase the separation of the particles forthegiven positions.

Formula:

The gravitational force acting on a body, $F\left(x\right)=\frac{G{m}_1{m}_2}{x^2}$ (i)

We integratethefunction of force equation (i) and substitute the given values to get the potential energy function as follows:

$$\begin{gathered} {\int }_0^{U}dU={\int }_{\infty }^{x}F\left(x\right)dx \\ U\left(x\right)={\int }_{\infty }^x\frac{G{m}_1{m}_2}{x^2}dx \\ =G{m}_1{m}_2{\left[-\frac{1}{x}\right]}_{\infty }^x \\ =-\frac{G{m}_1{m}_2}{x^2} \end{gathered}$$

Hence, the corresponding potential energy equation is$-\frac{G{m}_1{m}_2}{x^2}$ .

Work required increasing the separation of the particles from $x=x_1$to $x=x_1+d$by integrating the same force equation as follows:

$$\begin{gathered} W={\int }_{x_1}^{x_1+d}\frac{G{m}_1{m}_2}{x^2}dx \\ =G{m}_1{m}_2{\left[-\frac{1}{x}\right]}_{x_1}^{x_1+d} \\ =-G{m}_1{m}_2\left[\frac{1}{x_1+d}-\left(\frac{1}{x_1}\right)\right] \\ =G{m}_1{m}_2\left[\frac{1}{x_1}-\frac{1}{x_1+d}\right] \\ =\frac{G{m}_1{m}_{2}d}{x_1\left(x_1+d\right)} \end{gathered}$$

Hence, the value of the work is$\frac{G{m}_1{m}_{2}d}{x_1\left(x_1+d\right)}$ .