91Ó°ÊÓ

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

b) Assumption, $U\left(x\right)→0$for$x→∞$

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:

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

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:

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

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