Question

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$\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathit{d}$?

Short answer

a) Corresponding potential energy function U(x) is$-\frac{G{m}_1{m}_2}{x}$

b) Work required increasing the separation of the particles from$x=x_1$ to$x=x_1+d$ is$\frac{G{m}_1{m}_2}{x_1\left(x_1+d\right)}$ .

Step-by-step solution

The given data

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$x\to \infty$

Understanding the concept of gravitational force

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)

a) Calculation of corresponding potential energy function

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$-\frac{G{m}_1{m}_2}{x^2}$ .

b) Calculation of the required work to increase the separation of the particles

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$\frac{G{m}_1{m}_{2}d}{x_1\left(x_1+d\right)}$ .