Question

For motion near the surface of the earth, we usually assume that the gravitational force on a mass m is

$\mathit{F}\mathbf{=}\mathbf{-}\mathit{m}\mathit{g}\mathit{k}$

but for motion involving an appreciable variation in distance r from the center of the earth, we must use

$\mathit{F}\mathbf{=}\mathbf{-}\frac{\mathbf{C}}{{\mathbf{r}}^{\mathbf{2}}}{\mathit{e}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{C}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{r}}{\left|r\right|}\mathbf{=}\mathbf{-}\frac{\mathbf{C}}{{\mathbf{r}}^{\mathbf{3}}}\mathit{r}$

where C is a constant. Show that both these F’s are conservative, and find the potential for each.

Short answer

Both F’s are conservative

Step-by-step solution

Given Information

The force field are mentioned below.

$\begin{array}{rcl}F& =& -\frac{C}{r^2}{e}_r\\ & =& \frac{C}{r^2}\frac{r}{\left|r\right|}\\ & =& -\frac{C}{r^3}r\\ F& =& -mg\text{k}\\ & & \end{array}$

Definition of conservative force and scalar potential

A force is said to be conservative if $\mathbf{\nabla }\mathbf{\times }\mathit{F}\mathbf{=}\mathbf{0}$.

The scalar potential is independent of the path. The scalar potential is the sum of potential in all the 3 dimensions calculated separately.

The formula for the scalar potential is$\mathit{W}\mathbf{=}\mathbf{\int }\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$ .

Use ∇×F to check conservative

The force is said to be conservative if$\nabla \times F=0$.

$\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& -mg\end{array}$

$$\begin{gathered} i\left(0-0\right)-j\left(0-0\right)+k\left(0-0\right) \\ 0 \end{gathered}$$

Solve for other equation.

$\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{-Cx}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}& \frac{-Cy}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}& \frac{-Cz}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}\end{array}$

$\begin{array}{rcl}\left(\nabla \times F\right)i& =& \left(\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}-\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}\right)\\ & =& 0\end{array}$

$\begin{array}{rcl}\left(\nabla \times F\right)j& =& \left(\frac{3Cxz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}-\frac{3Cxz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}\right)\\ & =& 0\end{array}$

$\begin{array}{rcl}\left(\nabla \times F\right)k& =& \left(\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}-\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}\right)\\ & =& 0\end{array}$

Hence, the equations are conservative.