In Problems 64 through 66 you are given the equation(s) used to solve a problem. For each of these, you are to
a. Write a realistic problem for which this is the correct equation(s).
b. Draw a pictorial representation.
c. Finish the solution of the problem.

$64.\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.68\times {10}^{26}\mathrm{kg}\right)}{r^2}=\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2}$

The equation is $\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.68\times {10}^{26}\mathrm{kg}\right)}{r^2}=\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2}$

The equation given as

$\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.68\times {10}^{26}\mathrm{kg}\right)}{r^2}=\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2}$

We can see that calculation of two gravity is equated.

acceleration due to gravity on Earth = acceleration due to gravity on Saturn

The formula used is

$g=\frac{GM}{r^2}$

Where

G =6.67 x 10^-11 Nm²/ kg²
Mass of Earth, Me=5.98 x 10²⁴ kg
Mass of Saturn, Ms =5.68 x 10²⁶ kg
Radius of earth, Re=6.37 x 10⁶ m

So we can find the distance from the center of Saturn where acceleration due to gravity is same as the acceleration due to gravity on the surface of earth.

The equation given as below

$\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.68\times {10}^{26}\mathrm{kg}\right)}{r^2}=\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2}$

The pictorial representation is as below

The equation is

$\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.68\times {10}^{26}\mathrm{kg}\right)}{r^2}=\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2}$

Solve the given equation for r as below

$$\begin{gathered} \frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.68\times {10}^{26}\mathrm{kg}\right)}{{\mathrm{r}}^2}=\frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2} \\ {r}^2=\frac{\left(5.68\times {10}^{26}\mathrm{kg}\right){\left(6.37\times {10}^6\mathrm{m}\right)}^2}{\left(5.98\times {10}^{24}\mathrm{kg}\right)} \\ r=6.21\times {10}^7\mathrm{m} \end{gathered}$$