91Ó°ÊÓ

The given expression is

$\frac{\left(6.67×{10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98×{10}^{24}\mathrm{kg}\right)(1000\mathrm{kg})}{r^2}=\frac{(1000\mathrm{kg})(1997\mathrm{m}/\mathrm{s}{)}^2}{r}$

The given equation is

$\frac{\left(6.67×{10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98×{10}^{24}\mathrm{kg}\right)(1000\mathrm{kg})}{r^2}=\frac{(1000\mathrm{kg})(1997\mathrm{m}/\mathrm{s}{)}^2}{r}$

which is in the form of

$\frac{G{M}_{E}m}{r^2}=\frac{m{v}^2}{r}$

Where

G = gravitational universal constants

m= mass

r= distance

M_E = mass of earth

So gravitational force is equal to centripetal force.

This equation can solve a problem .

Find the radius of orbit of 1000 kg object rotating earth with an speed of 1997 m/s.

The given equation is

$\frac{\left(6.67×{10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98×{10}^{24}\mathrm{kg}\right)(1000\mathrm{kg})}{r^2}=\frac{(1000\mathrm{kg})(1997\mathrm{m}/\mathrm{s}{)}^2}{r}$

The pictorial representation of the problem statement is as below

The equation given is

$\frac{\left(6.67×{10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98×{10}^{24}\mathrm{kg}\right)(1000\mathrm{kg})}{r^2}=\frac{(1000\mathrm{kg})(1997\mathrm{m}/\mathrm{s}{)}^2}{r}$

Solve the given equation to find the radius of orbit.

$$\begin{gathered} \frac{\left(6.67×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.98×{10}^{24}\mathrm{kg}\right)(1000\mathrm{kg})}{r^2}=\frac{(1000\mathrm{kg})(1997\mathrm{m}/\mathrm{s}{)}^2}{r} \\ r=\frac{\left(6.67×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.98×{10}^{24}\mathrm{kg}\right)}{(1997\mathrm{m}/\mathrm{s}{)}^2} \\ =1×{10}^8\mathrm{m} \end{gathered}$$

Radius of the orbit is 1x10⁸m