91Ó°ÊÓ

Altitude is $160\text{km}\left(\frac{{10}^3\text{″¾}}{1\text{ k³¾}}\right)=1.60×{10}^5\text{″¾}$.

Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

The formulae are as follows:

$\mathrm{R}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$

Here,$\text{R}$ is the radius of the planet and$\text{M}$ is the mass of the planet.

If r is the radius of the orbit, then the magnitude of the gravitational force acting on the satellite is given by,

$\frac{\text{GMm}}{{\text{r}}^{\text{2}}}$,

Here M is the mass of Earth and m is the mass of the satellite.

The magnitude of the acceleration of the satellite is given by,

$\frac{{\text{v}}^{\text{2}}}{\text{r}}$,

Here $\text{v}$is its speed.

Newton's second law yields,

$\frac{\text{GMm}}{{\text{r}}^{\text{2}}}=\frac{{\text{mv}}^{\text{2}}}{\text{r}}$

Since the radius of Earth is $6.37×{10}^6\text{″¾}$, therefore the orbit radius will be,

$\begin{array}{c}\mathrm{r}=\text{(6.37}×{\text{10}}^{\text{6}}\text{″¾}+\text{1.60}×{\text{10}}^5\text{″¾)}\\ =\text{6.53}×{\text{10}}^{\text{6}}\text{″¾}\end{array}$

The solution for$\text{v}$ is

$\begin{array}{c}\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\\ =\sqrt{\frac{\text{(}6.67×{10}^{-11}\text{N}â‹\dots {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{)(}5.98×{10}^{24}\text{ k²µ)}}{6.53×{10}^6\text{m}}}\\ =7.81×{10}^3\text{m/s}\end{array}$

Therefore, the linear speed is$7.81×{10}^3\text{m/s}$

Since the circumference of the circular orbit is $\text{2Ï€°ù}$, the period is,

$\begin{array}{c}\mathrm{T}=\frac{2\text{π}\mathrm{r}}{\mathrm{v}}\\ =\frac{2\text{π}(6.53×{10}^6\text{m})}{7.82×{10}^3\text{m/s}}\\ =5.25×{10}^3\text{s}\left(\frac{1\text{ }\min }{60\text{ }\mathrm{s}}\right)\\ =87.5\text{ }\min \end{array}$

Therefore, the period of revolution is $87.5\min$.