Question

(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of $\mathbf{\text{160km}}$above Earth’s surface?

(b) What is the period of revolution?

Short answer

1. The linear speed is$7.81\times {10}^{\text{3}}\text{m/s}$.
2. The period of revolution is$87.5\text{\hspace{0.17em}min}$

Step-by-step solution

Given

Altitude is $160\text{km}\left(\frac{{10}^3\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}km}}\right)=1.60\times {10}^5\text{\hspace{0.17em}m}$.

Determining the concept

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.

(a) Determining the linear speed

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 $6.37\times {10}^6\text{\hspace{0.17em}m}$, therefore the orbit radius will be,

$\begin{array}{c}\mathrm{r}=\text{(6.37}\times {\text{10}}^{\text{6}}\text{\hspace{0.17em}m}+\text{1.60}\times {\text{10}}^5\text{\hspace{0.17em}m)}\\ =\text{6.53}\times {\text{10}}^{\text{6}}\text{\hspace{0.17em}m}\end{array}$

The solution for$\text{v}$ is

$\begin{array}{c}\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\\ =\sqrt{\frac{\text{(}6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{)(}5.98\times {10}^{24}\text{\hspace{0.17em}kg)}}{6.53\times {10}^6\text{m}}}\\ =7.81\times {10}^3\text{m/s}\end{array}$

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

(b) Determining the period of revolution 

Since the circumference of the circular orbit is $\text{2πr}$, the period is,

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

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