Chapter 4, Problem 9P

What is the orbital velocity of an Earth satellite 42,230 km from Earth’s center? How long does it take to circle its orbit once?

To determine

(a)

The orbital velocity of Earth’s Satellite (time required to completes a circle around its orbit once.)

Answer to Problem 9P

Orbital velocity of Earth’s Satellite is $3.07\times {10}^3\text{m/s}$

Explanation of Solution

Given Info:

Distance of Earth’s Satellite from Earth center = $42,230km$

  $G=6.67\times {10}^{-11}{\text{m}}^{\text{3}}{\text{/s}}^{\text{2}}\text{kg} ;m_{earth}=5.98\times {10}^{24}\text{ kg}$

Formula Used:

1. $V=\sqrt{ar}=\sqrt{{\text{G m}}_{\text{Earth}}/r}$
2. $a=V^2/r$
3. $F_{Gravity}=G.m_{earth}.m_{sat}/r^2$
4. $\sum F=ma$

Calculation:

• Orbital velocity is a product of universal gravitational constant. Distance of satellite from earth centre and gravitational mass.
• To find the velocity, we need acceleration which keeps the object moving in a circle and we can find it from the force of gravity.
• Force of gravity cab be calculated from following :-

  $F_{Gravity}=G.m_{earth}.m_{sat}/r^2$

Where, G = Gravitational constant

m _earth = mass of earth

m _sat= mass of satellite

r = distance between earth and satellite

We know that, from Newton’s second Law, acceleration is,

  $\sum F=ma$

But in case of planet, only a gravitational force is acting on satellite. No other force is present hence summation sign is not needed. $$

∴ $m_{earth}.a=G.m_{earth}.m_{sat}/r^2$

∴ $asat=G.m_{earth}/r^2$ ...... (Dividing by m _sat on both sides)

Acceleration in the middle of circle to keep object in circular motion is given by,

a = $V^2$ /r

∴

  $V^2=a.r$

∴ $V=\sqrt{a.r}$

To find velocity of satellite, put acceleration of satellite in above formula,

∴ $V=\sqrt{G.\text{mearth}.r/r^2}$

∴ $V=\sqrt{G.\text{mearth}/r}$

We have,

  $\begin{array}{l}r=42230km=4.223x{10}^{4}km\hfill \\ =4.223x{10}^{4}x{10}^{3}m\hfill \\ =4.223x{10}^{7}m\hfill \end{array}$

∴ $V=\sqrt{6.67\times {10}^{-11}{m}^3/s^{2}kg)\left(5.98X{10}^{24} kg\right)/4.223x{10}^{7}m}$

∴ $V=3.0732X{10}^{3}m/s$

Conclusion:

Orbital velocity Earth ‘Satellite is $3.0732\times {10}^{3}m/s$

To determine

Time required completing a circle around its orbit once?

Answer to Problem 9P

Time required to complete a circle around one orbit is 8.

Explanation of Solution

Given info:

Distance of earth satellite from earth centre = 42230 km

Formula used:

  $c=2\pi r$

  $t=\frac{\text{c}}{\text{v}}$

Calculation:

Time required to complete a circle around one orbit of satellite is a ratio of distance travelled in one orbit to the velocity of satellite.

$t=\frac{\text{c}}{\text{v}}$

C is distance travelled in one orbit. It is nothing but the circumference of circular orbit.

  $\begin{array}{l}c=2\pi r\hfill \\ =2x\pi x\left(4.223x{10}^7\right)\hfill \\ =2.653x{10}^{8}m\hfill \end{array}$

Put value of c in

  $t=\frac{\text{c}}{\text{v}}$

∴

  $\begin{array}{l}T=2.653x108/3.0732x{10}^3\\ T=86326.95sec\\ T=8.632\times {10}^4\text{s}\end{array}$

Conclusion:

Time required for Satellite to complete one orbit is $8.632\times {10}^4\text{s}$