Chapter 7, Problem 61A

a.

To determine

Speed of satellite in orbit.

Answer to Problem 61A

Speed of satellite in orbit is $v=3.07\times {10}^3\text{m/s}$ .

Explanation of Solution

Given:

Orbital radius $r=4.23\times {10}^7\text{m}$

Formula Used:

Velocity of satellite is given by,

  $v=\sqrt{\frac{G{m}_E}{r}}$

Here, $G$ universal gravitation constant, $m_E$ is mass of Earth and $r$ is radius of earth.

Calculation:

As,

  $\begin{array}{l}G=6.67\times {10}^{-11}{\text{N-m}}^2/{\text{kg}}^{\text{2}}\\ m_E=5.97\times {10}^{24}\text{kg}\\ r=4.23\times {10}^7\text{m}\end{array}$

So,

  $\begin{array}{c}v=\sqrt{\frac{\left(6.67\times {10}^{-11}\right)\left(5.97\times {10}^{24}\right)}{4.23\times {10}^7}}\\ =3.07\times {10}^3\text{m/s}\end{array}$

Conclusion:

Therefore, the velocity of satellite is $3.07\times {10}^3\text{m/s}$ .

b.

To determine

Orbit period of satellite.

Answer to Problem 61A

Orbit period of satellite is $T=1\text{day}$

Explanation of Solution

Given:

Orbital radius $r=4.23\times {10}^7\text{m}$

Formula Used:

Orbit period is given by,

  $T=2\pi \sqrt{\frac{r^3}{G{m}_E}}$

Where, $r$ is radius, $G$ is gravitational constant and $m_E$ is mass of earth.

Calculation:

As,

So substituting the values,

  $\begin{array}{c}T=2\times 3.14\times \sqrt{\frac{{\left(4.23\times {10}^7\right)}^3}{\left(6.67\times {10}^{-11}\right)\left(5.97\times {10}^{24}\right)}}\\ =6.28\times \sqrt{1.9\times {10}^8}\\ =8.66\times {10}^4\text{s}\\ \text{=}\frac{8.66\times {10}^4}{\left(60\times 60\times 24\right)}\text{day}\\ \approx \text{1}\text{day}\end{array}$

Conclusion:

Therefore, the orbit period of satellite is $T=1\text{day}$ .