Chapter 7, Problem 29P

(a)

To determine

To Find: The expression for the satellite’s mechanical energy.

Answer to Problem 29P

  $-\frac{GMm}{2\left(R+r\right)}$

Explanation of Solution

Given:

A satellite of mass $m$ is in a circular earth orbit of radius $r$ .

Formula Used:

Kinetic energy is calculated as

  $KE=\frac{1}{2}m{v}^2$

Where,

  $m=$ Mass

  $v=$ velocity

The mechanical energy of the satellite is given as

  $\begin{array}{l}E=-\frac{GMm}{2\left(R+r\right)}\\ \left|E\right|=\left|KE\right|=\frac{1}{2}\left|PE\right|\end{array}$

Calculation:

Let mass of Earth be $M$

Let radius of Earth be $R$

Also, mass of satellite $=m$

circular orbit of satellite is of radius $=r$

Now, the tangential velocity of the satellite is given as

  $V=\sqrt{\frac{GM}{R+r}}$

Where,

  $G=$ Gravitational constant

Thus, the kinetic energy of a satellite in a circular orbit is calculated as

  $\begin{array}{l}KE=\frac{1}{2}m\times {\left(\sqrt{\frac{GM}{R+r}}\right)}^2\\ KE=\frac{1}{2}m\times \frac{GM}{R+r}\\ KE=\frac{GMm}{2\left(R+r\right)}\end{array}$

Also, the gravitational potential energy at infinity is considered to be zero, thus the potential energy at distance $\left(R+r\right)$ from the center of the earth is given as

  $PE=-\frac{GMm}{R+r}$

Now, the total mechanical energy of the satellite is given as

  $\begin{array}{l}E=KE+PE\\ E=\frac{GMm}{2\left(R+r\right)}+\left(-\frac{GMm}{\left(R+r\right)}\right)\\ E=-\frac{GMm}{2\left(R+r\right)}\end{array}$

Conclusion:

Thus, satellite’s mechanical energy is $-\frac{GMm}{2\left(R+r\right)}$

(b)

To determine

To Find: The energy and speed of satellite.

Answer to Problem 29P

The energy of satellite is $121.2\times {10}^{9}J$ and speed of satellite is $4.923km/s$

Explanation of Solution

Given:

A satellite of mass $m$ is in a circular earth orbit of radius $r$ , where $m=1.00\times {10}^{4}kg$ and $r=1.00\times {10}^{7}m$

Formula Used:

Work done by spring is calculated as

  $W_{Spring}=\frac{1}{2}K{x}^2$

Where,

  $K=$ Spring constant

  $x=$ Displacement

Calculation:

The mechanical energy of the satellite is calculated as

  $E=-\frac{GMm}{2\left(R+r\right)}$

Here,

  $\begin{array}{l}m=1.00\times {10}^{4}kg\\ r=1.00\times {10}^{7}m\\ M=5.972\times {10}^{24}kg\\ R=6378\times {10}^{3}m\\ G=6.67\times {10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}\end{array}$

Substitute the values:

The energy of satellite is

  $\begin{array}{l}E=-\frac{\left(6.67\times {10}^{-11}\right)\left(5.972\times {10}^{24}\right)\left(1.00\times {10}^4\right)}{2\left(6378\times {10}^3+1.00\times {10}^7\right)}\\ E=-\frac{\left(6.67\times {10}^{-11}\right)\left(5.972\times {10}^{24}\right)\left(1.00\times {10}^4\right)}{2\left(6378000+10000000\right)}\\ E=-\frac{\left(6.67\right)\left(5.972\right)\left(1.00\right)\times {10}^{28-11}}{2\left(16378000\right)}\\ E=-\frac{39.83324\times {10}^{17}}{32756000}\\ E=-0.1212\times {10}^{12}J\\ E=-121.2\times {10}^{9}J\end{array}$

Also, we know

  $\left|E\right|=\left|KE\right|$

Thus, speed of the satellite is calculated as

  $\begin{array}{l}\frac{1}{2}m{v}^2=121.2\times {10}^9\\ v^2=\frac{2\times 121.2\times {10}^9}{1.00\times {10}^4}\\ v^2=242.4\times {10}^5\\ v=4923m/s\\ v=4.923km/s\end{array}$

Conclusion:

Thus, energy of the satellite is $121.2\times {10}^{9}J$ and speed of the satellite is $4.923km/s.$