Chapter 7.2, Problem 14PP

(a)

To determine

Speed of the satellite.

Answer to Problem 14PP

The speed of satellite, $v=7.75\times {10}^3\text{ m/s}$

Explanation of Solution

Given:

Height, $h=2.25\times {10}^5\text{m}$

Increase in orbit, $d=24\text{ km = 2}\text{.4}\times {10}^4\text{m}$

Radius of earth, $r_E=6.38\times {10}^6\text{m}$

Formula used:

The speed is given as,

  $v=\sqrt{\frac{G{m}_E}{r}}$ ,

Where,

  $\begin{array}{l}G=6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\\ m_E=5.97\times {10}^{24}\text{kg}\end{array}$

Calculation:

The required radius,

  $\begin{array}{l}r=(h+d)+r_E\\ r=(2.25\times {10}^5+2.40\times {10}^4)+6.38\times {10}^6\\ r=6.63\times {10}^6\text{m}\end{array}$

Substituting the given values in velocity formula,

  $\begin{array}{l}v=\sqrt{\frac{6.67\times {10}^{-11}\times 5.97\times {10}^{24}}{6.63\times {10}^6}}\\ v=7.75\times {10}^3\text{m/s}\end{array}$

Conclusion:

Thus, the speed of satellite, $v=7.75\times {10}^3\text{ m/s}$ .

(b)

To determine

To identify: Whether the speed is faster or slower.

Answer to Problem 14PP

Orbital speed is slower than previous speed.

Explanation of Solution

Given:

Previous orbital speed, $v_p=7.76\times {10}^3\text{m/s}$

Previous time period, $T_p=89\text{ min}$

Formula used:

  $T=2\pi \sqrt{\frac{r^3}{G{m}_E}}$

Calculation:

From part (a), it is observed that the speed is slower than the previous speed.

Substituting all the values in above formula,

  $\begin{array}{l}T=2\pi \sqrt{\frac{{\left(6.63\times {10}^6\text{m}\right)}^3}{\left(6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\right)\left(5.97\times {10}^{24}\text{kg}\right)}}\\ T=4.69\times {10}^3\text{s}\\ T=79\text{ min}\end{array}$

So, the previous time period is compared with the current one, then

  $\begin{array}{l}=89\text{ min}-79\text{ min}\\ \text{=10 min}\end{array}$

Hence, the speed is slower than the previous speed by 10 minutes.

Conclusion:

Orbital speed is slower than previous speed.

(c)

To determine

To explain: The reason for slow speed.

Explanation of Solution

Given:

Previous orbital speed, $v_p=7.76\times {10}^3\text{m/s}$

Current orbital speed, $v=7.75\times {10}^3\text{ m/s}$

Formula used:

The speed is given as,

  $v=\sqrt{\frac{G{m}_E}{r}}$ ,

Where,

  $\begin{array}{l}G=6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\\ m_E=5.97\times {10}^{24}\text{kg}\end{array}$

As the orbital speed is inversely proportional to the square root of radius,

  $v\propto \frac{1}{\sqrt{r}}$

Hence, if we increase the radius, then the velocity will decrease and vice versa.

Conclusion:

Hence, due to dependence on radius the velocity becomes slower.