Chapter 7.2, Problem 21SSC

a.

To determine

Larger orbital period satellite.

Answer to Problem 21SSC

The satellite with 160 km orbital radius have larger period.

Explanation of Solution

Given:

Orbital radius of first satellite is $r_1=150\text{km}$

Orbital radius of second satellite is $r_2=160\text{km}$

Formula used:

The orbital period is given as

  $T=2\pi \sqrt{\frac{r^3}{Gm}}$

Here, $r$ is orbital radius, $G$ is gravitational constant and $m$ is earth’smass.

Calculation:

For first satellite, the orbital period is

  $T_1=2\pi \sqrt{\frac{r_1{}^3}{Gm}}$

For second satellite, the orbital period is

  $T_2=2\pi \sqrt{\frac{r_2{}^3}{Gm}}$

So,

  $\begin{array}{l}\frac{T_2}{T_1}=\frac{2\pi \sqrt{\frac{r_2{}^3}{Gm}}}{2\pi \sqrt{\frac{r_1{}^3}{Gm}}}=\frac{\sqrt{r_2{}^3}}{\sqrt{r_1{}^3}}\\ =\frac{\sqrt{{160}^3}}{\sqrt{{150}^3}}=1.1\\ \Rightarrow T_2=1.1T_1\end{array}$Conclusion:

Therefore, the orbital period of second satellite with radius 160 km is larger.

b.

To determine

Satellite having greater speed.

Answer to Problem 21SSC

The satellite having orbital radius 150 km have greater speed.

Explanation of Solution

Given:

Orbital radius of first satellite is $r_1=150\text{km}$

Orbital radius of second satellite is $r_2=160\text{km}$

Formula used:

The velocity is given as,

  $v=\sqrt{\frac{Gm}{r}}$

Where, $r$ is orbital radius, $G$ is gravitational constant and $m$ is earth’s mass.

Calculation:

For first satellite, the velocity is

  $v_1=\sqrt{\frac{Gm}{r_1}}$

For second satellite, the velocity is

  $v_2=\sqrt{\frac{Gm}{r_2}}$

So,

  $\begin{array}{l}\frac{v_1}{v_2}=\frac{\sqrt{\frac{Gm}{r_1}}}{\sqrt{\frac{Gm}{r_2}}}=\frac{\sqrt{r_2}}{\sqrt{r_1}}\\ =\frac{\sqrt{160}}{\sqrt{150}}=1.033\\ \Rightarrow v_1=1.033v_2\end{array}$

Conclusion:

Therefore, the first satellite with radius 150 km have higher velocity.