Chapter 9, Problem 43P

(a)

To determine

The radius of the orbit.

Answer to Problem 43P

The radius of the orbit is $\underset{\_}{2.66\times {10}^7\text{m}}$.

Explanation of Solution

The gravitational force acting on the satellite is equal centripetal force.

Write the expression for the force acting on the satellite.

  ${\displaystyle \sum F}=ma$        (I)

Here, $m$ is the mass of the satellite, $a$ is the acceleration of the satellite.

Equate the gravitational force, and centripetal force.

  $\frac{G{M}_{E}m}{r^2}=\frac{m{\upsilon }^2}{r}$        (II)

Here, $M_E$ is the mass of Earth, $r$ is the radius of Earth, $\upsilon$ is the speed of the satellite, $G$ is the gravitational constant.

Substitute, $\frac{2\pi r}{T}$ for $\upsilon$ in equation (II) and rewrite to obtain an expression for $r$.

  $\begin{array}{c}\frac{G{M}_{E}m}{r^2}=\frac{m}{r}{\left(\frac{2\pi r}{T}\right)}^2\\ G{M}_E{T}^2=4{\pi }^2{r}^3\\ r=\left(\frac{G{M}_E{T}^2}{4{\pi }^2}\right)\end{array}$        (III)

Here, $T$ is the time period.

Conclusion

Substitute, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $43080\text{s}$ for $T$ in equation (III) to find the value of $r$.

  $\begin{array}{c}r=\left(\frac{6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\times 5.98\times {10}^{24}\text{kg}\times {\left(43080\text{s}\right)}^2}{4{\left(3.14\right)}^2}\right)\\ =2.66\times {10}^7\text{m}\end{array}$

Therefore, the radius of the orbit is $\underset{\_}{2.66\times {10}^7\text{m}}$.

(b)

To determine

The speed of the satellite.

Answer to Problem 43P

The speed of the satellite is $\underset{\_}{3.87\times {10}^3\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for speed in terms of period.

  $\upsilon =\frac{2\pi r}{T}$        (IV)

Conclusion:

Substitute, $2.66\times {10}^7\text{m}$ for $r$, and $43080\text{s}$ for $T$ in equation (IV) to find the speed of the satellite.

  $\begin{array}{c}\upsilon =\frac{2\pi \left(2.66\times {10}^7\text{m}\right)}{43080\text{s}}\\ =3.87\times {10}^3\text{m}/\text{s}\end{array}$

Therefore, the speed of the satellite is $\underset{\_}{3.87\times {10}^3\text{m}/\text{s}}$.

(c)

To determine

The fractional change in the frequency due to time dilation.

Answer to Problem 43P

The fractional change in the frequency due to time dilation is $\underset{\_}{-8.34\times {10}^{11}}$.

Explanation of Solution

Write the expression for the frequency.

  $f=\frac{1}{T}$        (V)

Here, $f$ is the frequency.

Take the derivative of equation (V) on both sides.

  $df=-\frac{dT}{T^2}$        (VI)

Substitute, $f$ for $\frac{1}{T}$ in equation (VI).

  $\begin{array}{c}df=-f\left(\frac{dT}{T}\right)\\ \frac{df}{f}=-\frac{dT}{T}\end{array}$        (VII)

The fractional change in the frequency is equal to the fractional change in time period according to equation (VII).

Substitute, $\gamma \Delta t_p-t_p$ for $dT$, and $\Delta t_p$ for $T$ in equation (VII).

  $\begin{array}{c}\frac{df}{f}=-\frac{\gamma \Delta t_p-t_p}{\Delta t_p}\\ =-\left(\gamma -1\right)\end{array}$        (VIII)

Substitute, $\frac{1}{\sqrt{1-\left({\upsilon }^2/c^2\right)}}$ for $\gamma$ in equation (VIII).

  $\begin{array}{c}\frac{df}{f}=-\left(\frac{1}{\sqrt{1-\left({\upsilon }^2/c^2\right)}}-1\right)\\ \approx 1-\left[1+\frac{1}{2}\left({\upsilon }^2/c^2\right)\right]\\ =-\frac{1}{2}\left({\upsilon }^2/c^2\right)\end{array}$        (IX)

Here, $c$ is the speed of the light.

Conclusion:

Substitute, $3.87\times {10}^3\text{m}/\text{s}$ for $\upsilon$, and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (IX) to obtain the fractional change in frequency.

  $\begin{array}{c}\frac{df}{f}=-\frac{1}{2}\left({\left(3.87\times {10}^3\text{m}/\text{s}\right)}^2/{\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2\right)\\ =-8.34\times {10}^{-11}\end{array}$

Therefore, the fractional change in the frequency due to time dilation is $\underset{\_}{-8.34\times {10}^{11}}$.

(d)

To determine

The fractional change in frequency due to the change in position of the satellite.

Answer to Problem 43P

The fractional change in frequency due to the change in position of the satellite is $\underset{\_}{+5.29\times {10}^{-10}}$.

Explanation of Solution

Write the expression for the gravitational potential.

  $\Delta U_g=-\frac{G{M}_{E}m}{r}$        (X)

The fractional change in frequency due to the change in position of the satellite is.

  $\frac{\Delta f}{f}=\frac{\Delta U_g}{m{c}^2}$        (XI)

Conclusion:

Substitute, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$) for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, and $\frac{1}{2.66\times {10}^7\text{m}}-\frac{1}{6.37\times {10}^6\text{m}}$ for $r$ in equation (XI).

  $\begin{array}{c}\Delta U_g=-\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)m\frac{1}{2.66\times {10}^7\text{m}}-\frac{1}{6.37\times {10}^6\text{m}}\\ =\left(4.76\times {10}^7\text{J}/\text{kg}\right)m\end{array}$

Substitute, $\left(4.76\times {10}^7\text{J}/\text{kg}\right)m$ for $\Delta U_g$, and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (XI) to find fractional change in frequency due to the change in position of the satellite.

  $\begin{array}{c}\frac{\Delta f}{f}=\frac{\left(4.76\times {10}^7\text{J}/\text{kg}\right)m}{m{\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}\\ =+5.29\times {10}^{-10}\end{array}$

Therefore, the fractional change in frequency due to the change in position of the satellite is $\underset{\_}{+5.29\times {10}^{-10}}$.

(e)

To determine

The overall fractional change in the frequency.

Answer to Problem 43P

The overall fractional change in the frequency is $\underset{\_}{+4.45\times {10}^{-10}}$.

Explanation of Solution

The fractional change in frequency due to the change in position of the satellite is $+5.29\times {10}^{-10}$, and the fractional change in the frequency due to time dilation is $-8.34\times {10}^{11}$.

Hence the overall change in the frequency is.

  $-8.34\times {10}^{11}++5.29\times {10}^{-10}=+4.46\times {10}^{-10}$

Conclusion:

Therefore, the overall fractional change in the frequency is $\underset{\_}{+4.45\times {10}^{-10}}$.