Chapter 39, Problem 65P

(a)

To determine

The radius of the orbit of GPS satellite.

Answer to Problem 65P

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

Explanation of Solution

Write the expression for the gravitational force acting on the satellite.

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

Here, $F_g$ is the gravitational constant, $G$ is the gravitational constant, $M_E$ is the mass of Earth, $m$ is the mass of satellite, and $r$ is the radius of the orbit.

Write the expression for the centripetal force acting on the satellite.

  $F_c=\frac{m{v}^2}{r}$                                                                                                                 (II)

Here, $F_c$ is the centripetal force acting on the satellite, and $v$ is the velocity of satellite.

Write the expression for the velocity of the satellite.

  $v=\frac{2\pi r}{T}$                                                                                                                   (III)

Here, $T$ is the time period of revolution of satellite around Earth.

Use expression (III) in (II).

  $F_c=\frac{m}{r}{\left(\frac{2\pi r}{T}\right)}^2$                                                                                                       (IV)

Equate expressions (IV) and (I) and solve for $r$.

  $\begin{array}{c}\frac{G{M}_{E}m}{r^2}=\frac{m}{r}{\left(\frac{2\pi r}{T}\right)}^2\\ G{M}_E{T}^2=4{\pi }^2{r}^3\\ r={\left(\frac{G{M}_E{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}\end{array}$                                                                                              (V)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, and $11\text{h}\text{58}\text{min}$ for $T$ in equation (V) to find $r$.

  $\begin{array}{c}r=\left(\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^2{\text{/kg}}^{\text{2}}\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(11\text{h}\times \frac{3600\text{s}}{1\text{hr}}+58\text{min}\times \frac{60\text{s}}{1\text{min}}\right)}{4{\pi }^2}\right)\\ =2.66\times {10}^7\text{m}\end{array}$

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

(b)

To determine

The speed of the satellite.

Answer to Problem 65P

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

Explanation of Solution

Equation (III) gives the orbital speed of the satellite.

  $v=\frac{2\pi r}{T}$

Conclusion:

Substitute $2.66\times {10}^7\text{m}$ for $r$, and $11\text{h}\text{58}\text{min}$ for $T$ in the above equation to find the orbital speed.

  $\begin{array}{c}v=\frac{2\pi \left(2.66\times {10}^7\text{m}\right)}{11\text{h}\times \frac{3600\text{s}}{1\text{h}}+58\text{min}\times \frac{60\text{s}}{1\text{min}}}\\ =3.87\times {10}^3\text{m/s}\end{array}$

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

(c)

To determine

The fractional change in the frequency due to time dilation.

Answer to Problem 65P

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.

  $f=\frac{1}{T}$                                                                                                                     (VI)

Here, $f$ is the frequency of satellite.

Differentiate expression.

  $\begin{array}{c}df=-\frac{dT}{T^2}\\ =-f\frac{dT}{T}\\ \frac{df}{f}=-\frac{dT}{T}\end{array}$                                                                                                      (VII)

The small fractional decrease in the frequency received from the satellite is equal to the fractional increase in period of oscillator due to the time dilation.

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

Write the expression for Lorentz factor.

  $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$                                                                                                         (IX)

Use expression (IX) in (VIII).

  $\frac{df}{f}=1-\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$                                                                                                   (X)

Write the binomial expansion for $\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$.

  $\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=1+\frac{1}{2}{\left(\frac{v}{c}\right)}^2+\mathrm{...}$                                                                                     (XI)

Substitute expression (XI) in (X).

  $\begin{array}{c}\frac{df}{f}=1-\left(1+\frac{1}{2}{\left(\frac{v}{c}\right)}^2\right)\\ =-\frac{1}{2}{\left(\frac{v}{c}\right)}^2\end{array}$                                                                                             (XII)

Conclusion:

Substitute $3.87\times {10}^3\text{m/s}$ for $v$, and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (XII) to find $\frac{df}{f}$.

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

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 from Earth’s surface to its orbital position.

Answer to Problem 65P

The fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position is $\underset{\_}{5.29\times {10}^{-10}}$.

Explanation of Solution

Write the expression for the change in gravitational potential energy.

  $\Delta U_g=-G{M}_{E}m\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$                                                                                     (XIII)

Here, $\Delta U_g$ is the change in gravitational potential energy, $r_2$ is the final position, and $r_1$ is the initial position of the satellite.

Write the given expression for the fractional change in frequency.

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

Conclusion:

Substitute $5.98\times {10}^{24}\text{kg}$ for $M_E$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $2.66\times {10}^7\text{m}$ for $r_2$, and $6.37\times {10}^6\text{m}$ for $r_1$ in equation (XIV) to find $\Delta U_g$.

  $\begin{array}{c}\Delta U_g=-\left(6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{28}\text{kg}\right)m\left[\frac{1}{2.66\times {10}^7\text{m}}-\frac{1}{6.37\times {10}^6\text{m}}\right]\\ =\left(4.76\times {10}^7\text{J/kg}\right)m\end{array}$

Substitute $\left(4.76\times {10}^7\text{J/kg}\right)m$ for $\Delta U_g$, and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (XIV) to find $\frac{\Delta f}{f}$.

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

Therefore, the fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position is $\underset{\_}{5.29\times {10}^{-10}}$.

(e)

To determine

The overall fractional change in frequency due to both time dilation and gravitational blue shift.

Answer to Problem 65P

The overall fractional change in frequency is $\underset{\_}{4.46\times {10}^{-10}}$.

Explanation of Solution

Write the expression for the overall fractional change in frequency.

  $\Delta f=\frac{\Delta f}{f}+\frac{df}{f}$                                                                                                        (XV)

Here, $\Delta f$ is the overall fractional change in frequency.

Conclusion:

Substitute $5.29\times {10}^{-10}$ for $\frac{\Delta f}{f}$, and $-8.34\times {10}^{-11}$ for $\frac{df}{f}$ in equation (XV) to find $\Delta f$.

  $\begin{array}{c}\Delta f=-8.34\times {10}^{-11}+5.29\times {10}^{-10}\\ =4.46\times {10}^{-10}\end{array}$

Therefore, the overall fractional change in frequency is $\underset{\_}{4.46\times {10}^{-10}}$.