Chapter 39, Problem 39.65P

Review. A global positioning system (GPS) satellite moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The nonmilitary GPS signal is broadcast at a frequency of 1 575.42 MHz in the reference frame of the satellite. When it is received on the Earth’s surface by a GPS receiver (Fig. P38.41), what is the fractional change in this frequency due to time dilation as described by special relativity? (d) The gravitational “blueshift’’ of the frequency according to general relativity is a separate effect. It is called a blueshift to indicate a change to a higher frequency. The magnitude of that fractional change is given by

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

where U_g is the change in gravitational potential energy of an object–Earth system when the object of mass m is moved between the two points where the signal is observed. Calculate this fractional change in frequency due to the change in position of the satellite from the Earth’s surface to its orbital position. (e) What is the overall fractional change in frequency due to both time dilation and gravitational blueshift?

Figure P38.41

To determine

 The radius of the orbit of the GPS satellite.

Answer to Problem 39.65P

The radius of the orbit of the GPS satellite in which ir revolves around the earth is $2.67\times {10}^7\text{m}$.

Explanation of Solution

Given info: The time period of the satellite moving around earth in the circular orbit is $11\text{h 58min}$.

The value of force of gravitational constant $\left(G\right)$  is $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{kg}}^{\text{-2}}$.

The mass of earth is $5.98\times {10}^{24}\text{kg}$.

Explanation:

From Newton’s second law, the nature of the force between the earth and satellite system is the gravitational force between the earth and the satellite and the centripetal force between them. Both the forces should be equal so that the satellite can revolve in an orbit.

The formula to calculate gravitational force is

$F_g=\frac{G{M}_{e}m}{r^2}$ (1)

Here,

$F_g$ is the gravitational force.

$m$ is the mass of satellite.

$M_e$ is the mass of earth

$r$ is the distance between the earth and the satellite.

The speed of the satellite is

$v=\frac{2\pi r}{T}$

Here,

$T$ is the time period of the revolution of the satellite.

The time period of the satellite is,

$\begin{array}{c}T=11\text{h 58}\text{min}\\ =11\text{h +58}\text{min}\\ \text{=11}\text{h}\left(\frac{3600\sec }{1\text{h}}\right)+58\min \left(\frac{60\sec }{1\min }\right)\\ =43080\sec \end{array}$

The formula to calculate the centripetal force is

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

Here,

$F_c$ is the centripetal force

$v$ is the velocity of the orbiting satellite

Substitute $\frac{2\pi r}{T}$ for $v$ in above equation.

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

Equate equation (1) and (2)

$\begin{array}{c}{F}_g=F_c\\ \frac{G{M}_{e}m}{r^2}=\frac{m{\left(\frac{2\pi r}{T}\right)}^2}{r}\\ r={\left(\frac{G{M}_e{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}\end{array}$

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 $M_e$ $43080\sec$ for $T$ in the above equation.

$\begin{array}{l}r={\left(\frac{G{M}_e{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}\\ ={\left[\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{kg}}^{\text{-2}}\right)\left(5.98\times {10}^{24}\text{kg}\right){\left(43080\sec \right)}^2}{4{\pi }^2}\right]}^{\frac{1}{3}}\\ =2.67\times {10}^7\text{m}\end{array}$

Thus the radius of the orbit of the satellite is $2.67\times {10}^7\text{m}$.

Conclusion:

Therefore, the radius of the orbit of the GPS satellite in which ir revolves around the earth is $2.67\times {10}^7\text{m}$.

To determine

The speed of the orbiting satellite.

Answer to Problem 39.65P

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

Explanation of Solution

Explanation

The formula to calculate the speed of the satellite revolving around the earth in a circular orbit is,

$v=\frac{2\pi r}{T}$

Substitute $2.67\times {10}^7\text{m}$ for $r$ and $43080\sec$ for $T$ in the above equation.

$\begin{array}{c}v=\frac{2\pi \left(2.66\times {10}^7\text{m}\right)}{43080\text{s}}\\ =3.87\times {10}^3{\text{ms}}^{\text{-1}}\end{array}$

Thus the speed of the satellite is $3.87\times {10}^3{\text{ms}}^{\text{-1}}$.

Conclusion:

Therefore, speed of the satellite revolving around the earth in a circular orbit is $3.87\times {10}^3{\text{ms}}^{\text{-1}}$.

To determine

The fractional change in the frequency due the time dilation.

Answer to Problem 39.65P

The fractional change in the received signal frequency is $\left(-8.35\times {10}^{-11}\right)$.

Explanation of Solution

Explanation

Given info: The broadcast signal frequency of the GPS satellite is $1575.42\text{MHz}$.

The formula to calculate the frequency of any signal is,

$f=\frac{1}{T}$

Here,

$f$ is the frequency of the signal.

$T$ is the time period.

Differentiate the above equation.

$\begin{array}{c}df=-\frac{dT}{T^{{}^2}}\\ df=-\frac{1}{T}\frac{dT}{T}\\ df=-f\frac{dT}{T}\\ \frac{df}{f}=-\frac{dT}{T}\end{array}$ (3)

Thus, the fractional change in the frequency is the equal to fractional change in the time period.

The formula to calculate the fractional increase in time period is,

$\frac{dT}{T}=-\left(\gamma -1\right)$ (4)

Here,

$\gamma$ is the relativistic factor.

The formula to calculate the relativistic factor is,

$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Here,

$v$ is the velocity of the body.

$c$ is the speed of light.

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in equation (4).

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

Substitute $\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)$ for $\frac{dT}{T}$ in equation (3).

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

Take the Binomial expansion series expansion of the term ${\left(1-\frac{v^2}{c^2}\right)}^{-\frac{1}{2}}$.

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

Substitute $3.87\times {10}^3{\text{ms}}^{\text{-1}}$ for $v$ from part (b) solution and $3.00\times {10}^8{\text{ms}}^{\text{-1}}$ for $c$ in the above equation.

$\begin{array}{c}\frac{df}{f}=-\frac{1}{2}{\left(\frac{3.87\times {10}^3{\text{ms}}^{\text{-1}}}{3.00\times {10}^8{\text{ms}}^{\text{-1}}}\right)}^2\\ =-8.35\times {10}^{-11}\end{array}$

Thus the fractional change in frequency is $-8.35\times {10}^{-11}$.

Conclusion:

Therefore, fractional change in the received frequency is $-8.34\times {10}^{-11}$.

To determine

The magnitude of the fractional change in frequency in terms of due to gravitational blue shift.

Answer to Problem 39.65P

The fractional change in frequency due to the fractional blue shift is $+5.29\times {10}^{-10}$.

Explanation of Solution

Explanation

The formula to calculate the gravitational blue shift is,

$\frac{\Delta f}{f}=\frac{\Delta U_g}{m{c}^2}$ (5)

Here,

$\Delta U_g$ is the gravitational potential energy.

The formula to calculate the gravitational potential energy between the earth’s surface and the satellite orbit is

$\Delta U_g=-\frac{G{M}_{e}m}{\left(r-R_e\right)}$

Here,

${\mathrm{R}}_e$ is the radius of earth.

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 $M_e$, $2.67\times {10}^7\text{m}$ for $r$ and $6.37\times {10}^6\text{m}$ ${\mathrm{R}}_e$ for in the above equation.

$\begin{array}{c}\Delta U_g=-\left[\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{kg}}^{\text{-2}}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{\left(2.67\times {10}^7-6.37\times {10}^6\text{m}\right)}\right]m\\ =\left(4.76\times {10}^7{\text{Jkg}}^{\text{-1}}\right)m\end{array}$

Substitute $\left(4.76\times {10}^7{\text{Jkg}}^{\text{-1}}\right)m$ for $\Delta U_g$ in equation (5).

$\frac{\Delta f}{f}=\frac{\left(4.76\times {10}^7{\text{Jkg}}^{\text{-1}}\right)m}{m{c}^2}$

Substitute $3\times {10}^8{\text{ms}}^{\text{-1}}$ for $c$  in the above equation.

$\begin{array}{c}\frac{\Delta f}{f}=\frac{\left(4.76\times {10}^7{\text{Jkg}}^{\text{-1}}\right)m}{m{\left(3\times {10}^8{\text{ms}}^{\text{-1}}\right)}^2}\\ =+5.29\times {10}^{-10}\end{array}$

Thus the fractional change in frequency due to the gravitational blue shift is $+5.29\times {10}^{-10}$.

Conclusion:

Therefore, fractional change in frequency due to the gravitational blue shift is $+5.29\times {10}^{-10}$.

To determine

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

Answer to Problem 39.65P

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

Explanation of Solution

Explanation

The overall fractional change in the frequency is the sum of the both the fractional changes.

Thus the overall fractional change is the sum of the fractional change in the frequency due to the time dilation and fractional change in the frequency due to the gravitational blue shift.

The formula to calculate the overall fractional change is,

Overall fractional change = $=-\frac{1}{2}{\left(\frac{v}{c}\right)}^2+\frac{\Delta U_g}{m{c}^2}$

Substitute $-8.34\times {10}^{-11}$ for $-\frac{1}{2}{\left(\frac{v}{c}\right)}^2$ and $5.29\times {10}^{-10}$ for $\frac{\Delta U_g}{m{c}^2}$.in the above question

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

Conclusion:

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