Chapter 1, Problem 39P

An Earth satellite used in the Global Positioning System moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The satellite contains an oscillator producing the principal nonmilitary GPS signal. Its frequency is 1 575.42 MHz in the reference frame of the satellite. When it is received on the Earth’s surface, 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. The magnitude of that fractional change is given by

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

where ΔU_g/m is the change in gravitational potential energy per unit mass between the two points at which the signal is observed. Calculate this fractional change in frequency. (e) What is the overall fractional change in frequency? Superposed on both of these relativistic effects is a Doppler shift that is generally much larger. It can be a redshift or a blueshift, depending on the motion of a particular satellite relative to a GPS receiver (Fig. P1.39).

To determine

The radius of the orbit of the satellite.

Answer to Problem 39P

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

Explanation of Solution

Write the expression of Newton’s gravitational law

    $F=\frac{GMm}{r^2}$        (I)

Write the expression for centripetal force

    $F=\frac{m{v}^2}{r}$        (II)

Write the expression for the linear velocity of an object moving in circular path

    $v=\frac{2\pi r}{T}$        (III)

Here, $F$ is the force, $G$ is the universal gravitational constant, $M$ is the mass of Earth, $m$ is the mass of the satellite, and $r$ is the radius of the orbit.

Equating equation (I) and (II) and substitute equation (III)

$\begin{array}{c}\frac{GMm}{r^2}=\frac{m{v}^2}{r}\\ \frac{GMm}{r^2}=\frac{m}{r}{\left(\frac{2\pi r}{T}\right)}^2\\ GM{T}^2=4{\pi }^2{r}^3\\ r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}$

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{ kg}$ for $M$ and $43080\text{ s}$ for $T$ in the above equation to find the value of $r$

$\begin{array}{c}r={\left(\frac{6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}\left(5.98\times {10}^{24}\text{ kg}\right){\left(43080\text{ s}\right)}^2}{{\text{kg}}^{\text{2}}4{\pi }^2}\right)}^{1/3}\\ =2.66\times {10}^7\text{ m}\end{array}$

Conclusion:

Thus, the radius of the orbit of the satellite is $2.66\times {10}^7\text{ m}$.

To determine

The speed of the satellite.

Answer to Problem 39P

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

Explanation of Solution

Substitute $2.66\times {10}^7\text{ m}$ for $r$ and $43080\text{ s}$ for $T$ in equation (III) to find the value of $v$

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

Conclusion:

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

To determine

The fractional change in the frequency due to time dilation.

Answer to Problem 39P

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

Explanation of Solution

The fractional change in the frequency received on Earth is equal to the magnitude of the fractional increase in the period of moving oscillator due to time dilation.

Write the expression of fractional change in the frequency

    $\Delta f=-\left(\gamma -1\right)$        (IV)

Here, $\Delta f$ is the fractional change in the frequency and $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

Substitute $3.87\times {10}^3\text{m}/\text{s}$ for $v$ and $3.0\times {10}^8\text{m}/\text{s}$ for $c$ in equation (IV) and solve for $\Delta f$

$\begin{array}{c}\Delta f=-\left[\frac{1}{\sqrt{1-{\left(3.87\times {10}^3/3\times {10}^8\right)}^2}}-1\right]\\ =1-\left(1-\frac{1}{2}\left[-{\left(\frac{3.87\times {10}^3}{3\times {10}^8}\right)}^2\right]\right)\\ =-8.34\times {10}^{-11}\end{array}$

Conclusion:

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

To determine

The fractional change in the frequency.

Answer to Problem 39P

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

Explanation of Solution

Write the expression for the gravitational potential energy

    $U_g=-\frac{G{M}_{E}m}{r}$

Here, $U_g$ is the gravitational potential energy.

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{ kg}$ for $M$ and $\text{2}\text{.66}\times {\text{10}}^{\text{7}}\text{ m}$ for $r$ in the above equation to find the value of $U_g$

$\begin{array}{c}\Delta U_g=-\frac{6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}\left(5.98\times {10}^{24}\text{ kg}\right)m}{{\text{kg}}^{\text{2}}\text{ 2}\text{.66}\times {\text{10}}^{\text{7}}\text{ m}}+\frac{6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}\left(5.98\times {10}^{24}\text{ kg}\right)m}{{\text{kg}}^{\text{2}}\text{ 6}\text{.37}\times {\text{10}}^{\text{6}}\text{ m}}\\ =4.76\times {10}^7\text{J}/\text{kg}m\end{array}$

The fractional change in the frequency is $\frac{\Delta f}{f}=\frac{\Delta U_g}{m{c}^2}$.

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

Conclusion:

Thus, the fractional change in the frequency is $+5.29\times {10}^{-10}$.

To determine

The overall fractional change in the frequency.

Answer to Problem 39P

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

Explanation of Solution

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

Hence,

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

Conclusion:

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