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An Earth satellite used in the global positioning system moves in a circular orbit with period $11 \mathrm{~h} 58 \mathrm{~min}$. (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 $1575.42 \mathrm{MHz}$ 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 blue shift 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 $\Delta U_{\mathrm{g}}$ is the change in gravitational potential energy of an object-Earth system when the object of mass $m$ is moved between the two points 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 red shift or a blue shift, depending on the motion of a particular satellite relative to a GPS receiver

Solution

Verified$\textbf{Given}$

The period $T$ of the satellite's orbit is

$\begin{align*} T=11\ \text{h}\ 58\ \text{min} \end{align*}$

The mass of the Earth $M$ is

$\begin{align*} M=6\cdot 10^{24}\ \text{kg} \end{align*}$

The radius of the Earth $R$ is

$\begin{align*} R=6.4\cdot 10^6\ \text{m} \end{align*}$

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