A global positioning system (GPS) satellite moves in a circular orbit with period 11 h 58 min. (Assume the mass of the earth is 5.98 x 1024 kg, and the radius of the earth is 6.37 x 106 m.) (a) Determine the radius of its orbit (in m). m (b) Determine its speed (in m/s). m/s (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, 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 Af AU f mc² = where U, 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.
A global positioning system (GPS) satellite moves in a circular orbit with period 11 h 58 min. (Assume the mass of the earth is 5.98 x 1024 kg, and the radius of the earth is 6.37 x 106 m.) (a) Determine the radius of its orbit (in m). m (b) Determine its speed (in m/s). m/s (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, 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 Af AU f mc² = where U, 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.
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![### Educational Exercise on GPS Satellite Motion and Relativity
A global positioning system (GPS) satellite moves in a circular orbit with a period of 11 hours and 58 minutes. We are considering that the mass of the Earth is \(5.98 \times 10^{24} \text{ kg}\), and the radius of the Earth is \(6.37 \times 10^6 \text{ m}\).
#### Questions:
**(a)** Determine the radius of its orbit (in meters).
- **Radius (r):** \(\_\_\_\_\_\_\_\_\_ \text{ m}\)
**(b)** Determine its speed (in meters per second).
- **Speed (v):** \(\_\_\_\_\_\_\_\_\_ \text{ m/s}\)
**(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, what is the fractional change in this frequency due to time dilation as described by special relativity?
- \(\frac{\Delta f}{f} = \_\_\_\_\_\_\_\_\_\_\)
**(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:
\[
\frac{\Delta f}{f} = \frac{\Delta U_g}{mc^2}
\]
Where \(\Delta U_g\) 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.
- \(\frac{\Delta f}{f} = \_\_\_\_\_\_\_\_\_\_\)
**(e)** What is the overall fractional change in frequency due to both time dilation and gravitational blueshift?
- \(\frac{\Delta f}{f} = \_\_\_\_\_\_\_\_\_\_\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ac2863-162e-4262-bc0d-8a607c6a4bb3%2F73513a9b-5d2e-477e-9ae7-725b2e31c010%2F77xfcl8_processed.png&w=3840&q=75)
Transcribed Image Text:### Educational Exercise on GPS Satellite Motion and Relativity
A global positioning system (GPS) satellite moves in a circular orbit with a period of 11 hours and 58 minutes. We are considering that the mass of the Earth is \(5.98 \times 10^{24} \text{ kg}\), and the radius of the Earth is \(6.37 \times 10^6 \text{ m}\).
#### Questions:
**(a)** Determine the radius of its orbit (in meters).
- **Radius (r):** \(\_\_\_\_\_\_\_\_\_ \text{ m}\)
**(b)** Determine its speed (in meters per second).
- **Speed (v):** \(\_\_\_\_\_\_\_\_\_ \text{ m/s}\)
**(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, what is the fractional change in this frequency due to time dilation as described by special relativity?
- \(\frac{\Delta f}{f} = \_\_\_\_\_\_\_\_\_\_\)
**(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:
\[
\frac{\Delta f}{f} = \frac{\Delta U_g}{mc^2}
\]
Where \(\Delta U_g\) 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.
- \(\frac{\Delta f}{f} = \_\_\_\_\_\_\_\_\_\_\)
**(e)** What is the overall fractional change in frequency due to both time dilation and gravitational blueshift?
- \(\frac{\Delta f}{f} = \_\_\_\_\_\_\_\_\_\_\)
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