Global positioning satellite (GPS) receivers operate at the following two frequencies: \(L_1 = 1.57542 \, \text{GHz}\) and \(L_2 = 1.22760 \, \text{GHz}\).

(a) Show that when the radio frequency exceeds the plasma frequency (peak ionospheric plasma frequency < 10 MHz), the following relation for the group delay due to propagation through the plasma is given by:

\[
\tau = \frac{40.3 \, \text{TEC}}{f^2}
\]

where the group delay, \(\tau\), is measured in meters, TEC is the total electron content between the GPS receiver and the satellite (i.e., the column density of electrons measured in electrons/m\(^2\), where 1 TEC unit = \(10^{16}\) electrons/m\(^2\)), and the radio frequency is in Hz.

(b) Calculate the value of \(\tau\) in the case of 1 TEC unit (TECU) for both \(L_1\) and \(L_2\), and show that every excess of 10 cm on \(L_2 - L_1\) corresponds to 1 TECU of electron content.

Global positioning satellite (GPS) receivers operate at two distinct frequencies: L = 1.57542 GHz and L = 1.22760 GHz. The group delay caused by plasma propagation can be determined using the formula r = TEC/f^2, where r represents the group delay in meters, TEC is the total electron content in TECU (total electron content units), and f is the frequency in MHz.

However, this formula is only applicable when the radio frequency surpasses the peak ionospheric plasma frequency (which is less than 10 MHz).

To calculate the value of r for 1 TECU at both L and L2 frequencies, we can use the given equation r = 40.3 TEC/f^2.

For L1 with f = 1.57542 GHz, the formula becomes r = 244.9 / TECU. For L2 with f = 1.22760 GHz, the formula becomes r = 288.9 / TECU.

The frequency difference between L1 and L2 is ∆f = 347.82 MHz, and the excess number of wavelengths of L2 over L1 can be found using ∆N = ∆f / f1^2, where f1 is the frequency of L1.

In this case, ∆N equals 0.0722 wavelengths. Each excess of 10 cm on L2-L corresponds to 1 TECU of electron content. Thus, (0.0722 x 10^9) / (10 x 0.01) equals 72.2 TECU of electron content.

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