Question

a satellite in orbit around mars is operating on $2.25 \times 10^{10}$hz and is moving toward earth at $9.36 \times 10^{3}$m/s. how much higher in frequency is the transmitted signal due to the doppler shift? ? hz

Explanation:

Step1: Recall Doppler shift formula

For a source moving toward an observer, the frequency shift $\Delta f$ is given by:
$$\Delta f = f_0 \times \frac{v}{c}$$
where $f_0 = 2.25 \times 10^{10}\ \text{Hz}$, $v = 9.36 \times 10^3\ \text{m/s}$, $c = 3.00 \times 10^8\ \text{m/s}$ (speed of electromagnetic waves)

Step2: Substitute values into formula

$$\Delta f = 2.25 \times 10^{10} \times \frac{9.36 \times 10^3}{3.00 \times 10^8}$$

Step3: Calculate the result

First compute $\frac{9.36 \times 10^3}{3.00 \times 10^8} = 3.12 \times 10^{-5}$
Then:
$$\Delta f = 2.25 \times 10^{10} \times 3.12 \times 10^{-5} = 7.02 \times 10^{5}\ \text{Hz}$$

Answer:

$7.02 \times 10^{5}\ \text{Hz}$