Question

a radar beam with a frequency of 3.10 × 10¹⁰ hz is beamed at an asteroid. the reflected beam is shifted higher in frequency by 4.50 × 10⁶ hz. what is the speed of the asteroid toward the radar beam? ? m/s

Explanation:

Step1: Recall Doppler shift formula

For a reflecting source, the observed frequency shift $\Delta f$ is given by $\Delta f = \frac{2v}{c}f_0$, where $v$ is the speed of the asteroid, $c=3.00\times10^8$ m/s is the speed of light, and $f_0$ is the original frequency.

Step2: Rearrange to solve for $v$

$v = \frac{\Delta f \cdot c}{2f_0}$

Step3: Substitute given values

$\Delta f=4.50\times10^6$ Hz, $f_0=3.10\times10^{10}$ Hz, $c=3.00\times10^8$ m/s
$v = \frac{(4.50\times10^6) \cdot (3.00\times10^8)}{2 \cdot (3.10\times10^{10})}$

Step4: Calculate the result

First compute numerator: $(4.50\times10^6)(3.00\times10^8)=1.35\times10^{15}$
Denominator: $2(3.10\times10^{10})=6.20\times10^{10}$
$v=\frac{1.35\times10^{15}}{6.20\times10^{10}}\approx21.7$ m/s

Answer:

$21.7$ m/s