Question

Radar wave has frequency 8.1x$10^9$ H$_{2}$ the Reflected waves from aeroplane shows a frequency diff$^{n}$ 2.7x$10^3$ H$_{2}$ on Higher side what is the $O_{velocity}$ of aeroplane in line of site.

Answer 1

50 m/s

Answer 2

The radar emits a wave of frequency $f$ which is reflected by the aeroplane moving with velocity $v$ towards the radar. Due to the Doppler effect, the frequency of the wave received by the aeroplane is shifted, and the frequency of the reflected wave received back at the radar is shifted again. The total frequency shift $\Delta f = f'' - f$ for an object moving towards the radar is given by the formula:

$$\Delta f = f \left( \frac{c + v}{c - v} \right) - f = f \left( \frac{2v}{c - v} \right)$$

where $c$ is the speed of the radar wave (speed of light).

Since the velocity of the aeroplane $v$ is much smaller than the speed of light $c$ ($v \ll c$), we can use the approximation $c - v \approx c$. The formula simplifies to:

$$\Delta f \approx f \left( \frac{2v}{c} \right)$$

We can rearrange this formula to solve for the velocity $v$:

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

Substitute the given values: $f = 8.1 \times 10^9 \, \text{Hz}$, $\Delta f = 2.7 \times 10^3 \, \text{Hz}$, and $c = 3 \times 10^8 \, \text{m/s}$.

$$v = \frac{(2.7 \times 10^3 \, \text{Hz}) \times (3 \times 10^8 \, \text{m/s})}{2 \times (8.1 \times 10^9 \, \text{Hz})}$$ $$v = \frac{8.1 \times 10^{11}}{16.2 \times 10^9} \, \text{m/s}$$ $$v = 0.5 \times 10^{11-9} \, \text{m/s}$$ $$v = 0.5 \times 10^2 \, \text{m/s}$$ $$v = 50 \, \text{m/s}$$

The velocity of the aeroplane in the line of sight (towards the radar) is 50 m/s.