A-level Physics (Advancing Physics)/Radar and Triangulation/Worked Solutions

'''1. A radar pulse takes 8 minutes to travel to Venus and back. How far away is Venus at this time?'''

$$2d = ct = 3 \times 10^8 \times 8 \times 60 = 1.44 \times 10^{11} \mbox{ m}$$

$$d = 1.44 \times 10^{11}\times 0.5 = 7.2\times 10^{10} \mbox{ m}$$

2. Why can't a radar pulse be used to measure the distance to the Sun?

It would be impossible to pick up the reflected signal due to all the other signals coming from the Sun. Also, the signal would almost certainly be absorbed anyway. <> Regardless of (λ) wavelength, power density, or wavefront properties, the pulse would be absorbed with no reflection possible. Distances to pure energy sources are generally measured in terms of received light intensity, shifts of the light spectrum, and radio interferometry. The RF spectrum; and Laser (light) spectrum can be used to "listen" to radiation, but not bounce a pulse from an energy source having no true angle of incidence. Just as an observation, I will note that the sun can be seen on most radars either sunrise or sunset, usually when the sun is just above the horizon. But these receptions are unusable strobes (interference) and not a result of receiving a radar pulse from the sun. Radar technicians also use the sun as a "known" exact position to align the system to true north (and magnetic variations); this is called solar-boresighting and, again, only receives the radiation.

'''3. Radar is used to measure the velocity of a spacecraft travelling between the Earth and the Moon. Use the following data to measure this velocity:'''

First, calculate the distance of the spacecraft from the Earth at each time:

$$d_1 = \frac{c\Delta t_1}{2} = \frac{3.0 \times 10^{8} \times (45.51213 - 45.31213)}{2} = 30,000\mbox{ km}$$

$$d_2 = \frac{c\Delta t_2}{2} = \frac{3.0 \times 10^{8} \times (46.52785 - 46.32742)}{2} = 30,064.5\mbox{ km}$$

Next, calculate the distance the spacecraft has travelled between the two pulses:

$$\Delta d = d_2 - d_1 = 30,064.5 - 30,000 = 64.5\mbox{ km}$$

Now, calculate the time elapsed between the transmission of the two pulses:

$$\Delta t = t_\beta - t_\alpha = 46.32742 - 45.31213 = 1.01529\mbox{ s}$$

Finally, divide the distance the spacecraft has travelled between the two pulses by the time between the transmission of the two pulses, to give the average velocity of the spacecraft in that interval of time:

$$v = \frac{\Delta d}{\Delta t} = \frac{64.5}{1.01529} \approx 63.5 \mbox{ kms}^{-1} $$

'''4. The angles between the horizontal and a star are measured at midnight on January 1 as 89.99980&deg; and at midnight on June 1 as 89.99982&deg;. How far away is the star?'''

$$d = \frac{2r\tan{a}\tan{b}}{\tan{a} + \tan{b}} = \frac{2 \times 150 \times 10^9 \times \tan{89.9998}\tan{89.99982}}{\tan{89.9998} + \tan{89.99982}} = 4.52 \times 10^{13}\mbox{ km}$$

5. Why can't triangulation be used to measure the distance to another galaxy?

The difference between the two angles becomes so tiny that we don't have good enough equipment to measure it.