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

Radar and triangulation are two relatively easy methods of measuring the distance to some celestial objects. Radar can also be used to measure the velocity of a celestial object relative to us.

Radar
Essentially, radar is a system which uses a radio pulse to measure the distance to an object. The pulse is transmitted, reflected by the object, and then received at the site of the transmitter. The time taken for all this to happen is measured. This can be used to determine the distance to a planet or even the velocity of a spaceship.

Distance
The speed of electromagnetic waves (c) is constant in a vacuum: 3 x 108 ms−1. If we fire a pulse of radio waves to a planet within the Solar System, we know that:

$$d = ct$$

where d is the distance to the object, and t is the time taken for the pulse to travel there and back from the object. However, the pulse has to get both there and back, so:

$$2d = ct$$

$$d = \frac{ct}{2}$$

where d is the distance to the object, and t is the time taken for the pulse to return.

Velocity
The velocity of an object can be found by firing two radar pulses at an object at different times. Two distances are measured. When asked to calculate the relative velocity of an object in this way, use the following method:

1. Calculate the distance to the object at both times:

$$d_1 = \frac{c\Delta t_1}{2}$$

$$d_2 = \frac{c\Delta t_2}{2}$$

2. Calculate the distance the object has travelled between the two pulses. This is the difference between the two distances previously calculated:

$$\Delta d = d_2 - d_1$$

3. Calculate the time between the transmission (or reception, but not both) of the two pulses:

$$\Delta t = t_2 - t_1$$

4. Divide the distance calculated in step 2 by the time calculated in step 3 to find the average velocity of the object between the transmission of the two pulses:

$$v = \frac{\Delta d}{\Delta t}$$

Triangulation


We know that the Earth is, on average, about 150 Gm away from the Sun. If we measure the angle between the vertical and the light from a nearby star 6 months apart (i.e. on opposite sides of the Sun), we can approximate the distance from the Solar System to the star.

Let r be the radius of the Earth's orbit (assumed to be constant for simplicity's sake), a and b be the angles to the star (from the horizontal) when the Earth is on either side of the Sun, and let d be the perpendicular distance from the plane of the Earth's orbit to the star, as shown in the diagram on the right. By simple trigonometry:

$$2r = \frac{d}{\tan{a}} + \frac{d}{\tan{b}} = \frac{d(\tan{a} + \tan{b})}{\tan{a}\tan{b}}$$

Therefore:

$$d = \frac{2r\tan{a}\tan{b}}{\tan{a} + \tan{b}}$$

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

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

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:

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

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

/Worked Solutions/