Satellites, which play an increasing role in the information age, are controlled by the gravitational force - Leaving Cert Physics - Question 6 - 2019

The relationship between the period $T$ and radius of orbit $r$ of a satellite is given by: $T^2 \propto r^3$ This means that the square of the orbital period is directly proportional to the cube of the radius of its orbit.

Infrared radiation has a longer wavelength compared to visible light. The wavelength of infrared radiation generally ranges from about 700 nm to 1 mm, while visible light ranges from approximately 400 nm to 700 nm.

Infrared radiation can be detected in a school laboratory using a heat-sensitive device, such as a thermopile or a thermogram. Additionally, an infrared detector or sensor can measure the heat emitted from objects that radiate infrared wavelengths. Instruments like infrared cameras can visualize the heat emitted by sources, allowing for an indirect observation of infrared radiation.

The period of METEOSAT 11, which is in geostationary orbit, is 24 hours. This means it takes 24 hours for the satellite to complete one full orbit around the Earth, allowing it to remain fixed relative to a specific point on the Earth's surface.

To calculate the height $h$ above the Earth's surface for a geostationary satellite:

Using the formula for orbital radius: $T = 2 \pi \sqrt{\frac{r^3}{GM}}$ Since $T = 24$ hours, converting this to seconds gives: $T = 24 \times 3600 = 86400\,s$ Now, rearranging: $r = \frac{(T^2 G M)^{1/3}}{2\pi}$ where $G \approx 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}$ and $M = 6.0 \times 10^{24} kg$. Calculating:

1. Calculate $r$: $r \approx 4.224 \times 10^7 m$.
2. To find height $h$, we need the radius of the Earth: $$h = r - 6400 \times 10^3 \approx h \approx 3.56 \times 10^6 m \approx 35600 km.$

For a global positioning satellite orbiting the Earth:

(i) Its speed is 14000 km/h. Converting this to m/s: $v = \frac{14000 \times 1000}{3600} \approx 3889 m/s$

Now, using the formula: $F = \frac{G m M}{r^2}$
Setting $F = \frac{mv^2}{r}$ we can equate and solve for $r$:

$r = \frac{(G M)}{v^2} \approx 2650 \times 10^3 m$

(ii) To find angular velocity $\, \omega$: $\omega = \frac{v}{r} \approx \frac{3889}{2650 \times 10^3} \approx 1.47 \times 10^{-3} rad/s$

Using the radius of orbit: $r \approx 2650 \times 10^3 m$ The distance signal travels is approximately 2r: $distance \approx 2 \times 2650 \times 10^3 m \approx 5300 km$ Signal speed is light speed $c \approx 3 \times 10^8 m/s$. Now, time $t$ is: $t = \frac{distance}{speed} = \frac{5300 \times 10^3 m}{3 \times 10^8 m/s} \approx 0.018 s \approx 18 ms$.

Satellites remain in orbit due to the balance between gravitational forces and their inertia. As a satellite moves forward at high speeds, the gravitational pull from the Earth continuously acts upon it, pulling it towards the Earth. However, because the satellite has a horizontal velocity, it continues moving forward while being pulled down, creating a curved path that results in orbiting. If the satellite were to slow down or lose speed, the gravitational pull would eventually draw it closer to Earth, leading to orbital decay and potential re-entry.