What is the angular speed of a satellite in a geostationary orbit around the Earth? A B C D - AQA - A-Level Physics - Question 13 - 2020 - Paper 2

To find the angular speed ($\omega$) of a geostationary satellite, we can use the formula:

$\omega = \frac{2\pi}{T}$

where $T$ is the period of the satellite. For a geostationary satellite, the period is equal to 24 hours, which is the same as the rotational period of the Earth. Hence,

$T = 24 \text{ hours} = 86400 \text{ seconds}$

Substituting this value into the formula:

$\omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \text{ rad s}^{-1}$

Thus, the correct answer that approximates this value is:

B (7.3 x 10⁻⁵ rad s⁻¹).