Satellites N and F have the same mass and are circular orbits about the same planet - AQA - A-Level Physics - Question 11 - 2019 - Paper 2

The angular speed ( \omega ) is given by ( \omega = \frac{v}{r} ) where ( v ) is the orbital speed. Since satellite F has a larger radius, its angular speed is lower than that of satellite N.

The kinetic energy (KE) of a satellite is given by: $KE = \frac{1}{2}mv^2$ In circular orbits, the orbital speed is related to the radius. Satellite F, with a larger radius, has a lower speed, resulting in lower kinetic energy compared to N.

The orbital period (T) can be calculated using Kepler's third law: $T = 2\pi \sqrt{\frac{r^3}{GM}}$ As the orbital radius increases, the orbital period increases as well. Thus, the orbital period for F is greater than that for N.