Satellites N and F have the same mass and move in circular orbits about the same planet - AQA - A-Level Physics - Question 13 - 2022 - Paper 2

The orbital speed is given by the equation $v = \sqrt{\frac{GM}{r}}$ Since N is at a smaller radius, its speed will be higher than F's. Therefore, the speed is not smaller for N.

The kinetic energy is given by $KE = \frac{1}{2}mv^2$ Since the speed of N is greater, its kinetic energy will also be greater than that of F. Thus, this is not the correct answer.

The orbital period can be calculated using Kepler's third law: $T = 2\pi\sqrt{\frac{r^3}{GM}}$ Since N has a smaller radius, it will have a smaller period because the period decreases with decreasing radius. Therefore, the orbital period is smaller for N than for F.