A geostationary satellite

Problem statement

Since the 1960s, geostationary satellites have made telecommunications, weather forecasting and Earth observation possible. A geostationary satellite moves around the planet in a circular orbit in the plane of the equator. Its period of revolution equals the Earth's period of rotation, that is, it takes 24 h to complete one turn around the Earth, so it always stays above the same point of the planet. The figure shows the geostationary orbit of a satellite of mass $m = 1.2\cdot10^{3}$ kg around the Earth. The mass of the Earth is $M = 5.97\cdot10^{24}$ kg.

[Figure: the Earth with its N–S axis, the circular orbit in the plane of the equator and the satellite; a dimension line marks 36,000 km between the Earth and the satellite.]

a) List the four fundamental forces of nature and briefly describe how each of them acts (1.2 marks).

b) Calculate the gravitational force with which the Earth attracts the satellite, given that the universal gravitation constant is $G = 6.67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. You may neglect the radius of the Earth. Is the force you calculate attractive or repulsive? Give the result in scientific notation (0.8 marks).

c) The satellite now stops being geostationary and takes two days to go around the Earth. Explain whether its angular velocity increases or decreases, and by what factor compared with the geostationary satellite (0.5 marks).

Source: Madrid university entrance exam (PAU, Spain) · 2025-2026 · Ordinary session · General Science · Question 3A. Official exam set by the public universities of the Community of Madrid, to whom it belongs; translated from Spanish and reproduced with its source in order to explain it for teaching purposes.