A brief note on Satellites and Weightlessness . Read it before a test. Try each question at the end before you open its answer.
The big ideas Orbital speed : gravity supplies the pull to the centre, G M m r 2 = m v 2 r \dfrac{GMm}{r^2} = \dfrac{mv^2}{r} r 2 GM m = r m v 2 , so v = G M r v = \sqrt{\dfrac{GM}{r}} v = r GM . The satellite's own mass cancels. Period : T = 2 π r 3 G M T = 2\pi\sqrt{\dfrac{r^3}{GM}} T = 2 π GM r 3 ; higher orbits are slower and take longer. Energies : K = G M m 2 r K = \dfrac{GMm}{2r} K = 2 r GM m , U = − 2 K U = -2K U = − 2 K , E = − K E = -K E = − K . A bound satellite has negative total energy. Geostationary : period one day, above the equator, moving east, about 36 000 km 36\,000\ \text{km} 36 000 km up. Polar : low, over the poles, and the Earth turns beneath it. Weightlessness : in orbit, the craft and everything in it fall together, so nothing presses on anything. Gravity is still there. The facts at a glance Orbit radius × 4 \times\,4 × 4 Changes by Speed v v v × 1 2 \times\,\tfrac12 × 2 1 Period T T T × 8 \times\,8 × 8 Kinetic energy K K K × 1 4 \times\,\tfrac14 × 4 1 Total energy E E E × 1 4 \times\,\tfrac14 × 4 1 , so it rises toward zero
A worked example A 450 kg 450\ \text{kg} 450 kg weather satellite circles 1500 km 1500\ \text{km} 1500 km up. Use G M = g R 2 = 4.01 × 10 14 m 3 / s 2 GM = gR^2 = 4.01 \times 10^{14}\ \text{m}^3/\text{s}^2 GM = g R 2 = 4.01 × 1 0 14 m 3 / s 2 and R = 6400 km R = 6400\ \text{km} R = 6400 km .
Radius. r = 6400 km + 1500 km = 7900 km r = 6400\ \text{km} + 1500\ \text{km} = 7900\ \text{km} r = 6400 km + 1500 km = 7900 km . Speed and period. v = 4.01 × 10 14 m 3 / s 2 7.9 × 10 6 m = 7.13 km/s v = \sqrt{\dfrac{4.01 \times 10^{14}\ \text{m}^3/\text{s}^2}{7.9 \times 10^{6}\ \text{m}}} = 7.13\ \text{km/s} v = 7.9 × 1 0 6 m 4.01 × 1 0 14 m 3 / s 2 = 7.13 km/s , and T = 2 π r v = 116.1 min T = \dfrac{2\pi r}{v} = 116.1\ \text{min} T = v 2 π r = 116.1 min . Total energy. E = − G M m 2 r = − 1.14 × 10 10 J E = -\dfrac{GMm}{2r} = -1.14 \times 10^{10}\ \text{J} E = − 2 r GM m = − 1.14 × 1 0 10 J . Practice with new numbers
A program to try The program works out the weather satellite of the worked example.
Read it, then run it.
Change the height until the period is 2 2 2 hours. How high is that orbit?
Mistakes to avoid The height for r r r . In every orbit formula, r = R + h r = R + h r = R + h . A heavier satellite needing more speed. The mass cancels; every satellite at that height moves at the same speed. A positive total energy. A satellite in orbit has E = − K E = -K E = − K , below zero. Raising an orbit makes it faster. It ends up slower, though it took energy to get there. No gravity in orbit. At 300 km 300\ \text{km} 300 km up, g g g is still about 90 % 90\% 90% of its value on the ground. Check yourself Does a heavier satellite need a greater speed for the same orbit? Stuck? Show a hint Show the answerA satellite's orbit radius is made 4 4 4 times larger. What happens to its period? Stuck? Show a hint Show the answerA satellite has 3.0 × 10 9 J 3.0 \times 10^{9}\ \text{J} 3.0 × 1 0 9 J of kinetic energy. What are its potential and total energies? Stuck? Show a hint Show the answerWhy must a geostationary satellite be above the equator? Stuck? Show a hint Show the answerWhat does a spring balance read when an astronaut stands on it in an orbiting station? Stuck? Show a hint Show the answer