Satellites and their energy
Falling around the Earth
Imagine a cannon on a very tall mountain, above the air. Fire a ball sideways. It curves down and lands some distance away.
Fire it faster and it lands farther. But the Earth is round, so the ground also curves away beneath the ball.
At just the right speed, the ball falls exactly as fast as the ground curves away. It keeps falling, but never lands. It is in orbit. This is a satellite.
The Moon is a natural satellite of the Earth. India’s communication and weather satellites are artificial ones, launched by ISRO.
For a circular orbit at height h, the radius is R plus h. Gravity provides the centripetal force: m v squared over R plus h equals G M m over R plus h, squared.
The satellite’s mass cancels. The orbital speed is v equals the square root of G M over R plus h. Higher orbits are slower.
The period is the distance round, 2π times R plus h, divided by v. Square it and you get T squared proportional to R plus h, cubed: Kepler’s third law again.
Just above the surface, v is the square root of g R, about 7.9 kilometres per second, and one orbit takes about 85 minutes.
Energy of an orbiting satellite
A satellite has kinetic energy because it moves, and potential energy because it is in the Earth’s gravity. Let us find both. Write r for R plus h.
Kinetic energy is half m v squared. With v squared equal to G M over r, this gives K equals G M m over 2r. Watch the blue bar rise above zero.
Potential energy, with zero at infinity, is U equals minus G M m over r. The orange bar hangs below zero, exactly twice as long as the blue one.
Add them. The total energy is E equals minus G M m over 2r. It is negative, and exactly equal to minus K.
Negative total energy means the satellite is bound to the Earth. It cannot escape on its own. To set it free, we must supply G M m over 2r more energy.
Now raise the orbit. All three bars shrink towards zero. The satellite moves more slowly, but its total energy goes up, from more negative to less negative.
So lifting a satellite to a higher orbit needs energy, even though it ends up with less kinetic energy. The extra goes into potential energy.
Try it: raise the orbit
- Drag the height slider, or pick ISS, GPS or Geostationary.
- The Earth and orbit are drawn to scale. The motion is sped up the same amount for every orbit.
- Compare the bars: U is always −2K and E is always −K.
- Space pauses the satellite.
Quick check
A satellite in a circular orbit has a kinetic energy of 5 GJ. What is its total energy?
- +5 GJ
- −5 GJ
- −10 GJ
- Zero
Answer
B. For a circular orbit K = GMm/2r and E = −GMm/2r, so E = −K = −5 GJ. The potential energy is U = −2K = −10 GJ, and K + U = 5 − 10 = −5 GJ.
Higher orbit
A satellite is moved from a low circular orbit to a higher circular orbit. Which statement is correct?
- Its speed increases and its period decreases
- Its speed decreases and its period increases
- Its speed and period both increase
- Its speed stays the same because there is no friction in space
Answer
B. v = √(GM/r) gets smaller as r grows. The path is also longer, so the period T = 2πr/v gets longer. Compare the ISS (about 7.7 km/s, roughly 90 min) with a geostationary satellite (about 3.1 km/s, 24 h).
Summary
- Orbital speed:
v = √(GM / (R + h)) - Period:
T = 2π √((R + h)³ / GM) - Energies (r = R + h):
K = GMm/2r, U = −GMm/r, E = −GMm/2r - Close to the surface:
v ≈ √(gR) ≈ 7.9 km/s, T ≈ 85 min
- A satellite is always falling towards the Earth; its sideways speed makes it miss. Gravity supplies the centripetal force.
- Orbital speed and period do not depend on the satellite’s mass. Higher orbits are slower and take longer.
- In a circular orbit E = −K and U = 2E. The negative total energy means the satellite is bound to the Earth.
- Moving to a higher orbit raises the total energy (it becomes less negative), so it needs energy from the rocket.