Why a satellite does not simply fall straight down
A satellite is continuously accelerated toward Earth by gravity, but it also has sideways motion. In a suitable orbit its direction changes as fast as the planet’s curved surface falls away beneath it. It is in free fall rather than being supported by an outward “antigravity” force. The simplest model is a satellite in a circular orbit around a spherical Earth.
What you will learn
- Distinguish altitude h from orbital radius r measured from Earth's centre.
- Derive ideal circular orbital speed and period.
- Explain the counterintuitive result that higher circular orbits have lower speed but longer period.
- Calculate a real-world-style low Earth orbit and test the result in the Orbit Explorer.
Gravity provides centripetal acceleration
For a small satellite of mass m at distance r from Earth's centre, the gravitational force magnitude is GMₑm/r², where Mₑ is Earth's mass. An object moving in a circle at speed v requires centripetal acceleration v²/r. Equating them cancels satellite mass and yields the orbital speed formula. Here μ = GMₑ is Earth's standard gravitational parameter.
The altitude h is measured above the chosen Earth-radius reference, not from Earth's centre. Using h in place of r is a common and serious mistake. The formulas are valid for an ideal circular orbit; elliptical trajectories require different treatment.
Worked example: a satellite 400 km above Earth
Use mean Earth radius Rₑ ≈ 6,371 km and μ ≈ 3.9860 × 10¹⁴ m³/s². At h = 400 km, orbital radius is r = 6,771 km = 6.771 × 10⁶ m. Hence v ≈ √(3.9860 × 10¹⁴ / 6.771 × 10⁶) ≈ 7.673 km/s. Orbital period T ≈ 2πr/v ≈ 5,545 s ≈ 92.4 minutes. The speed is approximately 27,620 km/h; this is an ideal calculation, not a live measurement of a particular satellite.
What changes when altitude rises?
As r grows, gravity weakens and the circular orbital speed decreases as 1/√r. But the orbital circumference increases as r, so the time per orbit grows as r³ᐟ². A satellite in a larger circular orbit moves more slowly and travels farther per revolution, resulting in a much longer period. The angular speed also decreases.
These relationships express Kepler’s third-law scaling for orbits around the same dominant central body in the Newtonian two-body approximation.
Energy in a circular orbit
With the conventional zero of gravitational potential energy at infinite separation, specific gravitational potential energy is −μ/r and specific kinetic energy is v²/2 = μ/(2r). The total specific mechanical energy is −μ/(2r). It is negative for a bound circular orbit. Raising a satellite to a higher stable circular orbit requires adding energy even though the final circular speed is lower. This apparent contradiction disappears when potential and kinetic energies are considered together.
Guided virtual investigation
- Open the Circular Orbit Explorer and set altitude to 400 km.
- Record orbital speed and period; compare with 7.67 km/s and approximately 92.4 min.
- Increase altitude in stages while keeping the central body as Earth.
- Predict whether speed and period each increase or decrease before looking at the readouts.
- Plot T² against r³ using consistent SI units; the ideal data should follow a straight-line relationship.
- Explain why the orbit animation is schematic and may not be drawn to the same scale as the numerical calculations.
What this model does not include
Atmospheric drag, Earth's oblateness, gravitational effects of the Moon and Sun, changes in reference frames, spacecraft thrust and orbital perturbations are omitted. The model treats Earth as spherical and non-rotating. A real geostationary orbit also requires the correct direction, near-zero inclination, circularity and a period equal to Earth's sidereal rotation period; altitude alone is not sufficient to describe any real orbit's ground track.
Typical misconceptions
- “There is no gravity in orbit.” Gravity is precisely what bends the path.
- “A satellite of greater mass needs greater speed for the same orbit.” Not in the test-particle approximation; mass cancels.
- “Higher circular orbits are faster.” They are slower, even though getting there requires energy.
- “Orbital radius equals altitude.” Radius includes Earth's radius plus altitude.
Practice and explanations
- If circular orbit radius becomes four times as large, what happens to speed? Answer: speed halves.
- With radius four times as large, period becomes how many times greater? Answer: 4³ᐟ² = 8 times.
- What provides centripetal force? Answer: Earth's gravitational attraction.
- A 400 km altitude satellite: can we use r = 400,000 m in the speed formula? Answer: no; use Earth radius plus 400,000 m.
- If satellite mass doubles at the same circular orbital radius, ideal speed changes by what factor? Answer: it does not change.
Learning and exam connection
This lab links gravitation, circular motion, conservation of energy and Kepler’s laws. In NCERT/JEE/NEET solutions, draw the planet, mark r from its centre and distinguish the formulas for circular speed, escape speed and the speed of a spacecraft while transferring between orbits. Predict the trend before trusting a numerical output.
Frequently asked questions
Does gravity vanish for an orbiting satellite?
No. Gravity supplies its centripetal acceleration.
Why do higher circular orbits have lower speeds?
The circular-orbit formula v = √(μ/r) decreases with increasing radius.
Is orbital radius the same as altitude?
No. Orbital radius is Earth radius plus altitude.
Why does a higher circular orbit take longer?
Its path is longer while its circular orbital speed is lower.
Further reading and references
References are for further independent learning; explanations and examples on this page are written for NANDAX.