A circular orbit is the special case of orbital motion where gravity supplies exactly the centripetal force needed to keep an object at constant radius and constant speed, so setting GMm/r² = mv²/r gives the orbital speed v = √(GM/r).
A circular orbit is what happens when an object (a satellite, a moon, a planet) moves around a much more massive body at a constant distance and constant speed. There is nothing mysterious going on. A circular orbit is just uniform circular motion where the centripetal force happens to be gravity. The satellite is constantly falling toward the central body, but it's moving sideways fast enough that it falls around the body instead of into it.
That one idea generates everything you need on the exam. Set Newton's law of gravitation equal to the centripetal force requirement, GMm/r² = mv²/r, and you can solve for the orbital speed v = √(GM/r). Notice the satellite's own mass m cancels, which is why a bolt and a space station at the same radius orbit at the same speed. From v you can build the period (T = 2πr/v), the kinetic energy, and the total mechanical energy E = -GMm/(2r). Every formula about circular orbits is just that one force equation rearranged.
Circular orbits live in Topic 7.2, Orbits of Planets and Satellites, in the gravitation unit (Unit 7). The unit asks you to apply Newton's law of universal gravitation to orbiting bodies, and the circular orbit is the cleanest version of that. It's where two big course threads finally merge. The circular motion machinery from Unit 2 (centripetal acceleration, net force pointing toward the center) and the energy framework from Unit 3 (kinetic plus potential energy) both get reused here with gravity plugged in as the force. The circular case is also your entry point to Kepler's third law, since squaring the period relation for a circular orbit gives T² = (4π²/GM)r³ directly. If you can derive v = √(GM/r) from scratch, you've basically unlocked half of Unit 7.
Keep studying AP® Physics C: Mechanics Unit 7
Centripetal Force (Unit 2)
A circular orbit is uniform circular motion with gravity playing the role of centripetal force. Nothing new is added in Unit 7 except which force points toward the center. If you can do a ball-on-a-string problem, you can do a satellite problem by swapping tension for GMm/r².
Kepler's Third Law (Unit 7)
Kepler's third law, T² ∝ r³, falls right out of the circular orbit equations. Set gravitational force equal to centripetal force, write v as 2πr/T, and rearrange. Deriving Kepler's third law from a circular orbit is a classic FRQ-style derivation, so practice it until it's automatic.
Mechanical Energy (Unit 3)
For a circular orbit, the energies lock into fixed ratios. Kinetic energy is GMm/(2r), potential energy is -GMm/r, so total energy is E = -GMm/(2r), exactly half the potential energy and the negative of the kinetic. The negative total energy is the signature of a bound orbit, and it's the quantity you track when a satellite changes orbits or escapes.
Elliptical Orbit (Unit 7)
An elliptical orbit is the general case and a circular orbit is the special case where the ellipse's two foci sit on top of each other. In an ellipse, speed and radius both change while energy and angular momentum stay constant. In a circle, speed and radius are constant too, which is what makes the math so clean.
Circular orbits show up in two main flavors. Multiple-choice questions love proportional reasoning, asking how v, T, or E changes if the radius doubles or the central mass quadruples (for example, v scales as 1/√r, so doubling r divides speed by √2). Free-response questions tend to ask for derivations. The greatest hits include deriving v = √(GM/r) from Newton's second law, deriving Kepler's third law for a circular orbit, finding the total mechanical energy of a satellite, and computing the energy needed to move between two circular orbits or to escape entirely. The single move the exam rewards over and over is writing GMm/r² = mv²/r as your starting line and showing the algebra cleanly. Watch the classic traps too, like using mg instead of GMm/r² far from a surface, or dropping the negative sign on gravitational potential energy.
In a circular orbit, the radius and speed never change, so kinetic and potential energy are each individually constant. In an elliptical orbit, the satellite speeds up near the central body and slows down far from it, so kinetic and potential energy trade back and forth while total mechanical energy and angular momentum stay fixed. The v = √(GM/r) formula works only for circles. Use it on an ellipse and you'll get the wrong answer, because an elliptical orbit has no single radius.
A circular orbit is uniform circular motion where gravity is the centripetal force, so the master equation is GMm/r² = mv²/r.
Orbital speed is v = √(GM/r), which depends only on the central mass and the radius, not on the satellite's mass.
Total mechanical energy in a circular orbit is E = -GMm/(2r), and the negative sign tells you the satellite is gravitationally bound.
Bigger orbits are slower and longer. Speed decreases as 1/√r while the period grows, following Kepler's third law T² = (4π²/GM)r³.
The formulas v = √(GM/r) and E = -GMm/(2r) apply only to circular orbits, so check the orbit shape before plugging in.
Almost every circular orbit FRQ starts the same way, by setting the gravitational force equal to mv²/r and solving for whatever the question asks.
It's the path of an object moving around a massive body at constant radius and constant speed, where gravity provides exactly the centripetal force required. The defining equation is GMm/r² = mv²/r, which gives orbital speed v = √(GM/r).
Yes. Even though its speed is constant, its direction changes continuously, so it has a centripetal acceleration of magnitude v²/r (equal to GM/r²) pointing toward the central body. Constant speed is not the same as zero acceleration.
No. When you set GMm/r² equal to mv²/r, the satellite's mass m cancels, leaving v = √(GM/r). Only the central body's mass and the orbital radius matter, which is why all objects at the same radius orbit at the same speed.
A circular orbit has constant radius and constant speed, while an elliptical orbit has changing radius and speed (fastest at closest approach, slowest at the far point). The shortcut formulas v = √(GM/r) and E = -GMm/(2r) only work for circular orbits.
Because gravitational potential energy, U = -GMm/r, is negative and larger in magnitude than the kinetic energy, GMm/(2r). The total comes out to E = -GMm/(2r), and that negative value means the satellite is bound. You'd have to add energy to push it to escape, where E = 0.
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