Kepler's Third Law states that the square of an orbit's period is proportional to the cube of its semi-major axis (T² ∝ a³); in AP Physics C, you derive it for circular orbits by setting gravitational force equal to centripetal force, getting T² = 4π²a³/(GM) where M is the central mass.
Kepler's Third Law is the relationship between how long an orbit takes and how big it is. Formally, the square of the orbital period is proportional to the cube of the semi-major axis of the orbit, written T² ∝ a³. Bigger orbits don't just take longer because the path is longer. They also take longer because the object moves more slowly, since gravity is weaker farther out. The third law captures both effects in one clean equation.
In AP Physics C: Mechanics, this isn't a fact you memorize from astronomy class. It's something you derive. For a circular orbit of radius r, set the gravitational force equal to the centripetal force (GMm/r² = mv²/r), substitute v = 2πr/T, and solve. You get T² = (4π²/GM)r³. Two things to notice. First, the orbiting object's mass m cancels out completely, so a pebble and a space station at the same radius have the same period. Second, the proportionality constant 4π²/GM depends only on the central mass M. That means the law lets you compare any two satellites of the same body, or work backward from a measured period and radius to find the mass of the thing being orbited.
Kepler's Third Law lives in Topic 7.2, Orbits of Planets and Satellites, in the gravitation unit of AP Physics C: Mechanics. It's the payoff of the whole unit because it ties Newton's law of universal gravitation to circular motion from earlier in the course. The exam loves this derivation precisely because it tests whether you can connect force laws to kinematics rather than just plug into a formula. It's also one of the most historically satisfying results in the course. Kepler found the T²-a³ pattern empirically from planetary data, and Newton later showed it falls right out of an inverse-square force law. When you do that derivation on the exam, you're recreating one of physics' greatest hits.
Gravitational Force (Unit 7)
Kepler's Third Law is what Newton's law of gravitation predicts when you apply it to orbits. The exponent 3 in T² ∝ r³ comes directly from the inverse-square nature of gravity, so the law is really evidence that gravity falls off as 1/r².
Orbital Period (Unit 7)
The period T is the quantity Kepler's Third Law actually solves for. Once you know the radius and the central mass, T = 2π√(r³/GM) hands you the period without ever tracking the satellite's position over time.
Circular Orbit (Unit 7)
The AP derivation assumes a circular orbit so that gravitational force supplies a constant centripetal force, mv²/r. Circular motion from Unit 3 is the bridge that turns Newton's gravitation into Kepler's law.
Semi-Major Axis (Unit 7)
For elliptical orbits, the 'a' in T² ∝ a³ is the semi-major axis, not the radius. A circle is just the special case where the semi-major axis equals the radius, which is why the circular-orbit derivation generalizes.
Expect Kepler's Third Law in two flavors. Multiple-choice questions love ratio reasoning, such as asking how the period changes if the orbital radius quadruples (answer: T increases by a factor of 4^(3/2) = 8). You should be fast with the 3/2-power scaling without re-deriving anything. Free-response questions tend to ask for the derivation itself. A classic setup gives you a satellite in circular orbit and asks you to derive an expression for the period in terms of G, M, and r, or to use a measured period to find the mass of a planet. The non-negotiable skill is starting from GMm/r² = mv²/r and substituting v = 2πr/T cleanly. Watch for the common point-loser of using the orbiting object's mass instead of the central mass in the final expression. No released FRQ needs you to quote the law by name, but the derivation pattern shows up regularly in gravitation free-response questions.
Kepler's Second Law (equal areas in equal times) describes how an object's speed varies within a single elliptical orbit, and it's really a statement of angular momentum conservation. Kepler's Third Law compares different orbits to each other, relating each orbit's period to its size. Second law: one orbit, speed changes. Third law: many orbits, period scales with a³ to the 1/2... more precisely, T ∝ a^(3/2).
Kepler's Third Law states T² ∝ a³, meaning the square of the orbital period is proportional to the cube of the semi-major axis (or radius, for a circular orbit).
You derive it on the AP exam by setting gravitational force equal to centripetal force, which gives T² = 4π²r³/(GM) for a circular orbit.
The orbiting object's mass cancels in the derivation, so the period depends only on the orbital radius and the central body's mass M.
Because the constant 4π²/GM is the same for everything orbiting one central body, the law lets you compare satellites or compute the central mass from a measured T and r.
For quick scaling questions, remember T ∝ r^(3/2), so quadrupling the radius makes the period 8 times longer.
The law applies to elliptical orbits too if you use the semi-major axis a in place of the circular radius r.
It's the relationship T² ∝ a³, where T is the orbital period and a is the semi-major axis of the orbit. For a circular orbit, the full expression is T² = 4π²r³/(GM), with M being the mass of the central body.
You don't need to memorize the final formula, but you do need to derive it. Set GMm/r² = mv²/r, substitute v = 2πr/T, and solve for T². The exam rewards the derivation, and the equation sheet gives you the pieces (Newton's gravitation and circular motion) to rebuild it.
No. The satellite's mass m cancels when you set gravitational force equal to centripetal force, so the period depends only on the orbital radius and the central mass M. A 1 kg probe and a 400,000 kg station at the same radius have identical periods.
The second law says a single orbiting body sweeps out equal areas in equal times, which is conservation of angular momentum in disguise. The third law compares different orbits, saying larger orbits have longer periods according to T² ∝ a³.
The period increases by a factor of 2^(3/2), which is about 2.83. This comes straight from T ∝ r^(3/2), and this kind of ratio reasoning is a classic multiple-choice question.
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