Orbital radius (R) is the distance from the center of a central body (like a planet or star) to a satellite in circular orbit. On AP Physics 1, it connects to orbital period through Kepler's third law, T² = (4π²/GM)R³, where gravity alone provides the centripetal acceleration.
Orbital radius is the distance R measured from the center of the central body (not its surface) out to a satellite moving in a circular orbit. That "from the center" detail matters constantly on the exam. If a satellite orbits at an altitude h above a planet of radius r_p, the orbital radius is r_p + h, not h.
In the AP Physics 1 CED, orbital radius lives in Topic 2.9 (Circular Motion). The big idea (EK 2.9.B.1) is that for a circular orbit, gravity is the only force, so it alone provides the centripetal acceleration. Setting gravitational force equal to mass times centripetal acceleration and solving gives the derived equation T² = (4π²/GM)R³. Notice what's in that equation and what isn't. The period depends on the orbital radius R and the central body's mass M, but the satellite's own mass cancels out. A bowling ball and a space station at the same R have the same period.
Orbital radius is the variable that drives everything in Topic 2.9, Unit 2 (Force and Translational Dynamics). Learning objective 2.9.B asks you to describe circular orbits using Kepler's third law, and that law is fundamentally a statement about how period scales with orbital radius (T² ∝ R³). Learning objective 2.9.A is also in play, since centripetal acceleration a_c = v²/r uses the orbital radius as the radius of the circular path. If you can track how changing R changes speed, acceleration, and period, you've mastered the most-tested skill in this topic. AP questions love ratio reasoning here, like "radius quadruples, what happens to the period?" The answer comes straight from the R³ scaling, no calculator gymnastics needed.
Keep studying AP® Physics 1 Unit 2
Kepler's Third Law (Unit 2)
T² = (4π²/GM)R³ is the orbital radius's job description. Kepler's third law is really just Newton's second law applied to an orbit, with gravity supplying the centripetal force. Bigger R means a much longer period, since T grows like R to the 3/2 power.
Orbital Period (Unit 2)
Period and orbital radius are two sides of the same equation. If a question gives you one, it's usually testing whether you can find the other. Quadrupling R multiplies T by 8, because 4^(3/2) = 8.
Tangential Speed (Unit 2)
A satellite's speed is set by its orbital radius, not chosen freely. Setting gravity equal to mv²/R shows that satellites in larger orbits actually move slower. That feels backwards until you remember gravity is weaker farther out, so less speed is needed to stay in orbit.
Centripetal Acceleration (Unit 2)
The orbital radius is the r in a_c = v²/r. For orbits, that centripetal acceleration comes entirely from gravitational attraction (EK 2.9.B.1), which is why you can derive Kepler's third law in the first place.
Orbital radius shows up in two main ways. First, ratio-style multiple choice. A typical stem says the period drops from 24 hours to 6 hours and asks by what factor the radius must change, or gives three planets with radii in a 1 : 4 : 9 ratio and asks for their periods. The move is always the same. Take T² ∝ R³, set up the ratio, and solve without plugging in G or M. Second, derivation-style free response. The 2018 exam gave a spacecraft of mass m in a circular orbit of radius R around Earth and expected you to set gravitational force equal to centripetal force and solve for speed or period. Watch for the classic trap where a planet's physical radius and a satellite's orbital radius both appear in the same problem (like "a satellite orbits at 3R from the center, where R is the planet's radius"). Keep those two distances straight and label them differently in your work.
Orbital radius is measured from the central body's center to the satellite. Planetary radius is the size of the planet itself. They're both called R in different problems, which is exactly why AP questions mix them, like a satellite orbiting at 3R where R is the planet's radius. If a problem gives altitude above the surface, add the planet's radius to get the orbital radius before using Kepler's third law.
Orbital radius is measured from the center of the central body, so a satellite's orbital radius equals the planet's radius plus its altitude above the surface.
Kepler's third law, T² = (4π²/GM)R³, says the square of the period is proportional to the cube of the orbital radius.
The orbital period depends on the orbital radius and the central body's mass M, but not on the satellite's mass, which cancels out of the derivation.
For circular orbits, gravity is the only force, and it alone provides the centripetal acceleration toward the center (EK 2.9.B.1).
Satellites at larger orbital radii move slower and take much longer to complete an orbit, since T scales as R to the 3/2 power.
On ratio problems, quadrupling the orbital radius multiplies the period by 8, because 4^(3/2) = 8.
It's the distance R from the center of a central body to a satellite in circular orbit. It appears in Topic 2.9 and connects to orbital period through Kepler's third law, T² = (4π²/GM)R³.
No. The planet's radius is the size of the planet itself, while orbital radius is measured from the planet's center out to the satellite. If a satellite orbits 400 km above Earth's surface, its orbital radius is Earth's radius plus 400 km.
No. The satellite's mass cancels when you set gravitational force equal to centripetal force, so any object at the same orbital radius around the same central body has the same period and speed.
The period increases by a factor of 2^(3/2), which is about 2.83. That comes straight from T² ∝ R³, the ratio reasoning AP multiple choice questions test constantly.
Setting gravity equal to mv²/R shows that speed decreases as orbital radius increases. Satellites farther out move slower because gravity is weaker there, so less speed is needed to maintain the circular path.
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