Orbits of planets and satellites — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Circular and elliptical orbits, Kepler's three laws of planetary motion, gravitational centripetal force balance, bound/unbound orbit classification, orbital energy, escape velocity, and orbital period/speed calculations.

You should already know: Newton's law of universal gravitation. Centripetal acceleration for uniform circular motion. Conservation of mechanical energy and angular momentum.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Orbits of planets and satellites?

An orbit is the curved path a celestial object (planet, moon, artificial satellite, or comet) follows around a second much more massive body, held exclusively by gravitational attraction between the two masses. For AP Physics C: Mechanics, this topic falls under Unit 7 Gravitation, which accounts for 12–18% of the total exam score per the official Course and Exam Description (CED). Orbit problems appear in both multiple choice (MCQ) and free response (FRQ) sections, often combining gravitation with circular motion, energy, and angular momentum concepts from earlier units.

Notation conventions used here: $M$ = mass of the central body (assumed stationary because $M\gg m$, where $m$ = mass of the orbiting body), $r$ = center-to-center distance between the two bodies, $a$ = semi-major axis of an elliptical orbit, $G=6.67\times 10^{-11}{\text{ Ncdotpm}}^2/{\text{kg}}^2$ = universal gravitational constant. AP C: Mech only tests two-body Keplerian orbits with a dominant central mass, so complex multi-body orbital interactions are not covered on the exam.

2. Circular Orbits and Centripetal Force Balance

For a circular orbit, gravitational force from the central mass provides exactly the centripetal force required to keep the orbiting body moving at constant speed along a circular path. Because $M\gg m$, we can ignore the small acceleration of the central body around the shared center of mass, simplifying the derivation significantly.

Starting from Newton's second law for circular motion: the net force equals gravitational force, which equals mass times centripetal acceleration: $$F_{net}=F_g=m{a}_c$$ Substitute $F_g=\frac{GMm}{r^2}$ and $a_c=\frac{v^2}{r}$ to get: $$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$$ The orbiting mass $m$ cancels out entirely, meaning orbital speed is independent of the orbiting body's mass: $$v=\sqrt{\frac{GM}{r}}$$ This key result tells us that for circular orbits, orbital speed decreases as orbital radius increases: farther-out satellites always move slower than closer satellites. We can also derive orbital period $T$ (time to complete one full orbit) by substituting $v=\frac{2\pi r}{T}$ to get: $$T^2=\frac{4{\pi }^2{r}^3}{GM}$$ which is Kepler's third law for the special case of circular orbits.

Worked Example

A small artificial satellite orbits Mars at a height of 200 km above Mars' surface. Mars has a mass of $6.42\times 10^{23}$ kg and radius $3.39\times 10^6$ m. What is the satellite's orbital speed?

1. First, calculate the center-to-center orbital radius: $r=R_{\text{Mars}}+h=3.39\times 10^6+2.00\times 10^5=3.59\times 10^6$ m.
2. For circular orbits, use the force-balance result $v=\sqrt{\frac{GM}{r}}$.
3. Substitute values: $\frac{GM}{r}=\frac{(6.67\times 10^{-11})(6.42\times 10^{23})}{3.59\times 10^6}\approx 1.19\times 10^7$.
4. Take the square root: $v=\sqrt{1.19\times 10^7}\approx 3450$ m/s, or 3.45 km/s.

Exam tip: Always confirm that $r$ is the distance between centers of mass, not just height above the surface of the central body. If given altitude, add the central body's radius to get $r$ before plugging into any formula.

3. Kepler's Laws of Planetary Motion

Kepler derived three empirical laws from observational data before Newton developed gravitational theory, and Newton's law of universal gravitation confirms all three laws for two-body orbits:

1. Law of Orbits: All planets move in elliptical orbits with the central body (e.g. the Sun) at one focus of the ellipse. For an ellipse, eccentricity $e$ describes how "stretched" the orbit is: $e=0$ is a perfect circle, $0<e<1$ is a bound elliptical orbit, and $e\ge 1$ is an unbound orbit. The closest distance to the central focus (perihelion for solar orbits, perigee for Earth orbits) is $r_p=a(1-e)$, and the farthest distance (aphelion/apogee) is $r_a=a(1+e)$, where $a$ is the semi-major axis (half the longest diameter of the ellipse).
2. Law of Areas: A line joining the orbiting body and the central body sweeps out equal areas in equal time intervals. This is a direct consequence of conservation of angular momentum: gravity acts along the line connecting the two bodies, so it exerts zero torque, meaning angular momentum $L$ is constant. Speed is higher at smaller $r$ (closest approach) and lower at larger $r$ (farthest approach).
3. Law of Periods: The square of the orbital period is proportional to the cube of the semi-major axis, or $T^2=\frac{4{\pi }^2{a}^3}{GM}$, which generalizes the circular orbit result (where $a=r$ for circles).

Worked Example

An asteroid orbits the Sun in an elliptical orbit with perihelion distance 1 AU and aphelion distance 7 AU (1 AU = Earth's semi-major axis, Earth's orbital period = 1 year). Find (a) the semi-major axis of the asteroid, (b) the asteroid's orbital period, and (c) the ratio of the asteroid's perihelion speed to aphelion speed.

1. (a) By definition, $a=\frac{r_p+r_a}{2}=\frac{1+7}{2}=4$ AU.
2. (b) Use Kepler's third law for two objects orbiting the same Sun: $\frac{T_1^2}{a_1^3}=\frac{T_2^2}{a_2^3}$. Substitute Earth's values: $\frac{(1\text{ yr}{)}^2}{(1\text{ AU}{)}^3}=\frac{T^2}{(4\text{ AU}{)}^3}⟹T^2=64⟹T=8$ years.
3. (c) Use conservation of angular momentum: at perihelion and aphelion, velocity is perpendicular to the radius vector, so $m{r}_p{v}_p=m{r}_a{v}_a⟹\frac{v_p}{v_a}=\frac{r_a}{r_p}=\frac{7}{1}=7$. The perihelion speed is 7 times the aphelion speed.

Exam tip: When comparing orbits of objects around the same central mass, you can use any units for $T$ and $a$ as long as they are consistent, so no unit conversion is needed, which saves significant time on MCQs.

4. Orbital Energy and Escape Velocity

For all bound orbits (circular or elliptical, $0\le e<1$), total mechanical energy is always negative, because gravitational potential energy is negative (we define $U=0$ at $r\to \infty$) and has twice the magnitude of the kinetic energy of the orbit.

For circular orbits, substitute $v^2=\frac{GM}{r}$ into kinetic energy: $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$. Gravitational potential energy is $U=-\frac{GMm}{r}$, so total energy is: $$E=K+U=-\frac{GMm}{2r}$$ This result generalizes to elliptical orbits by replacing $r$ with the semi-major axis $a$: $E=-\frac{GMm}{2a}$, so total energy only depends on the semi-major axis, not the eccentricity of the orbit.

An unbound orbit ($E\ge 0$) means the object can escape the central body's gravity entirely, reaching $r\to \infty$ with non-negative kinetic energy. Escape velocity $v_{esc}$ is the minimum speed needed at radius $r$ to escape, which occurs when total energy is exactly zero (the object has zero kinetic energy at infinity): $$\frac{1}{2}m{v}_{esc}^2-\frac{GMm}{r}=0⟹v_{esc}=\sqrt{\frac{2GM}{r}}=\sqrt{2}{v}_{circular}$$ Escape velocity is always larger than the speed of a circular orbit at the same radius.

Worked Example

What is the escape velocity from low Earth orbit, 400 km above Earth's surface? Earth's mass is $5.97\times 10^{24}$ kg, radius is $6.37\times 10^6$ m.

1. Calculate center-to-center radius: $r=R_E+h=6.37\times 10^6+4.00\times 10^5=6.77\times 10^6$ m.
2. Use the escape velocity formula $v_{esc}=\sqrt{\frac{2G{M}_E}{r}}$.
3. Substitute values: $\frac{2G{M}_E}{r}=\frac{2(6.67\times 10^{-11})(5.97\times 10^{24})}{6.77\times 10^6}\approx 1.18\times 10^8$.
4. Take the square root: $v_{esc}=\sqrt{1.18\times 10^8}\approx 10900$ m/s, or 10.9 km/s.

Exam tip: Escape velocity depends only on speed, not direction: any object with at least $v_{esc}$ will escape, regardless of direction, as long as it does not collide with the central body.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using height above the surface $h$ as $r$ in orbital speed or escape velocity calculations. Why: Problems often give altitude, and students confuse height above ground with center-to-center distance, which is what all gravitational formulas require. Correct move: Always write $r=R_{\text{central}}+h$ at the start of any problem where altitude is given, and explicitly confirm your value for $r$ before substituting.
• Wrong move: Claiming total mechanical energy is positive for a bound elliptical orbit. Why: Students remember kinetic energy is positive and forget gravitational potential energy is negative and has a larger magnitude for bound orbits. Correct move: For any closed, bound orbit ($e<1$), always remember $E=-\frac{GMm}{2a}<0$, so total energy is always negative.
• Wrong move: Using the proportional form of Kepler's third law ($T^2\propto a^3$) to compare periods of objects orbiting different central masses. Why: The proportionality only holds when the central mass $M$ is the same, since the constant of proportionality depends on $M$. Correct move: Only use the proportional form for objects orbiting the same central body; always use the full formula $T^2=\frac{4{\pi }^2{a}^3}{GM}$ when central masses differ.
• Wrong move: Using the semi-minor axis instead of semi-major axis in Kepler's third law or total energy calculations. Why: Students mix up the definitions of the two axes for ellipses. Correct move: Always use $a=\frac{r_p+r_a}{2}$ to get the semi-major axis if you are given periapsis and apoapsis distances, this avoids confusion about axis definitions.
• Wrong move: Writing escape velocity as $v_{esc}=\sqrt{\frac{GM}{r}}$, the same as circular orbital speed. Why: Students forget the factor of 2 from the energy derivation, mixing up the two formulas. Correct move: Always remember $v_{esc}=\sqrt{2}{v}_{\text{circular}}$ at the same radius, so escape velocity is always larger than circular orbital speed.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A satellite of mass $m$ is in a circular orbit of radius $r$ around a planet of mass $M$ ($M\gg m$). What is the total mechanical energy of the satellite in this orbit, with $U=0$ defined at $r\to \infty$? (A) $-\frac{GMm}{r}$ (B) $-\frac{GMm}{2r}$ (C) $\frac{GMm}{2r}$ (D) $\frac{GMm}{r}$

Worked Solution: For a circular orbit, kinetic energy is found from force balance: $K=\frac{1}{2}m{v}^2=\frac{1}{2}m\left(\frac{GM}{r}\right)=\frac{GMm}{2r}$. Gravitational potential energy is $U=-\frac{GMm}{r}$ with the standard zero-at-infinity convention. Total energy is $E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}$, which is negative as expected for a bound orbit. The correct answer is B.

Question 2 (Free Response)

A 1000 kg satellite is moved from a circular orbit of radius $2r_0$ to a circular orbit of radius $3r_0$ around Earth (mass $M$, $M\gg m$). (a) Derive an expression for the change in kinetic energy of the satellite in terms of $G$, $M$, $m$, and $r_0$. (b) Derive an expression for the change in total mechanical energy of the satellite. (c) Explain why energy must be added to the satellite to move it to a higher orbit, even though its orbital speed decreases.

Worked Solution: (a) Kinetic energy for a circular orbit is $K=\frac{GMm}{2r}$. The change is $\Delta K=K_f-K_i=\frac{GMm}{2(3r_0)}-\frac{GMm}{2(2r_0)}=\frac{GMm}{6r_0}-\frac{GMm}{4r_0}=-\frac{GMm}{12r_0}$. Kinetic energy decreases by $\frac{GMm}{12r_0}$. (b) Total energy for a circular orbit is $E=-\frac{GMm}{2r}$. The change is $\Delta E=E_f-E_i=-\frac{GMm}{2(3r_0)}-\left(-\frac{GMm}{2(2r_0)}\right)=-\frac{GMm}{6r_0}+\frac{GMm}{4r_0}=+\frac{GMm}{12r_0}$. Total energy increases by $\frac{GMm}{12r_0}$. (c) While kinetic energy decreases, gravitational potential energy increases by twice the magnitude of the kinetic energy decrease: $\Delta U=U_f-U_i=-\frac{GMm}{3r_0}-\left(-\frac{GMm}{2r_0}\right)=+\frac{GMm}{6r_0}=2|\Delta K|$, so the total change in energy is positive, requiring input energy from the satellite's engines.

Question 3 (Application / Real-World Style)

The dwarf planet Pluto has a mass of $1.30\times 10^{22}$ kg and radius of $1.19\times 10^6$ m. What is the escape velocity from the surface of Pluto? How does this compare to the speed of a typical amateur bullet (≈ 1000 m/s), and what does this mean for a hypothetical lander mission that wants to return sample from Pluto's surface?

Worked Solution: Escape velocity from the surface is $v_{esc}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\times 10^{-11})(1.30\times 10^{22})}{1.19\times 10^6}}\approx \sqrt{1.46\times 10^6}\approx 1200$ m/s. The escape velocity of Pluto (~1200 m/s) is only slightly higher than the speed of a typical amateur bullet (~1000 m/s). This means that even a small rocket with relatively low thrust can achieve the speed needed to escape Pluto's gravity for a sample return mission, requiring far less fuel than escaping a larger planet like Earth.

7. Quick Reference Cheatsheet

8. What's Next

This chapter lays the foundation for all orbital dynamics problems in AP Physics C: Mechanics, which are common multi-point FRQ topics that draw on multiple units. Immediately after mastering this topic, you will extend gravitational energy concepts to extended bodies and gravitational potential, and learn to solve problems involving orbital transfers and slingshot maneuvers that build directly on the force balance and energy relationships you learned here. Without understanding the core relationships for planetary and satellite orbits, these more complex problems will be impossible to solve correctly. This topic also connects to the broader course theme of central force motion and energy conservation, tying together earlier units on circular motion and angular momentum.

• Gravitational potential energy
• Conservation of angular momentum
• Uniform circular motion