Orbital Motion of Planets and Satellites

AP Physics 1· AP Physics 1 CED — Circular Motion and Gravitation· 14 min read

1. Gravity as Centripetal Force for Circular Orbits★★☆☆☆⏱ 4 min

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For any uniform circular orbit around a stationary central mass (we assume the central mass is far larger than the orbiting mass, so it does not accelerate significantly), the only force acting on the orbiting body is gravitational attraction from the central body, which points directly toward the center of the orbit. This means gravity is exactly the centripetal force required to keep the object moving in a circle.

G \frac{M m}{r^2} = m \frac{v^2}{r} = m \frac{4\pi^2 r}{T^2}

Where $M$ = mass of the central body, $m$ = mass of the orbiting body, $r$ = orbital radius (distance between centers of mass), $G$ = universal gravitational constant, $v$ = orbital speed, and $T$ = orbital period. A key result emerges immediately: the orbiting mass $m$ cancels out of the equation entirely. This means orbital speed and period do not depend on the mass of the satellite/planet, only on the central mass and orbital radius. Simplifying gives two core formulas:

v = \sqrt{\frac{GM}{r}} \quad \text{and} \quad T^2 = \frac{4\pi^2 r^3}{GM}

📐 Worked Example

A 500 kg artificial satellite is placed into a circular orbit 400 km above Earth’s surface. Earth has a mass of $6.0 \times 1 0^{24}$ kg, and Earth’s radius is 6400 km. What is the orbital speed of the satellite?

1
First calculate the total orbital radius $r$ , measured from Earth’s center to the satellite:
$r = R_{\text{Earth}} + h = 6400\ \text{km} + 400\ \text{km} = 6800\ \text{km} = 6.8 \times 10^6\ \text{m}$
2
Recall that orbital speed is $v = \frac{GM}{r}$ , and note that the satellite mass of 500 kg cancels out, so it is not needed for the calculation.
3
Substitute values for $GM$ :
$GM = (6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2)(6.0 \times 10^{24}\ \text{kg}) \approx 4.0 \times 10^{14}\ \text{m}^3/\text{s}^2$
4
Simplify to find orbital speed:
$v = \sqrt{\frac{4.0 \times 10^{14}}{6.8 \times 10^6}} \approx \sqrt{5.88 \times 10^7} \approx 7.7 \times 10^3\ \text{m/s}$

Exam tip:

Always add the radius of the central body to the satellite’s height above the surface to get the correct orbital radius $r$ ; forgetting this step is the most common error on these problems.

2. Kepler's Three Laws of Planetary Motion★★★☆☆⏱ 4 min

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Kepler derived three empirical laws of planetary motion from observational data decades before Newton developed his law of universal gravitation. Newton later confirmed that Kepler’s laws are a natural consequence of inverse-square gravitational attraction, and they remain core conceptual tools for AP Physics 1. AP Physics 1 almost exclusively tests uniform circular orbits for calculation, while elliptical orbits are only addressed qualitatively via Kepler’s laws, with no requirement to calculate eccentricity or other ellipse parameters.

Law of Orbits: All planets move in elliptical orbits with the Sun at one of the two foci. A circular orbit is just a special case of an ellipse where both foci overlap at the center.
Law of Equal Areas: A line connecting a planet to the Sun sweeps out equal areas in equal time intervals. This comes from conservation of angular momentum, so planets move faster at perihelion (closest point to the Sun) and slower at aphelion (farthest point).
Law of Periods: The square of an orbit’s period is proportional to the cube of the orbit’s semi-major axis. For circular orbits, the semi-major axis equals the orbital radius $r$ , so $T^{2} \propto r^{3}$ . For all objects orbiting the same central mass, this gives the ratio $\frac{T _{1}^{2}}{r _{1}^{3}} = \frac{T _{2}^{2}}{r _{2}^{3}}$ .

📐 Worked Example

Earth orbits the Sun with a period of 1 year and an average orbital radius of 1 AU (astronomical unit). The dwarf planet Ceres has an average orbital radius of 2.77 AU. What is Ceres’ orbital period in years?

1
Both objects orbit the same central mass (the Sun), so Kepler’s third law ratio applies:
$\frac{T_E^2}{r_E^3} = \frac{T_C^2}{r_C^3}$
2
Rearrange to solve for $T_{C}$ :
$T_C = T_E \left(\frac{r_C}{r_E}\right)^{3/2}$
3
Substitute values: $T_{E} = 1 yr$ , $\frac{r _{C}}{r _{E}} = 2.77$ , so $T_{C} = (2.77)^{3/2} = 2.77 \times 2.77$
4
Calculate and get the final period:
$T_C \approx 2.77 \times 1.66 \approx 4.6\ \text{years}$

Exam tip:

For ratio problems using Kepler’s third law, you can keep units like AU and years as long as they are consistent for both objects; no unit conversion is needed, which saves valuable time on MCQs.

3. Energy in Orbital Motion★★★★☆⏱ 3 min

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For any bound circular orbit, we can calculate the total mechanical energy as the sum of kinetic energy $K$ and gravitational potential energy $U$ , with the standard convention that $U = 0$ at infinite distance from the central body. We know $K = \frac{1}{2} m v^{2}$ and $U = - G \frac{M m}{r}$ .

Substitute $v^{2} = \frac{GM}{r}$ into the kinetic energy formula: $K = \frac{1}{2} m (\frac{GM}{r}) = \frac{GM m}{2 r}$ . This gives the useful relationship that $K = - \frac{1}{2} U$ for any circular orbit. Adding kinetic and potential energy gives total energy:

E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}

Key conceptual results: (1) Total energy is negative for bound orbits, meaning the orbiting body does not have enough energy to escape the central body’s gravity. (2) If total energy is zero or positive, the orbit is unbound, and the object will escape, never returning. (3) When a satellite moves to a higher orbit (larger $r$ ), total energy increases (becomes less negative), even though orbital speed and kinetic energy decrease, because potential energy increases more than kinetic energy decreases.

📐 Worked Example

A satellite is moved from a circular orbit of radius $r$ to a new circular orbit of radius $2 r$ . How does the total mechanical energy of the satellite change?

1
Write the formula for total energy of a circular orbit:
$E = -\frac{GMm}{2r}$
2
Calculate initial energy:
$E_1 = -\frac{GMm}{2r}$
3
Calculate final energy for the new orbit:
$E_2 = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}$
4
Compare the two values: $- \frac{GM m}{4 r}$ is greater than $- \frac{GM m}{2 r}$ , because it is less negative.
5
Conclusion: The total mechanical energy of the satellite increases by $\frac{GM m}{4 r}$ .

Exam tip:

Never drop the negative sign for gravitational potential energy or total orbital energy; the sign is what distinguishes bound orbits from unbound escape trajectories.

4. AP-Style Concept Check★★★☆☆⏱ 3 min

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Test your understanding with these AP-style practice questions:

Two satellites, Satellite A of mass $m$ and Satellite B of mass $2 m$ , orbit Earth in circular orbits of the same radius $r$ . Which of the following correctly compares their orbital speeds and periods?
- A. $v_{A} = v_{B}$ and $T_{A} = T_{B}$
- B. $v_{A} < v_{B}$ and $T_{A} < T_{B}$
- C. $v_{A} > v_{B}$ and $T_{A} < T_{B}$
- D. $v_{A} = v_{B}$ and $T_{A} < T_{B}$
Reveal answer
A —
Orbiting mass cancels out of all orbital speed and period formulas, so equal radius gives equal speed and equal period, regardless of satellite mass.

5. Common Pitfalls

Wrong move:

Using only the height of the satellite above the surface as the orbital radius $r$ , instead of adding the central body's radius.

Why:

Problems almost always state height above the surface, not distance from the center, so students forget the definition of orbital radius.

Correct move:

Always confirm whether the given distance is from the center or the surface; if it is from the surface, add the central body's radius before plugging into any orbital formula.

Wrong move:

Trying to use the orbiting mass $m$ to solve for orbital speed or period when it is not given.

Why:

Students forget to cancel $m$ during the force balance, because they are used to keeping all variables in Newton's second law.

Correct move:

Always cancel the orbiting mass early in the derivation; orbital speed and period never depend on the orbiting mass for circular orbits.

Wrong move:

Claiming a higher orbit satellite has higher orbital speed and more kinetic energy than a lower orbit satellite.

Why:

Students confuse total energy with kinetic energy; higher orbits have higher total energy, leading to the incorrect assumption that kinetic energy is also higher.

Correct move:

Remember $v = GM / r$ : as $r$ increases, $v$ decreases, so kinetic energy decreases even as total energy increases.

Wrong move:

Claiming astronauts in orbit are weightless because there is no gravity in orbit.

Why:

Popular media often repeats this misinformation, so students internalize the wrong explanation.

Correct move:

Explain that astronauts are weightless because they are in constant free fall alongside their spacecraft; gravity still provides the centripetal force to keep them in orbit.

Wrong move:

Forgetting Kepler's second law applies to elliptical orbits only, and claiming speed varies in a circular orbit.

Why:

Students generalize the equal area rule to circular orbits where distance from the center is constant.

Correct move:

Remember that in a circular orbit, distance from the center is constant, so speed is constant, in line with both Kepler's laws and uniform circular motion rules.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Gravitational Centripetal Force Balance	$G \frac{M m}{r ^{2}} = m \frac{v ^{2}}{r} = m \frac{4 π ^{2} r}{T ^{2}}$	Only for uniform circular orbits; $M$ = central mass, $r$ = orbital radius from center
Orbital Speed	$v = \frac{GM}{r}$	Independent of orbiting mass; $v$ decreases as $r$ increases
Kepler's Third Law (Ratio Form)	$\frac{T _{1}^{2}}{r _{1}^{3}} = \frac{T _{2}^{2}}{r _{2}^{3}}$	Only for objects orbiting the same central mass; any consistent units work
Kepler's Third Law (Quantitative)	$T^{2} = \frac{4 π ^{2} r ^{3}}{GM}$	Gives $T$ in seconds for SI units of $r$ , $G$ , $M$
Kinetic Energy (Circular Orbit)	$K = \frac{GM m}{2 r}$	Equal to $- U /2$ , where $U$ is gravitational potential energy
Gravitational Potential Energy	$U = - \frac{GM m}{r}$	Zero potential defined at infinity; negative for all bound orbits
Total Mechanical Energy	$E = - \frac{GM m}{2 r}$	Negative = bound orbit, $\geq 0$ = unbound (escape) orbit
Kepler's Second Law	Equal area swept in equal time	Implies faster speed at smaller orbital distances
Kepler's First Law	Orbits are ellipses with central body at one focus	Circular orbits are a special case of ellipses

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Compare satellite orbital speeds
2022 · FRQ
Kepler third law ratio problem

What's Next

This topic lays the foundational framework for understanding all gravitational motion, which connects directly to subsequent topics in AP Physics 1. Immediately after mastering orbital motion in Unit 3, you will move to torque and rotational motion in Unit 4, where you will extend the conservation of angular momentum you used qualitatively for Kepler's second law to quantitative problems of rotating systems. Without mastering the relationship between gravity and centripetal force here, you will struggle to connect inverse-square forces to circular motion in rotational dynamics and energy conservation problems later in the course. Orbital motion also reinforces the core theme of energy conservation, which is tested across all units of AP Physics 1.