Orbits of planets and satellites

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Gravitation· 14 min read

1. Circular Orbits and Centripetal Force Balance★★☆☆☆⏱ 4 min

📘 Definition

Orbit

$M$ = central mass, $m$ = orbiting mass, $r$ = center-to-center distance

Curved path of a celestial or artificial object held exclusively by gravitational attraction to a much more massive central body, assumed stationary because $M ≫ m$ .

Example:

An artificial satellite orbiting Earth, Earth orbiting the Sun

For a circular orbit, gravitational force from the central body provides exactly the centripetal force required to maintain constant speed along the circular path. We can ignore acceleration of the central body around the shared center of mass due to the large difference in mass.

🔬 Derivation

Goal:

Derive orbital speed and period for a circular orbit

Starting from:

Newton's second law for circular motion

1
Net gravitational force equals mass times centripetal acceleration:
2
$F_{net} = F_g = m a_c$
3
Substitute expressions for gravitational force and centripetal acceleration:
4
$\frac{G M m}{r^2} = \frac{m v^2}{r}$
5
Orbiting mass $m$ cancels out, giving the final expression for orbital speed:
6
$v = \sqrt{\frac{G M}{r}}$
7
Substitute $v = \frac{2 π r}{T}$ to solve for orbital period:
8
$T^2 = \frac{4 \pi^2 r^3}{G M}$

Result:

Orbital speed is independent of the orbiting body's mass, and decreases as orbital radius increases.

📐 Worked Example

A small artificial satellite orbits Mars at a height of 200 km above Mars' surface. Mars has a mass of $6.42 \times 1 0^{23}$ kg and radius $3.39 \times 1 0^{6}$ m. What is the satellite's orbital speed?

1
First calculate the center-to-center orbital radius by adding Mars' radius and the satellite's altitude:
2
$r = R_{\text{Mars}} + h = 3.39 \times 10^6 + 2.00 \times 10^5 = 3.59 \times 10^6 \text{ m}$
3
Use the force-balance result for circular orbital speed:
4
$v = \sqrt{\frac{G M}{r}}$
5
Substitute values to calculate the fraction under the square root:
6
$\frac{(6.67 \times 10^{-11})(6.42 \times 10^{23})}{3.59 \times 10^6} \approx 1.19 \times 10^7$
7
Take the square root to get final speed:
8
$v = \sqrt{1.19 \times 10^7} \approx 3450 \text{ m/s} = 3.45 \text{ km/s}$

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

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

Law of Orbits: All planets move in elliptical orbits with the central body at one focus of the ellipse. Eccentricity $e$ describes orbit shape: $e = 0$ = perfect circle, $0 < e < 1$ = bound elliptical orbit, $e \geq 1$ = unbound orbit. Closest distance (perihelion/perigee) is $r_{p} = a (1 - e)$ , farthest distance (aphelion/apogee) is $r_{a} = a (1 + e)$ , where $a$ = semi-major axis.
Law of Areas: A line joining the orbiting body and central body sweeps out equal areas in equal time intervals. This is a direct consequence of conservation of angular momentum: gravity exerts zero torque, so $L$ is constant, meaning speed is higher at smaller $r$ .
Law of Periods: The square of the orbital period is proportional to the cube of the semi-major axis: $T^{2} = \frac{4 π ^{2} a ^{3}}{GM}$ , which generalizes the circular orbit result (where $a = r$ ).

📐 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) Calculate semi-major axis from perihelion and aphelion distances:
2
$a = \frac{r_p + r_a}{2} = \frac{1 + 7}{2} = 4 \text{ AU}$
3
(b) Use Kepler's third law ratio for objects orbiting the same central mass:
4
$\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3} \implies \frac{(1 \text{ yr})^2}{(1 \text{ AU})^3} = \frac{T^2}{(4 \text{ AU})^3} \implies T^2 = 64 \implies T = 8 \text{ years}$
5
(c) Use conservation of angular momentum (velocity is perpendicular to radius at apsides):
6
$m r_p v_p = m r_a v_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{7}{1} = 7$

3. Orbital Energy and Escape Velocity★★★☆☆⏱ 3 min

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

🔬 Derivation

Goal:

Derive total orbital energy and escape velocity

1
For circular orbits, substitute $v^{2} = \frac{GM}{r}$ into kinetic energy:
2
$K = \frac{1}{2} m v^2 = \frac{G M m}{2 r}$
3
Add gravitational potential energy $U = - \frac{GM m}{r}$ to get total energy for circular orbits:
4
$E = K + U = - \frac{G M m}{2 r}$
5
For elliptical orbits, replace $r$ with the semi-major axis $a$ , so total energy only depends on $a$ , not eccentricity:
6
$E = - \frac{G M m}{2 a}$
7
Escape velocity is the minimum speed needed to escape gravity, with total energy zero at $r \to \infty$ :
8
$\frac{1}{2} m v_{esc}^2 - \frac{G M m}{r} = 0 \implies v_{esc} = \sqrt{\frac{2 G M}{r}} = \sqrt{2} v_{circular}$

Result:

Escape velocity is always larger than the speed of a circular orbit at the same radius, and is independent of the orbiting mass.

📐 Worked Example

What is the escape velocity from low Earth orbit, 400 km above Earth's surface? Earth's mass is $5.97 \times 1 0^{24}$ kg, radius is $6.37 \times 1 0^{6}$ m.

1
Calculate center-to-center radius:
2
$r = R_E + h = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6 \text{ m}$
3
Use the escape velocity formula:
4
$v_{esc} = \sqrt{\frac{2 G M_E}{r}}$
5
Substitute values:
6
$\frac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^6} \approx 1.18 \times 10^8$
7
Take the square root for final speed:
8
$v_{esc} = \sqrt{1.18 \times 10^8} \approx 10900 \text{ m/s} = 10.9 \text{ km/s}$

4. AP-Style Practice Worked Examples★★★★☆⏱ 3 min

📐 Worked Example

A satellite of mass $m$ is in a circular orbit of radius $r$ around a planet of mass $M$ ( $M ≫ m$ ). What is the total mechanical energy of the satellite in this orbit, with $U = 0$ defined at $r \to \infty$ ?

(A) $- \frac{GM m}{r}$ (B) $- \frac{GM m}{2 r}$ (C) $\frac{GM m}{2 r}$ (D) $\frac{GM m}{r}$

1
Calculate kinetic energy from force balance for circular orbits:
2
$K = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{G M m}{2 r}$
3
Gravitational potential energy with zero-at-infinity convention:
4
$U = - \frac{G M m}{r}$
5
Sum to get total energy:
6
$E = K + U = \frac{G M m}{2r} - \frac{G M m}{r} = - \frac{G M m}{2r}$
7
This is negative as expected for a bound orbit, so the correct answer is (B).

📐 Worked Example

A 1000 kg satellite is moved from a circular orbit of radius $2 r_{0}$ to a circular orbit of radius $3 r_{0}$ around Earth (mass $M$ , $M ≫ m$ ). (a) Derive an expression for the change in kinetic energy of the satellite. (b) Derive an expression for the change in total mechanical energy. (c) Explain why energy must be added to move to a higher orbit, even though orbital speed decreases.

1
(a) Kinetic energy for a circular orbit is $K = \frac{GM m}{2 r}$ . Calculate the change:
2
$\Delta K = K_f - K_i = \frac{G M m}{2(3 r_0)} - \frac{G M m}{2(2 r_0)} = - \frac{G M m}{12 r_0}$
3
Kinetic energy decreases by $\frac{GM m}{12 r _{0}}$ .
4
(b) Total energy for a circular orbit is $E = - \frac{GM m}{2 r}$ . Calculate the change:
5
$\Delta E = E_f - E_i = - \frac{G M m}{2(3 r_0)} - \left(- \frac{G M m}{2(2 r_0)}\right) = + \frac{G M m}{12 r_0}$
6
Total energy increases by $\frac{GM m}{12 r _{0}}$ .
7
(c) While kinetic energy decreases, gravitational potential energy increases by twice the magnitude of the kinetic energy decrease: $Δ U = + \frac{GM m}{6 r _{0}} = 2∣Δ K ∣$ . The total change in energy is positive, so energy input from the satellite's engines is required.

5. Common Pitfalls

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 the center-to-center distance required for all gravitational formulas.

Correct move:

Always write $r = R_{central} + h$ at the start of any problem where altitude is given, and 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 $E = - \frac{GM m}{2 a} < 0$ , 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 $T^{2} = \frac{4 π ^{2} a ^{3}}{GM}$ 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, which avoids confusion about axis definitions.

Wrong move:

Writing escape velocity as $v_{esc} = \frac{GM}{r}$ , 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 $v_{esc} = 2 v_{circular}$ at the same radius, so escape velocity is always larger than circular orbital speed.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Circular orbital speed	$v = \frac{GM}{r}$	Independent of orbiting mass $m$ ; $r$ = center-to-center distance
Kepler's Third Law (general)	$T^{2} = \frac{4 π ^{2} a ^{3}}{GM}$	$a$ = semi-major axis; applies to all bound orbits
Apsides radius	$r_{p} = a (1 - e), r_{a} = a (1 + e)$	$e$ = eccentricity; $e = 0$ = circle, $0 < e < 1$ = bound ellipse
Angular momentum at apsides	$r_{p} v_{p} = r_{a} v_{a}$	Holds because gravity exerts zero torque; $v ⊥ r$ at apsides
Total orbital energy	$E = - \frac{GM m}{2 a}$	Always negative for bound orbits; $a = r$ for circular orbits
Escape velocity	$v_{esc} = \frac{2 GM}{r}$	Minimum escape speed at distance $r$ ; $v_{esc} = 2 v_{circular}$
Kepler's Second Law	$\frac{d A}{d t} = \frac{L}{2 m} = constant$	Equal area in equal time; consequence of angular momentum conservation

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 · FRQ
Orbital energy change calculation
2022 · MCQ
Kepler's third law application
2021 · FRQ
Escape velocity derivation

What's Next

This topic lays the foundation for all orbital dynamics problems in AP Physics C: Mechanics, which are common multi-point FRQ topics that draw on concepts from multiple units. After mastering this core material, you will extend gravitational energy concepts to extended bodies and gravitational potential, and solve more complex problems involving orbital transfers and slingshot maneuvers that build directly on the force balance and energy relationships you learned here. 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. Mastery of these core relationships is required to solve more advanced gravitation problems correctly.