# Orbits of planets and satellites

> AP Physics C: Mechanics · Unit 7: Gravitation
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u7-orbits-of-planets-and-satellites/

This module covers circular and elliptical orbits of planets and satellites, Kepler's three laws, gravitational centripetal force balance, orbital energy, escape velocity, and bound/unbound orbit classification for AP Physics C: Mechanics.

**Prerequisites:** Newton's law of universal gravitation; [Uniform circular motion and centripetal acceleration](https://www.owlsprep.com/study/ap-physics-c-mech-u3-uniform-circular-motion/); [Conservation of mechanical energy and angular momentum](https://www.owlsprep.com/study/ap-physics-c-mech-u5-conservation-angular-momentum/)

## Learning objectives

- Apply centripetal force balance to calculate circular orbital speed and period
- Use Kepler's three laws to solve elliptical orbit problems
- Calculate total orbital energy and escape velocity for bound and unbound orbits
- Classify orbits based on total energy and eccentricity

## Circular Orbits and Centripetal Force Balance

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

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

*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:** 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 $v = \frac{2 \pi r}{T}$ to solve for orbital period:
8. $$T^2 = \frac{4 \pi^2 r^3}{G M}$$

*Conclusion:* 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 $6.42 \times 10^{23}$ 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 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. $$v = \sqrt{1.19 \times 10^7} \approx 3450 \text{ m/s} = 3.45 \text{ km/s}$$

> **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 substituting into any formula.

## 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 for two-body orbits with a dominant central mass.

1. **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.
2. **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 $r$.
3. **Law of Periods**: The square of the orbital period is proportional to the cube of the semi-major axis: $T^2 = \frac{4 \pi^2 a^3}{G M}$, 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. $$m r_p v_p = m r_a v_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{7}{1} = 7$$

> **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, saving time on MCQs.

## Orbital Energy and Escape Velocity

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:** Derive total orbital energy and escape velocity

1. For circular orbits, substitute $v^2 = \frac{G M}{r}$ into kinetic energy:
2. $$K = \frac{1}{2} m v^2 = \frac{G M m}{2 r}$$
3. Add gravitational potential energy $U = -\frac{G M m}{r}$ 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. $$\frac{1}{2} m v_{esc}^2 - \frac{G M m}{r} = 0 \implies v_{esc} = \sqrt{\frac{2 G M}{r}} = \sqrt{2} v_{circular}$$

*Conclusion:* 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 $5.97 \times 10^{24}$ kg, radius is $6.37 \times 10^6$ 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. $$v_{esc} = \sqrt{1.18 \times 10^8} \approx 10900 \text{ m/s} = 10.9 \text{ km/s}$$

> **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.

## AP-Style Practice Worked Examples

**Worked example:** 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) $- \frac{G M m}{r}$
(B) $- \frac{G M m}{2 r}$
(C) $\frac{G M m}{2 r}$
(D) $\frac{G M m}{r}$

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 $M$, $M \gg m$). (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 $K = \frac{G M m}{2 r}$. 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 $\frac{G M m}{12 r_0}$.
4. (b) Total energy for a circular orbit is $E = - \frac{G M m}{2 r}$. 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 $\frac{G M m}{12 r_0}$.
7. (c) While kinetic energy decreases, gravitational potential energy increases by twice the magnitude of the kinetic energy decrease: $\Delta U = + \frac{G M m}{6 r_0} = 2 |\Delta K|$. The total change in energy is positive, so energy input from the satellite's engines is required.

## Common pitfalls

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

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Circular orbital speed | $v = \sqrt{\frac{G M}{r}}$ | Independent of orbiting mass $m$; $r$ = center-to-center distance |
| Kepler's Third Law (general) | $T^2 = \frac{4 \pi^2 a^3}{G M}$ | $a$ = 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; $v \perp r$ at apsides |
| Total orbital energy | $E = - \frac{G M m}{2 a}$ | Always negative for bound orbits; $a=r$ for circular orbits |
| Escape velocity | $v_{esc} = \sqrt{\frac{2 G M}{r}}$ | Minimum escape speed at distance $r$; $v_{esc} = \sqrt{2} v_{\text{circular}}$ |
| Kepler's Second Law | $\frac{dA}{dt} = \frac{L}{2 m} = \text{constant}$ | Equal area in equal time; consequence of angular momentum conservation |

## 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.

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