# Orbital Motion

> A-Level Physics · CIE 9702
> Source: https://www.owlsprep.com/study/cie-9702-u17-orbital-motion/

This subtopic covers the physics of objects moving in stable orbits around larger celestial bodies. You will learn how gravitational force provides required centripetal force, and calculate key orbital parameters for satellites and planets.

**Prerequisites:** [Newton's law of universal gravitation](https://www.owlsprep.com/study/cie-9702-u17-newtons-law-gravitation/); [Centripetal force for circular motion](https://www.owlsprep.com/study/cie-9702-u06-circular-motion/)

## Learning objectives

- Derive the relationship between orbital speed, radius and centripetal force for circular orbits
- Distinguish between different classes of satellite orbits
- Calculate orbital parameters for satellites and planets
- Explain the properties and uses of geostationary orbits

## Gravitational Force as Centripetal Force

For any object in a stable circular orbit around a central mass $M$, the only significant force acting on the orbiting body is gravitational attraction. This force acts towards the center of the orbit, exactly providing the centripetal force required to maintain uniform circular motion.

**Stable Circular Orbit** — An orbit where gravitational force between the two bodies exactly equals the centripetal force needed to keep the orbiting body moving at constant radius and speed

*Example:* Most artificial communication satellites move in near-perfect circular orbits around Earth.

**Worked example:** A 1500 kg satellite orbits Earth at constant radius. Earth mass $M = 6.0 \times 10^{24} \text{ kg}$, orbital radius $r = 7.0 \times 10^6 \text{ m}$. Show that the gravitational force equals approximately 12,000 N. Use $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

1. Write Newton's law of gravitation for the force between Earth and satellite:

   $$F = \frac{GMm}{r^2}$$
2. Substitute the given values into the equation:

   $$F = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(1500)}{(7.0 \times 10^6)^2}$$
3. Calculate numerator and denominator to get final force:

   $$F = \frac{6.003 \times 10^{17}}{4.9 \times 10^{13}} \approx 12250 \text{ N} \approx 12000 \text{ N}$$
4. This gravitational force acts towards Earth's center, providing the required centripetal force for orbit.

## Deriving Orbital Speed and Period

**Derivation:** Derive expressions for orbital speed $v$ and orbital period $T$ for a circular orbit

*Starting from:* Equating gravitational force to centripetal force

1. Start with the force balance for orbiting mass $m$:

   $$\frac{GMm}{r^2} = \frac{mv^2}{r}$$
2. Cancel $m$ (mass of orbiting body) and simplify $r$:

   $$\frac{GM}{r} = v^2$$
3. Rearrange for orbital speed $v$:

   $$v = \sqrt{\frac{GM}{r}}$$
4. Relate speed to period: $v = \frac{2\pi r}{T}$. Substitute $v$:

   $$\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}$$
5. Rearrange to get Kepler's third law for circular orbits:

   $$T^2 = \frac{4\pi^2 r^3}{GM}$$

*Conclusion:* Orbital speed and period depend only on the mass of the central body $M$ and orbital radius $r$, not the mass of the orbiting body.

**Worked example:** Calculate the orbital speed of a satellite orbiting Mars at radius $r = 3.4 \times 10^6 \text{ m}$. Mars mass $M = 6.4 \times 10^{23} \text{ kg}$, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

1. Use the derived orbital speed relation:

   $$v = \sqrt{\frac{GM}{r}}$$
2. Calculate the product $GM$:

   $$GM = (6.67 \times 10^{-11})(6.4 \times 10^{23}) = 4.2688 \times 10^{13}$$
3. Divide by orbital radius:

   $$\frac{GM}{r} = \frac{4.2688 \times 10^{13}}{3.4 \times 10^6} \approx 1.2555 \times 10^7$$
4. Take the square root to get final speed:

   $$v = \sqrt{1.2555 \times 10^7} \approx 3500 \text{ m s}^{-1}$$

## Types of Satellite Orbits

Satellites are placed in different orbits depending on their intended use. Two common classes for Earth observation and communication are low Earth orbits (LEO) and geosynchronous orbits, classified by altitude and inclination relative to the equator.

- **Low Earth Orbit (LEO):** Altitudes 160 km to 2000 km above Earth's surface, orbital periods ~90 minutes, used for imaging and Earth observation.
- **Medium Earth Orbit (MEO):** Altitudes 2000 km to 35786 km, used for navigation systems like GPS.
- **Geosynchronous Orbit (GEO):** Altitude ~35786 km above Earth's surface, orbital period equal to 24 hours.

**Check your understanding**

Check your understanding:

1. What happens to orbital speed as orbital radius increases?

   - It increases
   - It decreases
   - It stays the same
   - It depends on satellite mass

   *Answer:* It decreases

   *Why:* From $v = \sqrt{GM/r}$, increasing $r$ decreases $v$, so speed decreases.

**Worked example:** Compare the orbital speed of a LEO satellite ($r = 7.0 \times 10^6$ m) and GEO satellite ($r = 4.2 \times 10^7$ m) around Earth ($M = 6.0 \times 10^{24}$ kg).

1. Calculate LEO orbital speed:

   $$v_{LEO} = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{7.0 \times 10^6}} \approx 7600 \text{ m s}^{-1}$$
2. Calculate GEO orbital speed:

   $$v_{GEO} = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{4.2 \times 10^7}} \approx 3100 \text{ m s}^{-1}$$
3. Result confirms that higher orbits have lower orbital speed, matching the derived relation.

## Geostationary Orbits

**Geostationary Orbit** — A special type of geosynchronous orbit that is circular, equatorial, and has a period equal to Earth's rotational period (24 hours). A satellite in this orbit remains fixed above the same point on the equator.

*Notation:* GEO

Geostationary orbits are ideal for communication and weather satellites because ground-based antennae do not need to track the moving satellite — they can point permanently at the fixed satellite position.

**Worked example:** Calculate the radius of a geostationary orbit around Earth. $M = 6.0 \times 10^{24} \text{ kg}$, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

1. Convert 24 hour period to SI units (seconds):

   $$T = 24 \times 60 \times 60 = 86400 \text{ s}$$
2. Rearrange Kepler's third law to solve for $r^3$:

   $$r^3 = \frac{GM T^2}{4 \pi^2}$$
3. Substitute all values:

   $$r^3 = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(86400)^2}{4 \pi^2} \approx 7.54 \times 10^{22} \text{ m}^3$$
4. Take the cube root to get radius:

   $$r \approx 4.2 \times 10^7 \text{ m} = 42000 \text{ km}$$

> **Exam tip:** CIE often asks to compare geostationary and polar orbits: polar orbits are low altitude, pass over poles, have ~90 minute periods, and are used for Earth imaging.

## Common pitfalls

- **Wrong:** Using height above Earth's surface as orbital radius $r$ instead of distance from Earth's center.
  - Why it fails: Orbital radius is always measured from the center of the central mass. For LEO this error changes results by ~10%, much more for high orbits.
  - Correct: Add the radius of the central body to the surface height to get total orbital radius $r$.
- **Wrong:** Including the mass of the orbiting body when calculating orbital speed or period.
  - Why it fails: The mass of the orbiting body cancels out during derivation, so orbital parameters do not depend on it for small orbiting masses.
  - Correct: Use only the mass of the central body $M$ in all orbital calculations.
- **Wrong:** Confusing geostationary and geosynchronous orbits.
  - Why it fails: All geostationary orbits are geosynchronous, but not all geosynchronous orbits are geostationary.
  - Correct: Remember geostationary orbits are equatorial and stay fixed over one point; geosynchronous only has a 24-hour period.
- **Wrong:** Equating gravitational potential energy to centripetal force.
  - Why it fails: Only gravitational force, not energy, provides the centripetal force for orbit. Mixing force and energy gives wrong equations.
  - Correct: Always start by equating gravitational force to centripetal force for orbital motion problems.
- **Wrong:** Forgetting to convert orbital period from hours to seconds when using SI units.
  - Why it fails: All CIE calculations require SI units, so mismatched units give incorrect results by orders of magnitude.
  - Correct: Always convert time to seconds, distance to meters, and mass to kilograms before calculation.

## Cheatsheet

| Quantity | Formula | Key Notes |
| --- | --- | --- |
| Orbital speed | $v = \sqrt{\frac{GM}{r}}$ | Independent of orbiting mass $m$ |
| Orbital period | $T^2 = \frac{4\pi^2 r^3}{GM}$ | Use $T$ in seconds always |
| Geostationary period | $T = 86400 \text{ s}$ | Equals Earth rotation period |
| Geostationary radius | $r \approx 4.2 \times 10^7 \text{ m}$ | From Earth's center |
| Typical LEO period | $T \approx 90 \text{ minutes}$ | Standard exam value |

## What's next

Understanding orbital motion is a foundational concept for gravitational physics that connects directly to broader topics like gravitational potential energy and escape velocity, which you will explore next. This subtopic also reinforces your earlier understanding of circular motion, and often appears in combined exam questions that test both topics. Orbital motion questions are very common in both Paper 1 multiple choice and Paper 2 structured questions, so mastering the derivation of orbital speed and Kepler's third law is critical for exam success. The concepts here are also applied to planetary motion problems that frequently feature in CIE A-Level Physics exams.

- [Oscillations](https://www.owlsprep.com/study/cie-9702-u18-overview/)
- [Simple Harmonic Motion](https://www.owlsprep.com/study/cie-9702-u18-simple-harmonic-motion/)
- [Energy in SHM](https://www.owlsprep.com/study/cie-9702-u18-energy-in-shm/)

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