Orbital Motion

A-Level Physics· Unit 17: Gravitational fields, Subtopic 5: Orbital motion· 25 min read

1. Gravitational Force as Centripetal Force★★☆☆☆⏱ 5 min

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.

📘 Definition

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 $M = 6.0 \times 1 0^{24} kg$ , orbital radius $r = 7.0 \times 1 0^{6} m$ . Show that the gravitational force equals approximately 12,000 N. Use $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ .

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.

2. Deriving Orbital Speed and Period★★★☆☆⏱ 8 min

🔬 Derivation

Goal:

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: $v = \frac{2 π r}{T}$ . Substitute $v$ :
$\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}$
5
Rearrange to get Kepler's third law for circular orbits:
$T^2 = \frac{4\pi^2 r^3}{GM}$

Result:

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 $r = 3.4 \times 1 0^{6} m$ . Mars mass $M = 6.4 \times 1 0^{23} kg$ , $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ .

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

3. Types of Satellite Orbits★★☆☆☆⏱ 6 min

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.

✓ Quick check

Check your understanding:

What happens to orbital speed as orbital radius increases?
- It increases
- It decreases
- It stays the same
- It depends on satellite mass
Reveal answer
1 —
From $v = GM / r$ , increasing $r$ decreases $v$ , so speed decreases.

📐 Worked Example

Compare the orbital speed of a LEO satellite ( $r = 7.0 \times 1 0^{6}$ m) and GEO satellite ( $r = 4.2 \times 1 0^{7}$ m) around Earth ( $M = 6.0 \times 1 0^{24}$ 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.

4. Geostationary Orbits★★★☆☆⏱ 7 min

📘 Definition

Geostationary Orbit

GEO

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.

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. $M = 6.0 \times 1 0^{24} kg$ , $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ .

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.

5. Common Pitfalls

Wrong move:

Using height above Earth's surface as orbital radius $r$ instead of distance from Earth's center.

Why:

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 move:

Add the radius of the central body to the surface height to get total orbital radius $r$ .

Wrong move:

Including the mass of the orbiting body when calculating orbital speed or period.

Why:

The mass of the orbiting body cancels out during derivation, so orbital parameters do not depend on it for small orbiting masses.

Correct move:

Use only the mass of the central body $M$ in all orbital calculations.

Wrong move:

Confusing geostationary and geosynchronous orbits.

Why:

All geostationary orbits are geosynchronous, but not all geosynchronous orbits are geostationary.

Correct move:

Remember geostationary orbits are equatorial and stay fixed over one point; geosynchronous only has a 24-hour period.

Wrong move:

Equating gravitational potential energy to centripetal force.

Why:

Only gravitational force, not energy, provides the centripetal force for orbit. Mixing force and energy gives wrong equations.

Correct move:

Always start by equating gravitational force to centripetal force for orbital motion problems.

Wrong move:

Forgetting to convert orbital period from hours to seconds when using SI units.

Why:

All CIE calculations require SI units, so mismatched units give incorrect results by orders of magnitude.

Correct move:

Always convert time to seconds, distance to meters, and mass to kilograms before calculation.

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Orbital speed	$v = \frac{GM}{r}$	Independent of orbiting mass $m$
Orbital period	$T^{2} = \frac{4 π ^{2} r ^{3}}{GM}$	Use $T$ in seconds always
Geostationary period	$T = 86400 s$	Equals Earth rotation period
Geostationary radius	$r \approx 4.2 \times 1 0^{7} m$	From Earth's center
Typical LEO period	$T \approx 90 minutes$	Standard exam value

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.

2022 · 22
Calculate geostationary orbit radius
2023 · 13
Compare orbital speeds of two satellites
2021 · 21
Derive orbital period relation

Going deeper

syllabusCIE 9702 Official SyllabusSection 17.3 Orbital motion

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.