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Derive $v = \sqrt{GM/r}$ for circular orbits

Apply Kepler's Third Law $T^2 \propto r^3$

Analyse geostationary orbits and their applications

Think First

ESTIMATE — The ISS orbits Earth at approximately 400 km altitude. Estimate its orbital period in minutes. Think about what you already know about low Earth orbit.

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Year 12 Physics Module 5: Advanced Mechanics 45 min Lesson 11 of 18 IQ2: Circular Motion

🛰️ Gravitational Orbits

Derive orbital velocity for satellites, apply Kepler's Third Law to planetary motion, and analyse geostationary orbits and their applications.

Deriving Orbital Velocity

The balance between gravity and centripetal motion

For a satellite in stable circular orbit, the only force acting is gravity — and this gravitational force provides exactly the centripetal force required.

Gravitational Orbits

∑

Equating forces

$$F_{\text{gravity}} = F_{\text{centripetal}}$$

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

$$v = \sqrt{\frac{GM}{r}}$$

r = centre-to-centre distance = R_planet + h

The satellite mass $m$ cancels out — orbital speed depends only on the planet's mass $M$ and the orbital radius $r$. A feather and a space station at the same altitude orbit at exactly the same speed (neglecting atmospheric drag).

Key point: Always state $r = R_{\text{Earth}} + h$ explicitly. The orbital radius is measured from the centre of the central body, not from its surface.

Worked Example — ISS Orbital Speed

Calculate the orbital speed of the International Space Station orbiting at 400 km altitude above Earth.

GIVEN

$h = 400 \times 10^3 \text{ m} = 0.400 \times 10^6 \text{ m}$, $R_E = 6.371 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$

FIND

Orbital speed $v$

METHOD

Use $r = R_E + h$, then $v = \sqrt{GM/r}$

$$r = 6.371 \times 10^6 \text{ m} + 0.400 \times 10^6 \text{ m} = 6.771 \times 10^6 \text{ m}$$

$$v = \sqrt{\frac{6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg}}{6.771 \times 10^6 \text{ m}}}$$

$$v = \sqrt{5.88 \times 10^7 \text{ m}^2\text{/s}^2} = 7.67 \times 10^3 \text{ m/s}$$

ANSWER: $v = 7.67 \text{ km/s}$ (approximately 27,600 km/h)

Kepler's Third Law from Newton's Law

Relating orbital period to orbital radius

Newton showed that Kepler's empirical Third Law follows directly from his law of universal gravitation combined with circular motion.

Starting again with gravitational force providing centripetal force, but this time expressing velocity in terms of period:

∑

Derivation

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

$$v = \frac{2\pi r}{T} \quad \Rightarrow \quad v^2 = \frac{4\pi^2 r^2}{T^2}$$

$$\frac{GMm}{r^2} = m \cdot \frac{4\pi^2 r^2}{T^2} \cdot \frac{1}{r}$$

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

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

$$\frac{T^2}{r^3} = \frac{4\pi^2}{GM} = \text{constant for any central body}$$

This means for all satellites orbiting the same central body, the ratio $T^2/r^3$ is identical. This is why we can compare the orbits of moons around Jupiter, or planets around the Sun, using a single constant.

HSC Tip: When doing Kepler's Third Law calculations, ensure your period is in seconds and radius in metres before computing $T^2/r^3$.

Worked Example — Verifying Kepler's Third Law for Mars

Mars has an orbital period of 687 days and orbits at an average distance of $2.28 \times 10^{11}$ m from the Sun. Verify that $T^2/r^3 \approx 2.97 \times 10^{-19} \text{ s}^2\text{/m}^3$.

GIVEN

$T = 687 \text{ days}$, $r = 2.28 \times 10^{11} \text{ m}$

FIND

Verify $T^2/r^3$

METHOD

Convert period to seconds, then compute $T^2/r^3$

$$T = 687 \times 24 \times 3600 \text{ s} = 5.936 \times 10^7 \text{ s}$$

$$T^2 = (5.936 \times 10^7)^2 = 3.524 \times 10^{15} \text{ s}^2$$

$$r^3 = (2.28 \times 10^{11})^3 = 1.186 \times 10^{34} \text{ m}^3$$

$$\frac{T^2}{r^3} = \frac{3.524 \times 10^{15} \text{ s}^2}{1.186 \times 10^{34} \text{ m}^3} = 2.97 \times 10^{-19} \text{ s}^2\text{/m}^3$$

ANSWER: $\dfrac{T^2}{r^3} = 2.97 \times 10^{-19} \text{ s}^2\text{/m}^3$ ✓ This constant applies to all bodies orbiting the Sun.

Geostationary Orbits

Satellites that stay fixed above one point on Earth

A geostationary satellite has an orbital period exactly equal to Earth's rotation period: 24 hours = 86,400 s. This means it appears to hover stationary above a fixed point on the equator — an incredibly useful property for communications and weather monitoring.

Finding the Geostationary Radius

We can calculate the required orbital radius by rearranging Kepler's Third Law with $T = 86{,}400 \text{ s}$:

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Geostationary radius derivation

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

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

$$r = \sqrt[3]{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2}}$$

$$r = 4.22 \times 10^7 \text{ m} = 42{,}200 \text{ km from Earth's centre}$$

The altitude above Earth's surface is:

$$h = r - R_E = 4.22 \times 10^7 \text{ m} - 6.371 \times 10^6 \text{ m} = 3.59 \times 10^7 \text{ m} \approx 35{,}900 \text{ km}$$

Critical requirement: All geostationary satellites must orbit at exactly 35,900 km altitude, directly above the equator, moving in the same direction as Earth's rotation. Any other orbit (e.g., inclined or polar) will cause the satellite to appear to drift north-south or east-west in the sky.

Applications of Geostationary Orbits

Communications: TV broadcast, telephone relays, internet backbone — ground antennas can remain fixed, pointing at one location in the sky
Weather monitoring: Geostationary weather satellites provide continuous coverage of the same region, enabling real-time storm tracking
Navigation aids: Some regional navigation systems use geostationary satellites

Limitation: Geostationary satellites cannot cover polar regions well (they sit on the equatorial plane, so visibility degrades at high latitudes). They also suffer from high signal latency due to the large distance — a signal round-trip takes approximately 240 ms.

∑

Key Formulas — Gravitational Orbits

$$v = \sqrt{\frac{GM}{r}}$$

Orbital velocity (m/s). $r = R + h$ from centre of central body.

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

Kepler's Third Law. $T^2/r^3$ = constant for all bodies orbiting the same central mass.

$$r = R + h$$

Centre-to-centre orbital radius. Always add planet radius to altitude.

$$T_{\text{geostationary}} = 24 \text{ h} = 86400 \text{ s}$$

Period matches Earth's rotation.

$$r_{\text{geostationary}} \approx 42{,}200 \text{ km} \text{ (from centre)}$$

$\approx 35{,}900 \text{ km}$ altitude above Earth's surface.

Common Misconceptions

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"Satellites need fuel to maintain orbit"

In a stable circular orbit in vacuum, no fuel is needed. Gravity alone provides the centripetal force continuously. Fuel is only needed for orbital adjustments or to counteract atmospheric drag at very low altitudes.

✗

"Heavier satellites orbit slower"

$v = \sqrt{GM/r}$ — the satellite's own mass $m$ cancels out of the equation entirely. Orbital speed depends only on the central body's mass and the orbital radius, not on the satellite's mass.

✗

"The Moon doesn't fall toward Earth because there's no gravity in space"

The Moon is falling toward Earth continuously. Its tangential velocity means it keeps "missing" the Earth — and that's exactly what an orbit is. There is significant gravity at the Moon's distance (~0.003 m/s²).

Real World: GPS Satellites

GPS satellites orbit at approximately 20,200 km altitude with a period of 11 hours 58 minutes — exactly half a sidereal day. This is not geostationary; they are in medium Earth orbit (MEO).

At least 4 satellites must be visible from any point on Earth for trilateration to determine position (latitude, longitude, and altitude). The GPS constellation has 31 operational satellites distributed across 6 orbital planes, ensuring global coverage at all times.

Each GPS satellite carries atomic clocks accurate to within nanoseconds. This precision is essential because light travels approximately 30 cm in 1 nanosecond — so every nanosecond of timing error becomes roughly 30 cm of positioning error. This is why GPS is a remarkable application of both orbital mechanics and special/general relativity (time dilation at orbital speeds and altitudes must be corrected for).

Activities

Q1. Calculate the orbital speed of a satellite orbiting Earth at 600 km altitude.

[2 marks]

Q2. Find the orbital period of a satellite orbiting at $r = 8.0 \times 10^6$ m from Earth's centre.

[2 marks]

Q3. Explain why geostationary satellites must orbit above the equator, not at some other latitude.

[2 marks]

Copy Into Books

Show notes to copy

Orbital velocity derivation:

$$F_g = F_c \Rightarrow \frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}}$$

where $r = R + h$ (centre-to-centre distance)

Kepler's Third Law:

$$T^2 = \frac{4\pi^2}{GM} r^3 \quad \text{or} \quad \frac{T^2}{r^3} = \text{constant}$$

Geostationary orbit:

$T = 24$ h = 86,400 s, $r \approx 42{,}200$ km from centre, $h \approx 35{,}900$ km altitude, must be equatorial

Revisit Your Estimate

Now that you've learned the method, revisit your initial estimate. The ISS orbits at $r = 6.771 \times 10^6$ m with speed $v = 7.67$ km/s. Using $T = 2\pi r / v$, the period is approximately 92 minutes. How close was your estimate?

Autosaved

Interactive: Orbit Calculator Interactive

Interactive: Orbital Mechanics Simulator

Key Terms

Orbital velocity

$v = \sqrt{GM/r}$ — speed needed for a stable circular orbit at radius $r$

Kepler's Third Law

$T^2 \propto r^3$ — the square of the period is proportional to the cube of the orbital radius

Geostationary orbit

Orbit with period = 24 h, altitude ~35,900 km, above the equator; satellite appears stationary

Centre-to-centre distance

$r = R + h$ — orbital radius measured from the centre of the central body

MEO / LEO

Medium Earth Orbit (e.g., GPS at 20,200 km) and Low Earth Orbit (e.g., ISS at 400 km)

Multiple Choice

1. The orbital speed of a satellite depends on:

A Its mass Incorrect — the satellite mass cancels out in the derivation.

B The planet's mass and orbital radius Correct! $v = \sqrt{GM/r}$ depends on $M$ (planet mass) and $r$ only.

C Its speed of launch Incorrect — once in stable orbit, the speed is determined by $r$ and $M$.

D Both masses Incorrect — only the central body's mass $M$ appears in the formula.

2. A geostationary satellite must have:

A Period = 12 hours Incorrect — 12 hours would make it appear to move across the sky.

B Period = 24 hours and equatorial orbit Correct! Both conditions are required for the satellite to appear stationary.

C Period = 48 hours Incorrect — the period must match Earth's rotation: 24 hours.

D Polar orbit Incorrect — polar orbits sweep over all latitudes; the satellite would move, not stay fixed.

3. If a satellite's orbital radius doubles, its period:

A Stays the same Incorrect — $T$ definitely changes with $r$.

B Doubles Incorrect — $T \propto r^{3/2}$, not $T \propto r$.

C Increases by $2\sqrt{2} \approx 2.83$ times Correct! From $T^2 \propto r^3$: if $r \to 2r$, then $T \to 2^{3/2}T = 2.83T$.

D Increases by $2^{3/2} = 2.83$ times The numerical value is right, but check the expression — C and D have the same value, but the question may be testing your reasoning. The key insight is $T \propto r^{3/2}$.

4. The centripetal force on an orbiting satellite is provided by:

A Its engines Incorrect — engines are not needed in a stable orbit (no fuel required).

B Gravitational attraction Correct! Gravity provides the centripetal force: $F_g = F_c$.

C Centrifugal force Incorrect — centrifugal force is a fictitious force in a rotating reference frame, not a real force.

D Solar radiation pressure Incorrect — solar radiation pressure is negligible compared to gravity for most satellites.

5. Two satellites orbit Earth at different radii. The one further out has:

A Higher speed Incorrect — $v = \sqrt{GM/r}$; larger $r$ means smaller $v$.

B Lower speed Correct! $v \propto 1/\sqrt{r}$ — the further out, the slower the satellite moves.

C Same speed Incorrect — speed depends on orbital radius; different radii mean different speeds.

D Shorter period Incorrect — larger orbits have longer periods ($T \propto r^{3/2}$).

Short Answer

Q1. The Hubble Space Telescope orbits at 547 km altitude. Calculate its orbital speed and period.

($G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_{\text{Earth}} = 5.97 \times 10^{24} \text{ kg}$, $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$)

[3 marks] Apply Band 4-5

Q2. Use Kepler's Third Law to find the orbital radius of a satellite with period 8.0 hours around Earth.

[3 marks] Apply Band 5-6

Q3. Assess the advantages and limitations of geostationary orbits for communication satellites compared to low Earth orbit (LEO) and medium Earth orbit (MEO) satellites.

[4 marks] Evaluate Band 6

Model Answers

Q1. Hubble Space Telescope — 3 marks

Step 1 — Find orbital radius (centre-to-centre):

$$r = R_E + h = 6.37 \times 10^6 \text{ m} + 5.47 \times 10^5 \text{ m} = 6.917 \times 10^6 \text{ m}$$

Step 2 — Calculate orbital speed:

$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.917 \times 10^6}} = 7.59 \times 10^3 \text{ m/s} = 7.59 \text{ km/s}$$

Step 3 — Calculate period:

$$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.917 \times 10^6}{7.59 \times 10^3} = 5726 \text{ s} = 95.4 \text{ min}$$

Award 1 mark for correct radius, 1 mark for correct speed, 1 mark for correct period with unit conversion.

Q2. Satellite with 8.0 h period — 3 marks

Step 1 — Convert period to seconds:

$$T = 8.0 \times 3600 = 28{,}800 \text{ s}$$

Step 2 — Use Kepler's Third Law rearranged:

$$r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (28800)^2}{4\pi^2} = 8.32 \times 10^{21} \text{ m}^3$$

Step 3 — Solve for r:

$$r = \sqrt[3]{8.32 \times 10^{21}} = 2.03 \times 10^7 \text{ m} = 20{,}300 \text{ km}$$

Altitude: $h = 2.03 \times 10^7 - 6.37 \times 10^6 = 1.39 \times 10^7 \text{ m} \approx 13{,}900 \text{ km}$

Award 1 mark for period conversion, 1 mark for correct substitution, 1 mark for final answer.

Q3. Assessment of geostationary vs LEO/MEO — 4 marks

Geostationary (GEO) — advantages:

Satellite appears fixed relative to ground — simple, cheap ground antennas that don't need tracking systems
Continuous coverage of one region — ideal for TV broadcast and regional communications
Only 3 satellites needed for near-global coverage

Geostationary — limitations:

High latency (~240 ms round-trip) — problematic for real-time applications like video calls and online gaming
Must orbit above equator — cannot provide good coverage for polar regions
Expensive to launch to high altitude (~35,900 km)
Weaker signal strength at greater distance requires more powerful transmitters

LEO (e.g., Starlink at 550 km): Low latency (~20 ms), strong signals, polar coverage possible — but requires thousands of satellites for continuous coverage, complex ground tracking needed.

MEO (e.g., GPS at 20,200 km): Compromise latency and coverage — moderate latency, fewer satellites needed than LEO, but still requires tracking and has higher latency than LEO.

Award marks for identifying at least 2 GEO advantages, 2 GEO limitations, and comparison with LEO/MEO. Must use "assess" language (weighing pros and cons).

Boss Battle

🎯 Orbital Mechanics Boss Battle

Test your orbital velocity and period calculation skills under pressure!

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Boss Battle: Gravitational Orbits

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