Year 12 Physics Module 5 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 11 of 18

Gravitational Orbits

A satellite stays in orbit because gravity continuously turns its velocity toward Earth. In this lesson, a circular-orbit model connects gravitational force to orbital speed, period and the practical differences between LEO, MEO and GEO.

Today's hook: The ISS completes an orbit in about 93 minutes even though Earth's gravity at that altitude is still strong. Why do its occupants feel weightless, and how can the same gravitational force determine both its speed and period?
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Connect and orient

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Two printable worksheets that build from foundations to exam-style application.

Build Foundations & guided practice Apply Application & data response
Before you read, predict

The ISS orbits Earth at approximately 400 km altitude. Estimate its orbital period in minutes.

What do you already know about low Earth orbit? Write down your estimate before working through the lesson, you will revisit it at the end.

Warm-up, for a satellite in stable circular orbit, what single force provides the centripetal acceleration?

Learning Intentions
goals

Know, Derive Orbital Velocity

  • Equate gravitational and centripetal force to derive $v = \sqrt{GM/r}$
  • State that $r = R_{\text{planet}} + h$ is the centre-to-centre radius

Understand, Circular Period Relation

  • Derive $T^2 = \frac{4\pi^2}{GM}r^3$ from Newton's law of gravitation
  • Apply $T^2/r^3 = \text{constant}$ to circular orbits around the same body

Can Do, Analyse Geostationary Orbits

  • Compare the properties and uses of LEO, MEO and GEO
  • Explain apparent weightlessness as shared free fall
Scan these before reading
vocab
Circular orbitA stable path where gravity provides exactly the centripetal force needed at that radius.
Orbital speed$v = \sqrt{GM/r}$, the speed required to maintain a circular orbit at radius $r$. Depends only on the central body's mass $M$ and the orbital radius.
Orbital periodThe time $T$ taken to complete one full orbit. Related to radius by Kepler's Third Law: $T^2 \propto r^3$.
Circular period relation$T^2/r^3 = 4\pi^2/(GM)$ for the circular-orbit model. General non-circular Kepler orbits are treated in Lesson 17.
Geostationary orbitAn equatorial, prograde orbit with period equal to Earth's sidereal rotation period (about 23 h 56 min; often approximated as 24 h). The satellite appears fixed from the ground.
Low Earth orbit (LEO)Orbits at altitudes of approximately 200–2000 km (e.g., ISS at 400 km). Shorter periods, lower latency than GEO.
Lesson boundary: Lesson 10 supplied $F_g=GMm/r^2$. Here it is applied to circular satellite orbits. Orbital energy is Lesson 12, escape velocity is Lesson 16, and general elliptical orbits plus Kepler's three laws are Lesson 17.
Misconceptions to fix
✗ Wrong: Satellites need fuel to maintain orbit.
✓ Right: In a stable circular orbit in vacuum, no fuel is needed. Gravity provides the centripetal force continuously. Fuel is only required for orbital adjustments or to fight atmospheric drag at very low altitudes.
✗ Wrong: Heavier satellites orbit more slowly.
✓ Right: $v = \sqrt{GM/r}$, the satellite's own mass cancels out. A feather and a space station at the same altitude orbit at exactly the same speed.
✗ Wrong: The orbital radius $r$ is measured from Earth's surface.
✓ Right: $r$ is always measured from the centre of Earth (or the central body). You must add Earth's radius $R_E$ to the altitude $h$: $r = R_E + h$.

Core Content

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Model gravity as the radial resultant force

Model conditions

Assume a circular orbit, a spherical central body of mass $M$, a satellite whose mass is negligible compared with $M$, no thrust or atmospheric drag, and radius $r$ measured centre-to-centre. Gravity is the inward resultant force; “centripetal force” is its radial role, not an additional force.

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Deriving Orbital Velocity
+5 XP

The balance between gravity and centripetal motion

A circular orbit is continuous free fall. The satellite's tangential velocity carries it forward while gravity accelerates it radially inward, continually changing the direction—not the magnitude—of its velocity in the ideal circular model.

Gravitational Orbits, satellite in circular orbit showing gravitational and centripetal force balance

A satellite in circular orbit: gravity provides the centripetal force. $r = R_E + h$ is measured from Earth's centre.

Equating forces

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

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

$\Rightarrow\quad v = \sqrt{\dfrac{GM}{r}}$

where $r = R_{\text{planet}} + h$ (centre-to-centre distance)

The satellite mass $m$ cancels out, orbital speed depends only on the planet's mass $M$ and the orbital radius $r$. This is why 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 Earth, not from its surface. Missing this step is the most common exam error in orbital calculations.

Worked example, ISS Orbital Speed

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

($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.371 \times 10^6$ m)

  1. Given. $h = 400 \times 10^3 \text{ m} = 4.00 \times 10^5 \text{ m}$.
  2. Find. Orbital speed $v$.
  3. Method. Find $r = R_E + h$, then $v = \sqrt{GM/r}$.
  4. Solve. $r = 6.371 \times 10^6 + 0.400 \times 10^6 = 6.771 \times 10^6 \text{ m}$.
  5. Solve. $v = \sqrt{\dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.771 \times 10^6}} = \sqrt{5.88 \times 10^7} = 7.67 \times 10^3 \text{ m/s}$.

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

Orbital velocity: $F_g = F_c \Rightarrow GMm/r^2 = mv^2/r \Rightarrow v = \sqrt{GM/r}$. Here $r = R_\text{planet} + h$ (centre-to-centre). Satellite mass cancels, orbital speed is independent of satellite mass. Higher orbit → lower speed.

Pause, copy the highlighted derivation and the $r = R + h$ rule into your book before moving on.

A satellite orbits at radius $r$ with speed $v$. If it moves to an orbit at radius $2r$, its new orbital speed is…

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Connect circular-orbit radius and period

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Circular-Orbit Period Relation
+5 XP

Relating orbital period to orbital radius

We just saw that $v = \sqrt{GM/r}$ gives orbital speed. That raises a question: how does the orbital period depend on radius? This card answers it → substitute $v = 2\pi r/T$ into $v = \sqrt{GM/r}$ to derive $T^2 = (4\pi^2/GM)r^3$.

For a circular orbit, expressing speed as circumference divided by period links the orbital radius and period. This is the circular special case of Kepler's Third Law.

Starting with $F_g = F_c$ but expressing velocity in terms of period $v = 2\pi r / T$:

Derivation for a circular orbit

$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, with $v = \dfrac{2\pi r}{T}$

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

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

$\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM} = \text{constant for all bodies orbiting the same mass}$

For circular satellites around the same central body, $T^2/r^3$ is constant. For a non-circular orbit, the general form uses semi-major axis $a$, not an instantaneous radius; that extension belongs in Lesson 17.

HSC tip

When calculating $T^2/r^3$, convert period to seconds and radius to metres first. Using mixed units (e.g., days and km) will give incorrect values for the constant.

Worked example 2, GPS circular-orbit period

A GPS satellite is modelled in a circular Earth orbit at $r=2.66\times10^7$ m. Calculate its period.

  1. Given. $r=2.66\times10^7$ m and $GM_E=3.986\times10^{14}$ m³ s$^{-2}$.
  2. Select. $T=2\pi\sqrt{r^3/(GM_E)}$.
  3. Substitute. $T=2\pi\sqrt{(2.66\times10^7)^3/(3.986\times10^{14})}$.
  4. Calculate. $T=4.31\times10^4$ s.
  5. Convert and interpret. $T=(4.31\times10^4)/3600=12.0$ h, consistent with a GPS satellite in MEO.

Circular-orbit period: $T^2 = (4\pi^2/GM)r^3$ and $T^2/r^3 = 4\pi^2/(GM)$ for satellites around the same central mass. Use seconds and metres. For non-circular orbits, Lesson 17 replaces $r$ with semi-major axis $a$.

Add the highlighted Kepler's Third Law and the rearrangement for $r$ to your notes before the check below.

For two satellites orbiting Earth, $T^2/r^3$ has the same value regardless of the satellite's mass.

Kepler's Third Law constant ($T^2/r^3$) is the same for satellites orbiting Earth and satellites orbiting Jupiter.

If a satellite's orbital radius doubles, its period increases by a factor of $2^{3/2} \approx 2.83$.

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Compare LEO, MEO and GEO

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Geostationary Orbits
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Satellites that stay fixed above one point on Earth

We just saw that $T^2/r^3 =$ constant for all Earth satellites. That raises a question: is there one special orbit where the satellite appears stationary from the ground? This card answers it → set $T = 86\,400$ s, solve $r = (GMT^2/4\pi^2)^{1/3} \approx 42\,200$ km; must be equatorial.

A geostationary satellite has a period equal to Earth's sidereal rotation period (about 23 h 56 min, or 86,164 s). Classroom calculations often use 24 h = 86,400 s, which gives essentially the same rounded orbital radius.

Finding the Geostationary Radius

Rearranging Kepler's Third Law with $T = 86{,}400 \text{ s}$:

Geostationary radius

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

$r = \left(\dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2}\right)^{1/3}$

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

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

Critical requirements

All geostationary satellites must orbit at exactly 35,900 km altitude, directly above the equator, moving in the same direction as Earth's rotation (west to east). Any other orbit, inclined, polar, or different altitude, will cause the satellite to drift across 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: Continuous coverage of the same region, enabling real-time storm tracking
  • Limitations: Poor coverage above ~70° latitude (too close to horizon); signal round-trip latency ~240 ms (satellite is far away)

Earth-orbit regions and uses

  • LEO (roughly 200–2000 km altitude): short periods and low communications latency; used for Earth observation, crewed spacecraft and low-latency constellations.
  • MEO (between LEO and GEO): wider coverage than LEO with moderate periods; navigation systems such as GPS operate near 20,200 km altitude.
  • GEO (35,786 km altitude): continuous view of one equatorial region; useful for fixed communications dishes and weather monitoring, but with greater latency and weak polar coverage.

Geostationary orbit: $T$ equals Earth's sidereal rotation period (86,164 s; often approximated as 86,400 s); $r \approx 42\,200$ km from Earth's centre; altitude $h \approx 35\,800$ km. Must be equatorial and prograde. Applications: TV broadcast, weather, fixed ground dish. Limits: weak polar coverage and greater latency.

Pause, copy the highlighted geostationary altitude and the three requirements into your book before moving on.

A geostationary satellite must be positioned…

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Investigate the circular-orbit model

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Key Formulas and Real-World Context
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Orbital mechanics formulae and GPS as a real-world application

We just saw the three orbit types: LEO, MEO and GEO. That raises a question: how do we consolidate the formulas and apply them to a real navigation system like GPS? This card answers it → GPS at 20,200 km altitude uses $T^2/r^3 = \text{const}$ and $v = \sqrt{GM/r}$ with atomic-clock timing.

Orbital velocity

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

$r = R + h$, from centre of central body. Units: $v$ in m/s, $r$ in m.

Kepler's Third Law

$T^2 = \dfrac{4\pi^2}{GM} r^3 \quad\Longleftrightarrow\quad \dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM}$

Constant for all bodies orbiting the same central mass $M$.

Geostationary values

$T_{\text{geo}} = 86{,}164 \text{ s}$ (sidereal day; $86{,}400$ s is the classroom approximation)

$r_{\text{geo}} \approx 42{,}200 \text{ km}$ from centre; altitude $\approx 35{,}900 \text{ km}$

Real world, GPS satellites

GPS satellites orbit at approximately 20,200 km altitude with a period of 11 h 58 min, exactly half a sidereal day. This is medium Earth orbit (MEO), not geostationary.

A receiver normally needs signals from at least four satellites to solve for three position coordinates and its clock offset. Atomic-clock timing makes distance estimates from signal travel time possible.

Interactive investigation

Use the Earth presets to compare ISS (LEO), GPS (MEO) and GEO. Record how increasing centre-to-centre radius changes speed and period. The display is a circular model and the drawing is not to scale.

Open calculator fullscreen ↗

Worked example, Orbital period from radius

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

  1. Given. $r = 8.0 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg.
  2. Find. Period $T$.
  3. Rearrange. $T^2 = \dfrac{4\pi^2}{GM} r^3 \Rightarrow T = 2\pi\sqrt{\dfrac{r^3}{GM}}$.
  4. Solve. $T = 2\pi\sqrt{\dfrac{(8.0 \times 10^6)^3}{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}} = 2\pi\sqrt{\dfrac{5.12 \times 10^{20}}{3.98 \times 10^{14}}} = 2\pi\sqrt{1.286 \times 10^6} = 7134 \text{ s} \approx 119 \text{ min}$.

Key orbital formulae: $v = \sqrt{GM/r}$ (higher orbit → lower speed); $T^2/r^3 = 4\pi^2/(GM)$ (constant for a given central body); $r$ from $T$: cube-root of $GMT^2/(4\pi^2)$. LEO ~400 km, GPS (MEO) ~20,200 km, GEO ~35,900 km.

Add the highlighted orbit altitudes and the key formulas to your notes before moving on.

Complete the formula: to find orbital radius from period, rearrange Kepler's Third Law to get $r = \left(\dfrac{GM T^2}{\_\_\pi^2}\right)^{1/3}$. The missing coefficient is _____.

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Consolidate the model and explain weightlessness

Wrap-up, Common Errors and Summary

Misconceptions, final check

✗ “Astronauts float because gravity is zero in low Earth orbit.”
✓ Gravity remains strong. The spacecraft and occupants accelerate together in orbital free fall, so the normal support force—and therefore apparent weight—is approximately zero.
✗ "The Moon doesn't fall toward Earth because there's no gravity in space."
✓ The Moon is continuously falling toward Earth. Its tangential velocity means it keeps "missing", that is precisely what an orbit is. Gravity at the Moon's distance is about 0.003 m/s², far from zero.
✗ "A satellite in LEO needs less speed to maintain its orbit than one in GEO."
✓ The opposite is true. $v = \sqrt{GM/r}$, a lower orbit (smaller $r$) requires a higher orbital speed. The ISS at 400 km travels at 7.67 km/s; a GEO satellite at 35,900 km travels at only about 3.07 km/s.

Copy into your books

Orbital Velocity

  • $F_g = F_c$: $GMm/r^2 = mv^2/r$
  • $v = \sqrt{GM/r}$
  • $r = R_{\text{planet}} + h$

Kepler's Third Law

  • $T^2 = (4\pi^2/GM)r^3$
  • $T^2/r^3 = 4\pi^2/GM$
  • Constant for same central body

Geostationary

  • $T = 86{,}400$ s (24 h)
  • $r \approx 42{,}200$ km from centre
  • Altitude $\approx 35{,}900$ km

Key Principle

  • Higher orbit → slower speed
  • Higher orbit → longer period
  • Satellite mass cancels out

Two satellites orbit Earth: satellite A at 400 km altitude, satellite B at 35,900 km altitude. Which statement is correct?

Three of these statements about a geostationary satellite are correct. Pick the odd one out.

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Respond using the circular-orbit model

Activity 1, Orbital Mechanics Drills
ApplyBand 4

Practise the key orbital mechanics calculations from this lesson

  1. State Kepler's Third Law in words and write its mathematical form. What are the SI units of $T^2/r^3$?
  2. Calculate the orbital speed of a satellite orbiting Earth at 600 km altitude. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.371 \times 10^6$ m)
  3. Find the orbital period of a satellite at $r = 8.0 \times 10^6$ m from Earth's centre.

Drill check, a satellite orbits at $r = 8.0 \times 10^6$ m. Its orbital speed (to 3 sig. fig., in km/s) is _____.

Activity 2, Geostationary Orbit Concept Check
UnderstandBand 5

Explain a key orbital mechanics principle with real-world application

A geostationary satellite appears motionless above one point on Earth. Explain why this requires a specific altitude (~35,900 km) and why the satellite must orbit above the equator. Use the formula $T^2 = 4\pi^2 r^3/(GM)$ in your explanation, and give one real-world application (e.g., TV broadcast).

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Apply gravitational-orbit reasoning independently

Quick recall, gravitational orbits
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 10 marks
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ApplyBand 4(3 marks) 1. The Hubble Space Telescope orbits at 547 km altitude. Calculate its orbital speed and period. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m)

1 mark: correct orbital radius with working · 1 mark: correct speed · 1 mark: correct period with unit conversion

ApplyBand 5(3 marks) 2. Use Kepler's Third Law to find the orbital radius of a satellite with an orbital period of 8.0 hours around Earth.

1 mark: convert T to seconds · 1 mark: correct substitution into $r^3 = GMT^2/(4\pi^2)$ · 1 mark: correct final answer in metres

EvaluateBand 6(4 marks) 3. Assess the advantages and limitations of geostationary orbits for communication satellites compared to low Earth orbit (LEO) satellites.

2 marks: at least 2 GEO advantages and 2 GEO limitations · 1 mark: comparison with LEO (latency, coverage, number of satellites) · 1 mark: uses "assess" language, weighing trade-offs

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1, Hubble Space Telescope (3 marks):

Step 1: $r = R_E + h = 6.37 \times 10^6 + 5.47 \times 10^5 = 6.917 \times 10^6 \text{ m}$ (1 mark).

Step 2: $v = \sqrt{GM/r} = \sqrt{(6.67 \times 10^{-11} \times 5.97 \times 10^{24})/(6.917 \times 10^6)} = 7.59 \times 10^3 \text{ m/s} = 7.59 \text{ km/s}$ (1 mark).

Step 3: $T = 2\pi r/v = 2\pi \times 6.917 \times 10^6 / 7.59 \times 10^3 = 5726 \text{ s} = 95.4 \text{ min}$ (1 mark).

Q2, Satellite with 8.0 h period (3 marks):

$T = 8.0 \times 3600 = 28{,}800 \text{ s}$ (1 mark).

$r^3 = GMT^2/(4\pi^2) = (6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (28800)^2)/(4\pi^2) = 8.32 \times 10^{21} \text{ m}^3$ (1 mark).

$r = (8.32 \times 10^{21})^{1/3} = 2.03 \times 10^7 \text{ m} = 20{,}300 \text{ km}$ (1 mark). Altitude: $h = 2.03 \times 10^7 - 6.37 \times 10^6 = 1.39 \times 10^7 \text{ m} \approx 13{,}900 \text{ km}$.

Q3, Assessment of GEO vs LEO (4 marks):

GEO advantages: Fixed position relative to ground, simple, cheap dish antennas (no tracking); continuous coverage of one region, ideal for TV broadcast and regional communications; only 3 satellites for near-global (non-polar) coverage.

GEO limitations: High latency (~240 ms round-trip), problematic for real-time video calls and gaming; no coverage of polar regions (equatorial plane); expensive to launch to 35,900 km altitude.

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

Award marks for balanced assessment. Must identify at least 2 advantages and 2 limitations, with comparison to LEO, using "assess" language (weighing trade-offs).

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Retrieve and reflect

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Arcade practice · Asteroid Blaster

Orbital mechanics practice, match the orbital parameters to the target. Hammers home the Kepler relationship between radius and period.

How did your thinking change?

At the start you were asked why the ISS occupants feel weightless even though the station completes an orbit in about 93 minutes and remains well within Earth's gravitational field.

Explain the complete causal chain: gravity is the radial resultant force; the ISS and occupants share the same free-fall acceleration; and the floor provides almost no normal support force. Then use $r=R_E+h$ and the circular-orbit equations to explain why the ISS has a short period compared with MEO and GEO satellites.

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