Gravitational Orbits
On 4 October 1957, the Soviet Union launched Sputnik 1 — the first artificial satellite — into an orbit ranging from 577 km (perigee) to 941 km (apogee), with a mean orbital radius of 7,252 km from Earth's centre. Its 96.2-minute period and 7.9 km/s orbital velocity were the first experimental verification of Kepler's Third Law ($T^2 \propto r^3$) using a human-made object.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
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 — Kepler's Third Law
- Derive $T^2 = \frac{4\pi^2}{GM}r^3$ from Newton's law of gravitation
- Apply $T^2/r^3 = \text{constant}$ to compare orbits around the same body
Can Do — Analyse Geostationary Orbits
- Calculate the geostationary orbital radius from $T = 86{,}400 \text{ s}$
- Explain the conditions and applications of geostationary orbits
Core Content
The balance between gravity and centripetal motion
Launch a ball at 1 km/s horizontally from a mountain: it curves down and hits the ground a few kilometres away. Launch it at 7.9 km/s and it curves downward at exactly the same rate Earth's surface curves away — it never hits the ground. Sputnik 1, launched on 4 October 1957 into an orbit with mean radius 7,252 km from Earth's centre, achieved exactly this speed. The force that keeps it in orbit is simply gravity, playing the role of centripetal force.
A satellite in circular orbit: gravity provides the centripetal force. $r = R_E + h$ is measured from Earth's centre.
$F_{\text{gravity}} = F_{\text{centripetal}}$
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$
$\Rightarrow\quad v = \sqrt{\dfrac{GM}{r}}$
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).
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.
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)
- Given. $h = 400 \times 10^3 \text{ m} = 4.00 \times 10^5 \text{ m}$.
- Find. Orbital speed $v$.
- Method. Find $r = R_E + h$, then $v = \sqrt{GM/r}$.
- Solve. $r = 6.371 \times 10^6 + 0.400 \times 10^6 = 6.771 \times 10^6 \text{ m}$.
- 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…
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 $T^2 = (4\pi^2/GM)r^3$.
Newton showed that Kepler's empirical Third Law follows directly from his law of universal gravitation combined with circular motion. By expressing speed in terms of period, we derive a relationship that holds for every satellite orbiting the same central body.
Starting with $F_g = F_c$ but expressing velocity in terms of period $v = 2\pi r / T$:
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, with $v = \dfrac{2\pi r}{T}$
$\dfrac{GM}{r^2} = \dfrac{4\pi^2 r}{T^2}$
$T^2 = \dfrac{4\pi^2}{GM} r^3$
$\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM} = \text{constant for all bodies orbiting the same mass}$
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.
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.
Mars has an orbital period of 687 days and orbits at $2.28 \times 10^{11}$ m from the Sun. Verify $T^2/r^3 \approx 2.97 \times 10^{-19}$ s²/m³.
- Given. $T = 687 \text{ days}$, $r = 2.28 \times 10^{11} \text{ m}$.
- Convert. $T = 687 \times 24 \times 3600 = 5.936 \times 10^7 \text{ s}$.
- Calculate. $T^2 = (5.936 \times 10^7)^2 = 3.524 \times 10^{15} \text{ s}^2$.
- Calculate. $r^3 = (2.28 \times 10^{11})^3 = 1.186 \times 10^{34} \text{ m}^3$.
- Verify. $T^2/r^3 = 3.524 \times 10^{15} / 1.186 \times 10^{34} = 2.97 \times 10^{-19} \text{ s}^2\text{/m}^3$ ✓
This constant applies to all planets and objects orbiting the Sun.
Kepler's Third Law: $T^2 = (4\pi^2/GM)r^3$; ratio $T^2/r^3 = 4\pi^2/(GM)$ = constant for all bodies orbiting the same mass $M$. Units: $T$ in seconds, $r$ in metres. To find $r$ from $T$: $r = (GMT^2/4\pi^2)^{1/3}$.
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 $2^{3/2} \approx 2.83$.
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 an orbital period exactly equal to Earth's rotation period (24 hours = 86,400 s). From the ground, it appears to hover motionless above a fixed equatorial point — an extraordinarily useful property for communications and weather monitoring.
Finding the Geostationary Radius
Rearranging Kepler's Third Law with $T = 86{,}400 \text{ s}$:
$r^3 = \dfrac{GMT^2}{4\pi^2}$
$r = \left(\dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2}\right)^{1/3}$
$r = 4.22 \times 10^7 \text{ m} = 42{,}200 \text{ km from Earth's centre}$
Altitude: $h = r - R_E = 4.22 \times 10^7 - 6.371 \times 10^6 \approx 3.59 \times 10^7 \text{ m} = 35{,}900 \text{ km}$
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)
Geostationary orbit: $T = 86\,400$ s (24 h); $r \approx 42\,200$ km from Earth's centre; altitude $h \approx 35\,900$ km. Must be equatorial, west-to-east. Applications: TV broadcast, weather — fixed ground dish. Limits: no polar coverage; ~240 ms latency.
Pause — copy the highlighted geostationary altitude and the three requirements into your book before moving on.
A geostationary satellite must be positioned…
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.
$v = \sqrt{\dfrac{GM}{r}}$
$r = R + h$ — from centre of central body. Units: $v$ in m/s, $r$ in m.
$T^2 = \dfrac{4\pi^2}{GM} r^3 \quad\Longleftrightarrow\quad \dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM}$
Constant for all bodies orbiting the same central mass $M$.
$T_{\text{geo}} = 24 \text{ h} = 86{,}400 \text{ s}$
$r_{\text{geo}} \approx 42{,}200 \text{ km}$ from centre; altitude $\approx 35{,}900 \text{ km}$
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.
At least 4 satellites must be simultaneously visible from any point on Earth for trilateration (latitude, longitude, and altitude). The GPS constellation has 31 operational satellites across 6 orbital planes.
Each GPS satellite carries atomic clocks accurate to within nanoseconds — because light travels ~30 cm per nanosecond, any timing error directly causes a positioning error of the same magnitude.
Find the orbital period of a satellite at $r = 8.0 \times 10^6$ m from Earth's centre.
- 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.
- Find. Period $T$.
- Rearrange. $T^2 = \dfrac{4\pi^2}{GM} r^3 \Rightarrow T = 2\pi\sqrt{\dfrac{r^3}{GM}}$.
- 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 _____.
Misconceptions — final check
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.
Practise the key orbital mechanics calculations from this lesson
- State Kepler's Third Law in words and write its mathematical form. What are the SI units of $T^2/r^3$?
- 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)
- 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 _____.
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).
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.
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).
Five timed questions on gravitational orbits. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
Enter the arenaOrbital mechanics practice — match the orbital parameters to the target. Hammers home the Kepler relationship between radius and period.
At the start you were asked about Sputnik 1, launched on 4 October 1957 by the Soviet Union into a mean orbit radius of 7,252 km from Earth's centre, with a 96.2-minute period and 7.9 km/s orbital speed.
Verify with the orbital velocity formula: $v = \sqrt{GM/r} = \sqrt{(6.674 \times 10^{-11} \times 5.97 \times 10^{24}) / (7.252 \times 10^6)} = \sqrt{3.98 \times 10^{14} / 7.252 \times 10^6} = \sqrt{5.49 \times 10^7} = 7.41 \text{ km/s}$. The small discrepancy from 7.9 km/s reflects Sputnik's elliptical orbit (577–941 km altitude), not circular. Kepler's Third Law: $T^2/r^3 = 4\pi^2/(GM) = 4\pi^2/(3.98 \times 10^{14}) = 9.91 \times 10^{-14} \text{ s}^2/\text{m}^3$ — a constant for all Earth satellites. Sputnik's launch was the first experimental confirmation of this. Has your understanding changed?