Kepler's third law and Newton's gravitational theory unite to explain everything from ISS orbits to GPS satellites to the motion of planets around the Sun. Master the equations that govern every object in space.
🛰️Derive and apply Kepler's third law: T² = 4π²r³/GM
🌐Define geostationary orbit and calculate its altitude
⚡Derive and apply escape velocity: v_esc = √(2GM/r)
⚖️Calculate orbital speed and relate it to gravitational field strength
🔋Calculate the total energy of a satellite in orbit: E_total = −GMm/2r
📡Compare geostationary and low Earth orbits for different applications
Derivation of Kepler's Third Law
For a satellite in a circular orbit of radius r around a central mass M, gravity provides the centripetal force:
This is Kepler's Third Law: the square of the orbital period is proportional to the cube of the orbital radius. It applies to any satellite (natural or artificial) orbiting the same central mass M.
Since GM is constant for a given central mass: T² ∝ r³, or equivalently r³/T² = GM/(4π²) = constant.
Kepler's Third Law: For objects orbiting the same central body, the ratio r³/T² is the same constant = GM/(4π²). This allows us to find M from orbital measurements (Cavendish didn't need to go to space!).
Orbital speed from the same equation: GM/r = v² → v = √(GM/r). Note v ∝ 1/√r — closer orbits are faster.
Geostationary Orbits
A geostationary satellite has these special properties:
Orbital period T = 24 hours = 86 400 s (same as Earth's rotation)
Orbits in the equatorial plane (directly above the equator)
Orbits in the same direction as Earth's rotation (west to east)
Appears stationary over a fixed point on the ground
Finding the geostationary orbit radius: T² = 4π²r³/(GM_E):
= ∛(7.536×10²²) = 4.22 × 10⁷ m from Earth's centre
Altitude above surface: h = r − R_E = 4.22×10⁷ − 6.37×10⁶ = 3.58 × 10⁷ m ≈ 35 800 km
Geostationary applications: TV broadcasting, weather satellites, communications. Disadvantage: large altitude means weak signal and time delay (~0.25 s one-way). For GPS/phone, low Earth orbit (LEO) satellites are preferred despite needing many satellites to give continuous coverage.
Orbit type
Altitude
Period
Applications
Low Earth (LEO)
200–2000 km
90–130 min
ISS, weather, GPS, imaging
Medium Earth (MEO)
2000–35 700 km
2–24 h
GPS constellation, navigation
Geostationary (GEO)
35 786 km
24 h
TV, weather, comms
Highly elliptical
Variable
Variable
High-latitude communications
Escape Velocity
The escape velocity is the minimum speed needed for a projectile to escape a planet's gravity without further propulsion. Using energy conservation:
At the surface (r = R): KE = ½mv², PE = −GMm/R. At infinity: KE = 0, PE = 0.
For escape: initial total energy ≥ 0 (so the object can reach infinity with KE ≥ 0):
½mv² − GMm/R ≥ 0 → v ≥ √(2GM/R)
v_esc = √(2GM/R) = √(2gR)
Since g = GM/R² → GM = gR²:
v_esc = √(2gR)
For Earth: v_esc = √(2 × 9.81 × 6.37×10⁶) = √(1.25×10⁸) = 11 180 m s⁻¹ ≈ 11.2 km s⁻¹
⚠️ Note: orbital speed v_orbit = √(GM/r) and escape speed v_esc = √(2GM/r). So v_esc = √2 × v_orbit ≈ 1.41 × v_orbit at the same radius. To escape from orbit requires 41% more speed than to remain in orbit.
A black hole is an object so dense that the escape velocity from its surface exceeds the speed of light c — hence nothing can escape, not even light.
Total Energy of a Satellite in Orbit
For a satellite of mass m in circular orbit of radius r:
KE: from GMm/r² = mv²/r → v² = GM/r → KE = ½mv² = GMm/(2r)
PE: E_p = −GMm/r
Total energy: E = KE + PE = GMm/(2r) − GMm/r = −GMm/(2r)
E_total = −GMm / (2r)
Key observations:
Total energy is negative — the satellite is bound.
|E_total| = ½|PE| (the total energy has half the magnitude of the PE).
KE = −E_total = GMm/(2r) (KE is always positive and equals |E_total|).
As r increases (moving to higher orbit), E_total increases (becomes less negative) — the satellite gains energy, paradoxically while slowing down! (KE decreases but total energy increases because PE increases by more.)
This explains why a satellite losing energy to atmospheric drag (r decreases) speeds up! As it descends to lower orbit, its total energy decreases (more negative) but it moves faster. Energy lost goes into heat from air resistance and the satellite's kinetic energy actually increases — seeming paradoxical but fully consistent with the virial theorem.
The Moon orbits Earth with period T = 27.3 days and orbital radius r = 3.84 × 10⁸ m. Use Kepler's third law to calculate Earth's mass.
1Convert period: T = 27.3 × 24 × 3600 = 2 358 720 s = 2.359 × 10⁶ s
2= √(2 × 1.257×10¹³ / 3.39×10⁶) = √(2 × 3.709×10⁶) = √(7.418×10⁶) = 2724 m s⁻¹ ≈ 5.03 km s⁻¹
v_esc(Mars) ≈ 5.03 km s⁻¹ — about 45% of Earth's escape velocity. This is why Mars lost much of its atmosphere over billions of years.
Q1. Planet X has twice Earth's orbital radius around the Sun. How does its orbital period compare to Earth's?
Q2. Which of the following is NOT a requirement for a geostationary satellite?
Q3. How does the orbital speed of a satellite change as its orbital radius increases?
Q4. The escape velocity is √2 times the orbital speed at the same radius. This is because:
Q5. The total energy of a satellite in circular orbit is −GMm/2r. As the orbit decays (r decreases), the total energy:
Challenge 1. Derive the expression for the total energy of a satellite E = −GMm/2r starting from expressions for KE and PE. Hence show that moving to a higher orbit requires energy input despite the satellite slowing down. Calculate the total energy change when moving a 500 kg satellite from r = 7.0×10⁶ m to r = 4.2×10⁷ m (geostationary). (M_E = 5.97×10²⁴ kg)
✓ For circular orbit: gravity = centripetal force: GMm/r² = mv²/r → v² = GM/r. KE = ½mv² = GMm/(2r). PE = −GMm/r. Total E = GMm/(2r) − GMm/r = GMm/(2r)(1−2) = −GMm/(2r). QED. As r increases: KE = GMm/(2r) decreases (satellite slows). PE = −GMm/r increases (becomes less negative). Net total energy E = −GMm/(2r) increases (becomes less negative). So despite KE decreasing, total energy increases — energy must be supplied by rocket motors. Energy change: ΔE = −GMm/(2r₂) − (−GMm/(2r₁)) = GMm/2 × (1/r₁ − 1/r₂) = (6.674×10⁻¹¹ × 5.97×10²⁴ × 500)/2 × (1/7.0×10⁶ − 1/4.2×10⁷) = (1.993×10¹⁷) × (1.429×10⁻⁷ − 2.381×10⁻⁸) = (1.993×10¹⁷) × (1.190×10⁻⁷) = 2.37×10¹⁰ J ≈ 23.7 GJ of energy must be supplied.
Challenge 2. Using T² = 4π²r³/(GM), derive an expression for the mass of the Sun given Earth's orbital period T = 365.25 days and orbital radius r = 1.50×10¹¹ m. Then find Jupiter's orbital period given its orbital radius is 7.78×10¹¹ m.
✓ M_Sun = 4π²r³/(GT²). T = 365.25 × 24 × 3600 = 3.156×10⁷ s. M_Sun = (4π² × (1.50×10¹¹)³)/(6.674×10⁻¹¹ × (3.156×10⁷)²) = (39.478 × 3.375×10³³)/(6.674×10⁻¹¹ × 9.960×10¹⁴) = (1.332×10³⁵)/(6.644×10⁴) = 2.005×10³⁰ kg ≈ 2.0×10³⁰ kg. ✓ Jupiter: using r³/T² = GM/(4π²) = constant for objects orbiting Sun. T_J² = T_E² × (r_J/r_E)³ = (3.156×10⁷)² × (7.78×10¹¹/1.50×10¹¹)³ = 9.96×10¹⁴ × (5.187)³ = 9.96×10¹⁴ × 139.5 = 1.389×10¹⁷. T_J = 3.727×10⁸ s ÷ (3.156×10⁷) = 11.81 years. (Actual: 11.86 years ✓)
Challenge 3 (Synoptic). A black hole has the Schwarzschild radius R_S = 2GM/c² — the radius at which the escape velocity equals the speed of light. Calculate R_S for the Sun (M = 2.0×10³⁰ kg). If the Sun were compressed to a sphere of radius R_S, what would its density be? Compare to nuclear density (~2×10¹⁷ kg m⁻³).
✓ R_S = 2GM/c² = 2×6.674×10⁻¹¹×2.0×10³⁰/(3×10⁸)² = 2.670×10²⁰/9×10¹⁶ = 2966 m ≈ 2.97 km. The Sun would need to be compressed into a sphere of radius ~3 km. Density: ρ = M/((4/3)πR_S³) = 2.0×10³⁰/((4/3)π×(2966)³) = 2.0×10³⁰/((4/3)π×2.609×10¹⁰) = 2.0×10³⁰/(1.094×10¹¹) = 1.83×10¹⁹ kg m⁻³. This is about 90 times nuclear density! Black holes are among the densest objects in the universe. (Note: the concept of "density" breaks down at the singularity — this is the average density within R_S.)