- Express the orbit equation and Kepler’s three laws mathematically
- Derive Kepler’s third law from Newton’s law of gravitation and use it to find the mass of a central body
- Calculate orbital (first cosmic) and escape (second cosmic) velocities using energy
- Solve problems on satellites and interplanetary transfers
We cannot put the Sun on a scale, yet we know its mass to better than 0.1%. With pencil and paper we can also work out how fast a rocket must go to leave the Earth, or how high a satellite must fly to orbit once in 24 hours. The key is celestial mechanics — Kepler’s geometric laws and Newton’s law of gravitation.
Kepler’s laws in mathematical form
First law: every planet moves on an ellipse with the Sun at one focus. In polar coordinates (angle measured from perihelion) the orbit equation is given below. The shape of the ellipse is set by the eccentricity e: e = 0 is a circle, e → 1 is a very elongated ellipse.
- rdistance from the planet to the Sun (focus), m
- asemi-major axis, m (or AU)
- eeccentricity, 0 ≤ e < 1 (dimensionless)
- θtrue anomaly — angle measured from perihelion
- rₚperihelion distance (closest to the Sun), θ = 0°
- rₐaphelion distance (farthest point), θ = 180°
For the Earth a ≈ 149.6 million km, e ≈ 0.0167: rₚ ≈ 147.1 million km (early January), rₐ ≈ 152.1 million km (early July).
Second law: the line joining a planet to the Sun sweeps out equal areas in equal times. This follows from conservation of angular momentum: gravity is a central force (directed at the Sun), so its torque about the Sun is zero and L = m·r·v⊥ stays constant. As a result a planet moves fastest at perihelion and slowest at aphelion.
- dA/dtdA/dtareal velocity, m²/s
- Langular momentum, kg·m²/s
- vₚ, vₐspeeds at perihelion and aphelion, m/s
Third law: the square of the period is proportional to the cube of the semi-major axis. Kepler found it from observations; Newton revealed the constant of proportionality — it depends on the mass of the central body. This is astronomy’s main way of measuring mass: the masses of planets, binary stars, stars with exoplanets and even galaxies are found this way.
- Torbital period, s
- asemi-major axis, m
- Ggravitational constant, 6.674 · 10⁻¹¹ N·m²/kg²
- M, mmasses of the central body and the orbiting body, kg (usually m ≪ M)
Newton’s law of gravitation and a derivation of Kepler’s law
- Fgravitational force, N
- m₁, m₂masses of the bodies, kg
- rdistance between the centres of the bodies, m
For a circular orbit, gravity provides the centripetal force: G·M·m / r² = m·v² / r. This gives v = √(GM/r). Since T = 2πr / v, we get T² = 4π²r² / v² = 4π²r³ / (GM) — Kepler’s third law for m ≪ M. For an ellipse the same result holds with a in place of r. Notice that the satellite’s own mass m cancels: the orbit does not depend on the satellite’s mass.
The Earth’s orbit has a semi-major axis a = 1.496 · 10¹¹ m and period T = 3.156 · 10⁷ s. Find the mass of the Sun.
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a³ = (1.496 · 10¹¹)³ ≈ 3.348 · 10³³ m³; 4π²a³ ≈ 1.322 · 10³⁵.
G·T² = 6.674 · 10⁻¹¹ · (3.156 · 10⁷)² ≈ 6.647 · 10⁴.
M ≈ 1.322 · 10³⁵ / 6.647 · 10⁴ ≈ 1.99 · 10³⁰ kg — about 333,000 times the Earth’s mass.
Cosmic velocities and energy
The total mechanical energy of a body in a gravitational field is the sum of its kinetic and potential energy (potential energy is taken as zero at infinity). To reach infinity the body needs total energy E ≥ 0. Setting E = 0 gives the escape (second cosmic) velocity, which is exactly √2 times the circular orbital speed. The speed at any point of any orbit is given by the vis-viva equation.
- Etotal mechanical energy, J; E < 0 — bound orbit, E ≥ 0 — the body escapes
- v₁circular orbital (first cosmic) velocity, m/s
- v₂escape (second cosmic) velocity, m/s
- GMfor the Earth 3.986 · 10¹⁴ m³/s²
- vspeed at distance r on the orbit, m/s
- asemi-major axis of the orbit, m (a = r for a circle)
The vis-viva equation: a = r gives v₁, a → ∞ gives v₂.
a) Find v₁ and v₂ at the Earth’s surface (R = 6371 km). b) Calculate the speed and period of the ISS flying 400 km up. (GM = 3.986 · 10¹⁴ m³/s²)
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v₂ = √2 · 7.91 ≈ 11.2 km/s.
b) r = R + h = 6371 + 400 = 6771 km = 6.771 · 10⁶ m.
v = √(3.986 · 10¹⁴ / 6.771 · 10⁶) ≈ 7.67 km/s (≈ 27,600 km/h).
T = 2πr / v = 2π · 6.771 · 10⁶ / 7673 ≈ 5540 s ≈ 92 min.
A geostationary satellite has a period equal to the Earth’s sidereal day: T = 86,164 s. Find its orbital radius and its height above the equator (Earth’s equatorial radius 6378 km).
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r = ∛(7.50 · 10²²) ≈ 4.216 · 10⁷ m ≈ 42,160 km.
h = r − R ≈ 42,160 − 6378 ≈ 35,800 km — the “about 36,000 km” from the earlier lesson.
On the most fuel-efficient route, a spacecraft follows an ellipse with perihelion at the Earth’s orbit (1 AU) and aphelion at Mars’s orbit (1.524 AU). How long does the trip take?
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Full period: T = a^(3/2) = 1.262^1.5 ≈ 1.418 years.
The trip is half the ellipse: t = T / 2 ≈ 0.709 years ≈ 259 days (about 8.5 months).
Where is it used?
- Space engineering — putting satellites into orbit, orbit corrections, interplanetary trajectories and gravity assists.
- Navigation and communication — choosing orbits for GPS, communication and weather satellites.
- Astrophysics — measuring the masses of binary stars, exoplanets and galaxies, and searching for dark matter.
- Planetary defence — predicting the paths of asteroids that approach the Earth.
Key points
- An orbit is an ellipse: r = a(1 − e²)/(1 + e·cos θ); rₚ = a(1 − e), rₐ = a(1 + e).
- The second law is conservation of angular momentum: vₚ·rₚ = vₐ·rₐ.
- T² = 4π²a³/(GM); in solar units T² = a³/M — the main way to measure mass.
- v₁ = √(GM/r), v₂ = √2·v₁; for the Earth 7.9 and 11.2 km/s.
- Vis-viva: v² = GM(2/r − 1/a); an orbit is bound when E < 0.
Check yourself
10 questions. Every correct answer earns XP.