In short

Kepler's three laws describe planetary orbits: (1) each planet traces an ellipse with the Sun at one focus; (2) the line from Sun to planet sweeps equal areas in equal times (a consequence of angular-momentum conservation); (3) the orbital period squared is proportional to the semi-major axis cubed: T^2 = \dfrac{4\pi^2}{GM}\,a^3. For circular orbits, derive the third law in two lines by setting gravity equal to centripetal force. A geostationary satellite sits at r \approx 42{,}164 km (altitude \approx 35{,}800 km) — the one radius where the period matches Earth's 23 h 56 min sidereal day. Elliptical and transfer orbits (like the Hohmann transfer Chandrayaan used to reach the Moon) are the same physics.

Look up. The Moon is a satellite, held in orbit by the Earth. Earth is a satellite of the Sun. The International Space Station is a satellite of Earth, zipping around once every 92 minutes — so low that you can see it with your naked eye just after sunset. A geostationary communications satellite, at 36,000 km, orbits with exactly the period of Earth's rotation, so from the ground it appears to hang motionless in the sky. Chandrayaan-3, in 2023, rode a carefully shaped elliptical trajectory out to the Moon and back. All of these orbits obey the same three rules.

Those rules are called Kepler's laws. They were found by patient observation — extracted from decades of raw planetary-position data — before Newton's theory of gravity existed. But once Newton wrote down F = GMm/r^2, the three laws became inevitable. Any inverse-square attractive force produces Kepler's three laws, always. They are not a special property of the Sun or of our solar system — they are the geometry of a 1/r^2 force.

This article derives all three, shows you how to compute orbital periods and altitudes, explains why a geostationary satellite is a specific altitude and not a choice, and sketches the transfer-orbit machinery ISRO uses to move a spacecraft between orbits cheaply.

Kepler's first law — every orbit is a conic section

First law. Every planet moves in an elliptical orbit with the Sun at one focus.

An ellipse is the set of points for which the sum of distances to two fixed points (the foci) is constant. For an orbit, the Sun (or more generally, the central mass) sits at one focus, not at the centre; the other focus is empty. The long axis of the ellipse (the major axis) has length 2a; a is called the semi-major axis. The short axis has length 2b. The ellipse's shape is described by its eccentricity \epsilon:

\epsilon = \frac{c}{a}, \quad \text{where } c = \text{distance from centre to focus.}

\epsilon = 0 gives a circle (foci coincide with the centre). \epsilon \to 1 gives a very elongated ellipse. \epsilon = 1 exactly gives a parabola (unbound); \epsilon > 1 gives a hyperbola (strongly unbound). All bound orbits have 0 \leq \epsilon < 1.

For the planets in our solar system, eccentricities are small — Earth's is 0.017, Mars's is 0.093. Their orbits look essentially circular to the naked eye. But Halley's Comet has \epsilon \approx 0.97, a spectacularly elongated ellipse that carries it from inside Venus's orbit out past Neptune.

Elliptical orbit with the Sun at one focus An ellipse with two marked foci, the Sun at the right focus and the empty focus on the left. The major axis of length 2a is horizontal; the semi-minor axis b is marked vertically. A planet at perihelion is at the right vertex; at aphelion at the left vertex. empty focus Sun perihelion aphelion 2a (major axis) b centre
The Sun sits at one focus (red, right); the other focus is empty (grey, left). The planet traces the elliptical boundary. **Perihelion** is the point of closest approach to the Sun (right vertex); **aphelion** the point of greatest distance (left vertex). The semi-major axis $a$ is half the major axis; $b$ is the semi-minor axis.

Why ellipses? The proof that all bound orbits under an inverse-square force are exact ellipses is a calculus exercise that integrates the two-body equation of motion. The result is called a conic-section orbit: the orbit is always a conic (circle, ellipse, parabola, or hyperbola), and the total mechanical energy E = KE + U decides which one:

Total energy E Orbit shape Bound?
E < 0 Ellipse (or circle if \epsilon = 0) Yes
E = 0 Parabola Marginal (just barely unbound)
E > 0 Hyperbola No

The full derivation — which uses the Laplace–Runge–Lenz vector, a hidden symmetry of the 1/r^2 force — is a second-year university topic. For now, take the first law as given and focus on the two that fall out quickly: the equal-areas law and the period-vs-radius law.

Kepler's second law — equal areas in equal times

Second law. The line joining the planet to the Sun sweeps out equal areas in equal intervals of time.

Picture a planet tracing its ellipse. At perihelion (closest to the Sun), it is moving fast. At aphelion (farthest from the Sun), it is moving slowly. How much slower? Exactly enough that the area swept per unit time is constant. A planet near perihelion, moving fast, sweeps a thin long sliver of area in one second. At aphelion, moving slowly, it sweeps a wider short wedge — same total area.

This is not an empirical coincidence. It is the conservation of angular momentum in disguise.

Derivation from angular momentum

Take the central mass M at the origin, the orbiting mass m at position \vec{r} moving with velocity \vec{v}. Its angular momentum about the origin is

\vec{L} = m\,\vec{r} \times \vec{v}.

The only force on m is gravity, pointing along -\vec{r} (toward M). So the torque about the origin is

\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times (\text{something along } -\hat{r}) = 0.

Why: the cross product of two parallel (or antiparallel) vectors is zero. Because gravity is a central force (it always points along the line between the two bodies), the torque it exerts about the central body is always zero.

Zero torque means angular momentum is conserved. So |\vec{L}| = mrv\sin\theta (where \theta is the angle between \vec{r} and \vec{v}) is constant throughout the orbit. See Conservation of Angular Momentum for the general theorem.

Now the area-sweep rate. In a small time dt, the planet moves through a tiny displacement d\vec{r} = \vec{v}\,dt, and the area swept out by \vec{r} is the area of the little triangle with sides \vec{r} and d\vec{r}:

dA = \tfrac{1}{2}|\vec{r} \times d\vec{r}| = \tfrac{1}{2}|\vec{r} \times \vec{v}|\,dt.

Divide by dt:

\frac{dA}{dt} = \tfrac{1}{2}|\vec{r} \times \vec{v}| = \frac{|\vec{L}|}{2m}.

Why: the magnitude of \vec{r} \times \vec{v} is the area of the parallelogram with sides \vec{r} and \vec{v}; half of that is the triangle. Substitute \vec{L} = m\vec{r}\times\vec{v} to get the final form.

Since L is conserved (no torque) and m is constant, dA/dt is constant. That is Kepler's second law, written as a differential statement:

\boxed{\; \frac{dA}{dt} = \frac{L}{2m} = \text{constant.} \;}

The rate at which the radius vector sweeps out area is literally the angular momentum per unit mass, up to a factor of 2. The second law is angular-momentum conservation.

The physical consequence

Because dA/dt is constant, the planet must move faster when r is small (so it can cover area at the same rate with a shorter sweep distance) and slower when r is large. This is exactly what happens: Earth moves at 30.3 km/s at perihelion in January and 29.3 km/s at aphelion in July.

A second consequence, less often stated but striking: because dA/dt is constant, the total area swept in one orbital period equals the area of the entire ellipse, \pi ab. So

T = \frac{\pi ab}{dA/dt} = \frac{2\pi ab \, m}{L}.

This gives you the orbital period in terms of a, b, and L. Feed Kepler's third law into this, and you get a complete relation between the orbit's geometry and its dynamics.

Equal areas swept in equal times An elliptical orbit with the Sun at the right focus. Two shaded sectors of the same area are shown: a long thin wedge near perihelion where the planet moves fast, and a shorter wider wedge near aphelion where the planet moves slowly. Sun equal area near perihelion (fast) equal area near aphelion (slow)
Two shaded sectors show how the planet sweeps out *equal areas* in equal time intervals. Near perihelion (right), the sector is a thin tall wedge — the planet covers a long arc in that time. Near aphelion (left), the sector is a squat wide one — the planet covers a shorter arc. Same area, same time.

Indian science footnote

Aryabhata, in the 5th century, already had a nearly correct picture of planetary motion — he knew the planets moved at varying speeds along their paths, though the exact geometric law was not formulated until Kepler. The Indian mathematical-astronomy tradition is one of the richest ancient contributions to the subject, and the "equal areas" pattern is, in a loose qualitative sense, visible in ancient planetary tables.

Kepler's third law — T² ∝ r³

Third law. The square of the orbital period is proportional to the cube of the semi-major axis:

T^2 \propto a^3.

This is the most useful of the three laws for everyday orbital calculations. Know the size of the orbit, and you know the period — instantly. It is also the easiest to derive, at least for the special case of a circular orbit, which is where we start.

Derivation for circular orbits

Take a satellite of mass m orbiting a much more massive body of mass M in a circle of radius r at speed v. Two pieces of physics:

Piece 1. The gravitational force provides the centripetal force.

\frac{GMm}{r^2} = \frac{mv^2}{r} \tag{1}

Why: the satellite is accelerating toward the centre (centripetal acceleration v^2/r), and the only force on it is gravity, directed toward the centre. So gravity = mass × centripetal acceleration. The mass m will cancel in a moment.

Solve for v^2:

v^2 = \frac{GM}{r}. \tag{2}

This is the orbital velocity formula — the speed at which a satellite must move in a circular orbit of radius r. (Derived in full in Escape Velocity and Orbital Velocity.)

Piece 2. The period is the circumference divided by the speed.

T = \frac{2\pi r}{v}.

Square both sides:

T^2 = \frac{4\pi^2 r^2}{v^2}.

Substitute v^2 = GM/r:

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

And there it is:

\boxed{\; T^2 = \frac{4\pi^2}{GM}\,r^3 \;}

Why: two lines of algebra, starting from Newton's law of gravity and the definition of centripetal force. The mass of the satellite cancels — the period depends only on the central mass M and the orbital radius r. A small satellite orbits with the same period as a large one, at the same altitude.

For an elliptical orbit, the result is the same with r replaced by the semi-major axis a:

T^2 = \frac{4\pi^2}{GM}\,a^3.

The proof for the elliptical case is beyond the scope of this article (it requires the conic-section machinery), but the circular case already contains the whole physical idea: T^2 scales as r^3, with a pre-factor involving only GM.

Using the third law

Write it as a ratio between two orbits around the same central body:

\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3.

This one equation lets you compute anything you want about an orbit given any other orbit around the same body. For example, take Earth's orbit (T = 1 year, a = 1 AU). Mars is at a = 1.524 AU. Its period should be

T_{\text{Mars}} = \sqrt{1.524^3} = \sqrt{3.540} \approx 1.881 \text{ years}.

The observed value is 1.881 years. Kepler's third law, verified to four decimals with a one-line calculation.

The Sun as the standard

For orbits around the Sun, it is useful to plug in Sun-mass units, so that 4\pi^2/GM_\odot becomes simply 1 \text{ yr}^2/\text{AU}^3. Then for any heliocentric orbit:

T \text{ (in years)} = a^{3/2} \text{ (in AU)}.

This is the form Kepler originally wrote down, before Newton ever derived it.

Geostationary orbit — a very specific altitude

A geostationary satellite is one that stays above the same point on Earth at all times. Since Earth rotates once per sidereal day (about 23 h 56 min 4 s, or 86,164 s — not exactly 24 hours, because the Sun's apparent motion includes Earth's orbital motion around the Sun), a geostationary satellite must orbit with exactly that period.

You cannot pick both the altitude and the period — Kepler's third law ties them together. So "geostationary" is not a choice but a calculation: find the radius r at which T = 86{,}164 s.

Step 1. Plug into the third law.

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

Use GM_\oplus = 3.986 \times 10^{14} m³/s² and T = 86{,}164 s:

r^3 = \frac{3.986 \times 10^{14}}{4\pi^2} \times (86{,}164)^2 = \frac{3.986 \times 10^{14}}{39.478} \times 7.424 \times 10^9
r^3 = (1.010 \times 10^{13}) \times (7.424 \times 10^9) = 7.497 \times 10^{22} \text{ m}^3
r = (7.497 \times 10^{22})^{1/3} \approx 4.216 \times 10^7 \text{ m} = 42{,}160 \text{ km}.

Step 2. Subtract Earth's radius to get the altitude.

h = r - R_\oplus = 42{,}160 - 6{,}371 = 35{,}789 \text{ km}.

So: a geostationary satellite must be placed at an altitude of about 35,800 km. Any other altitude gives a different period and the satellite drifts relative to the ground.

For that reason, the ring of geostationary orbit at 35,800 km is crowded. Every TV broadcast satellite, every weather satellite, every communications satellite trying to cover a fixed service area sits on that specific ring above Earth's equator. India's GSAT series, INSAT, and GSLV launches to GEO all target this altitude exactly.

One more constraint: the orbit plane must coincide with Earth's equator. A geostationary satellite over 15°N latitude would appear, from the ground, to bob north and south once a day. Only an equatorial orbit stays at a fixed sky point. So: altitude 35,800 km, inclination 0°, prograde direction (west to east). Everything about "geostationary" is determined.

Comparison of low Earth orbit, Moon's orbit, and geostationary orbit Earth at the centre with three concentric orbit paths: a low Earth orbit at about 400 km altitude shown as the innermost circle, geostationary orbit at 35800 km shown as a larger circle, and the Moon's orbit at 384400 km shown as the outermost circle for scale comparison. Earth LEO (T ≈ 92 min) GEO (T = 23 h 56 min) r = 42,160 km Moon's orbit (r ≈ 384,400 km) — 9× farther than GEO
Three orbit radii to scale (Earth's radius is exaggerated for visibility). The inner orbit is the ISS at $\sim$400 km altitude, period 92 min. The middle orbit is geostationary at 35,800 km altitude, period 23 h 56 min. The Moon (not shown on the same figure to scale) orbits nearly 9× farther than GEO.

Other specialised orbits

Kepler's third law pins down what period goes with what radius. A few standard choices:

Elliptical orbits and transfer trajectories

So far, everything has been circular. But most real orbits are ellipses of varying eccentricity, and the clever trick for moving a spacecraft cheaply between two orbits is a carefully designed ellipse called a Hohmann transfer.

The setup

Say a spacecraft is in a low, circular parking orbit around Earth at radius r_1, and you want to move it to a higher circular orbit at radius r_2. You cannot just turn sideways — gravity does not let you. The fuel-efficient way is:

  1. At the starting orbit, give the spacecraft a brief boost so its speed increases from the circular speed v_1 to some higher speed v_p. Its orbit becomes an ellipse whose perihelion (closest point to Earth) is r_1 and whose aphelion (farthest point) is r_2.
  2. Coast along the ellipse for half a period. No fuel burned — orbital motion is free.
  3. At aphelion, give another brief boost to raise the speed from the ellipse's aphelion speed v_a up to the circular speed v_2 for the new orbit. Now the spacecraft is in a circular orbit of radius r_2.

This is the Hohmann transfer orbit. It uses the bare minimum of fuel (two short burns) at the cost of a longer coasting time. It is what ISRO used for Mangalyaan's Mars Orbiter Mission and the lunar transfer legs of Chandrayaan-1, Chandrayaan-2, and Chandrayaan-3.

Hohmann transfer orbit between two circular orbits Earth at the origin with a small inner circular orbit (radius r1) and a larger outer circular orbit (radius r2). An ellipse connects a point on the inner orbit to the diametrically opposite point on the outer orbit, representing the Hohmann transfer. Arrows show the direction of motion and the two impulsive burns at the inner and outer transfer points. Earth r₁ (inner orbit) r₂ (outer orbit) burn 1 perihelion burn 2 aphelion
The Hohmann transfer ellipse (red) touches the inner circular orbit at perihelion (right) and the outer circular orbit at aphelion (left). **Burn 1** boosts the spacecraft from circular speed on the inner orbit into the elliptical transfer; **Burn 2** adds the final speed to circularise it at the outer radius. The two burns are the only fuel cost; the half-ellipse coast between them is free.

The transfer time

The transfer ellipse has semi-major axis a = (r_1 + r_2)/2. By Kepler's third law, its period is

T_{\text{transfer}} = 2\pi\sqrt{\frac{a^3}{GM}}.

The transfer itself is half an orbit, so

t_{\text{transfer}} = \tfrac{1}{2}T_{\text{transfer}} = \pi\sqrt{\frac{a^3}{GM}}.

For an Earth-to-Mars transfer (r_1 = 1 AU, r_2 = 1.524 AU, a = 1.262 AU), plugging in GM_\odot gives t_{\text{transfer}} \approx 259 days — the famous "about 9 months to Mars" result. For Mangalyaan in 2013, the actual transfer was 298 days, slightly longer than the pure Hohmann minimum because of additional manoeuvres.

Chandrayaan-3's orbital scheme was more elaborate. The Moon's orbit around Earth is much closer than Mars's to the Sun, so ISRO could use multiple small burns to progressively raise the apogee and eventually capture into lunar orbit — but every leg of that manoeuvre is an ellipse of the type you have been studying.

Worked examples

Example 1: Kepler's third law as a bridge — the ISS's altitude

The International Space Station orbits at roughly 400 km altitude and has a period of about 92 minutes. Use this to predict the altitude of a satellite with a 12-hour period (like a GPS satellite). Earth's radius is R_\oplus = 6371 km.

ISS and GPS orbital radii comparison Earth at the centre with two concentric circles representing the ISS orbit at radius 6771 km and the GPS orbit at radius about 26560 km. The GPS orbit is much larger. Earth ISS, T ≈ 92 min GPS, T = 12 h (find r)
ISS at low altitude has a short period. Kepler's third law gives the GPS orbit radius from a period ratio.

Step 1. Set up the ratio form of Kepler's third law.

\left(\frac{T_{\text{GPS}}}{T_{\text{ISS}}}\right)^2 = \left(\frac{r_{\text{GPS}}}{r_{\text{ISS}}}\right)^3

Why: both satellites orbit Earth, so GM is the same. Taking the ratio eliminates GM and leaves a pure geometric relation.

Step 2. Compute the ISS orbital radius.

r_{\text{ISS}} = R_\oplus + h_{\text{ISS}} = 6371 + 400 = 6771 \text{ km}

Step 3. Compute the period ratio.

T_{\text{ISS}} = 92 \text{ min}, \quad T_{\text{GPS}} = 12 \times 60 = 720 \text{ min}.
\frac{T_{\text{GPS}}}{T_{\text{ISS}}} = \frac{720}{92} = 7.826

Step 4. Solve for r_{\text{GPS}}.

\left(\frac{r_{\text{GPS}}}{r_{\text{ISS}}}\right)^3 = (7.826)^2 = 61.25
\frac{r_{\text{GPS}}}{r_{\text{ISS}}} = (61.25)^{1/3} \approx 3.942
r_{\text{GPS}} = 6771 \times 3.942 \approx 26{,}690 \text{ km}

Why: cube root of 61.25. A 3.94× ratio of radii gives a (3.94)^{3/2} = 7.83 ratio of periods — Kepler's third law in action.

Step 5. Altitude.

h_{\text{GPS}} = r_{\text{GPS}} - R_\oplus = 26{,}690 - 6371 \approx 20{,}320 \text{ km}

Result: A 12-hour orbit must have a radius of about 26,700 km, corresponding to an altitude of about 20,300 km. The actual GPS constellation orbits at 20,180 km — agreement to within 1%, the small difference due to using a rounded ISS period.

What this shows: Kepler's third law is a bridge. Measure any orbit's radius and period once, and you know them for every orbit around the same body. Knowing just that the ISS orbits at 400 km every 92 minutes is enough to predict the GPS altitude without any other input.

Example 2: The period of Chandrayaan's parking orbit

After its launch on 14 July 2023, Chandrayaan-3 was first placed in an elliptical Earth parking orbit with perigee (closest point) at 170 km altitude and apogee (farthest point) at 36,500 km altitude. What was the orbital period in this phase? Use GM_\oplus = 3.986 \times 10^{14} m³/s² and R_\oplus = 6371 km.

Elliptical parking orbit of Chandrayaan-3 Earth at the right focus of an elongated ellipse. Perigee at 170 km altitude is shown at the right vertex close to Earth; apogee at 36500 km altitude is at the left vertex far from Earth. Earth perigee (170 km) apogee (36,500 km) 2a = r_perigee + r_apogee
The parking orbit is a highly eccentric ellipse. The semi-major axis is half the sum of the perigee and apogee distances from Earth's centre.

Step 1. Compute the perigee and apogee distances from Earth's centre.

r_{\text{peri}} = 6371 + 170 = 6541 km = 6.541 \times 10^6 m

r_{\text{apo}} = 6371 + 36{,}500 = 42{,}871 km = 4.287 \times 10^7 m

Step 2. Compute the semi-major axis.

a = \frac{r_{\text{peri}} + r_{\text{apo}}}{2} = \frac{6.541 \times 10^6 + 4.287 \times 10^7}{2} = \frac{4.941 \times 10^7}{2} \approx 2.471 \times 10^7 \text{ m}

Why: the major axis of the ellipse is the straight-line segment from perigee to apogee, which passes through Earth's centre (the focus). Its length is the sum of the two distances from the focus to the vertices. The semi-major axis is half of that.

Step 3. Apply Kepler's third law.

T^2 = \frac{4\pi^2}{GM}\,a^3 = \frac{4\pi^2}{3.986 \times 10^{14}}\,(2.471 \times 10^7)^3
a^3 = (2.471)^3 \times 10^{21} = 15.07 \times 10^{21} = 1.507 \times 10^{22} \text{ m}^3
T^2 = \frac{4 \times 9.8696}{3.986 \times 10^{14}} \times 1.507 \times 10^{22} = \frac{39.478 \times 1.507 \times 10^{22}}{3.986 \times 10^{14}}
T^2 = \frac{5.951 \times 10^{23}}{3.986 \times 10^{14}} \approx 1.493 \times 10^9 \text{ s}^2
T \approx \sqrt{1.493 \times 10^9} \approx 3.864 \times 10^4 \text{ s}

Step 4. Convert to hours.

T \approx \frac{3.864 \times 10^4}{3600} \approx 10.73 \text{ hours} \approx 10 \text{ h } 44 \text{ min}.

Result: The parking orbit has a period of about 10.7 hours. This is what let ISRO plan the mission timeline: at each apogee pass, a small engine burn progressively raised the orbit by tens of thousands of kilometres, eventually stretching the apogee out to reach the Moon's distance. Each leg of the manoeuvre used Kepler's third law to predict when the spacecraft would return for the next burn.

What this shows: The third law does not care whether the orbit is circular or elliptical — it uses the semi-major axis a. For a circular orbit, a = r; for an ellipse, a is the average of perigee and apogee distances (from the centre, not the altitudes). The same formula covers both.

Common confusions

If you came here to compute orbital periods, geostationary altitudes, and transfer times, you have what you need. What follows is for readers who want the relation between total energy and orbit size, the full statement of Kepler's third law for elliptical orbits, and a glimpse of why the inverse-square law is uniquely special.

The vis-viva equation

Combine conservation of energy (E = \tfrac{1}{2}mv^2 - GMm/r) with the fact that for an elliptical orbit the total energy is related to the semi-major axis by

E = -\frac{GMm}{2a}.

(This is a result that falls out of the full orbit-derivation; I am quoting it here, and then deriving its consequences.) Solving for v in terms of r and a gives the vis-viva equation:

v^2 = GM\!\left(\frac{2}{r} - \frac{1}{a}\right).

This is a spectacularly useful formula. Plug in the current distance r and the orbit's semi-major axis a, and you immediately know the speed. Three special cases:

  • Circle (r = a): v^2 = GM/a. Matches the circular-orbit formula.
  • Perihelion (r = a(1 - \epsilon), where \epsilon is the eccentricity): v_{\text{peri}}^2 = GM \dfrac{1 + \epsilon}{a(1 - \epsilon)}.
  • Aphelion (r = a(1 + \epsilon)): v_{\text{apo}}^2 = GM \dfrac{1 - \epsilon}{a(1 + \epsilon)}.

The ratio of perihelion to aphelion speeds is

\frac{v_{\text{peri}}}{v_{\text{apo}}} = \frac{1 + \epsilon}{1 - \epsilon}.

Earth's \epsilon = 0.0167 gives v_{\text{peri}}/v_{\text{apo}} = 1.0340, a 3.4% speed variation over the year. Halley's \epsilon = 0.97 gives v_{\text{peri}}/v_{\text{apo}} = 65.7 — the comet moves 66 times faster near the Sun than it does at its farthest point.

Energy versus orbit size

For circular orbits, the total energy is

E = \tfrac{1}{2}mv^2 - \frac{GMm}{r} = \tfrac{1}{2}m \cdot \frac{GM}{r} - \frac{GMm}{r} = -\frac{GMm}{2r}.

And for elliptical orbits, E = -GMm/(2a), with a the semi-major axis. This is worth staring at:

  • The bigger the orbit, the less negative the energy — i.e., the system is less bound.
  • A circular orbit at radius r has total energy equal to the elliptical orbit with semi-major axis a = r — the orbit's shape doesn't enter the energy, only the semi-major axis does. Two orbits with the same a but different eccentricity have the same total energy.
  • Orbital energy is exactly half the potential energy (in magnitude), the other half being kinetic. This is a special case of the virial theorem for inverse-square forces.

Kepler's third law for elliptical orbits — the full version

For a two-body problem where both bodies have comparable masses (Earth and Sun, for example, or the two stars in a binary), the "central mass M" in T^2 = 4\pi^2 a^3/(GM) should be replaced by M_1 + M_2:

T^2 = \frac{4\pi^2}{G(M_1 + M_2)}\,a^3.

Here a is the semi-major axis of the relative orbit (one body as seen from the other). For a satellite around Earth (m \ll M), M_1 + M_2 \approx M_1 = M_\oplus and you recover the simpler form. For a binary star system, both masses must be included.

This refinement is how astrophysicists measure the masses of binary stars: observe the orbital period T and the orbit's semi-major axis a, and solve for M_1 + M_2.

Why inverse-square is special

The inverse-square law is not just any central force — it is one of only two power-law central forces that produce closed orbits (orbits that come back to the same point). The other is a Hooke's-law spring force \vec{F} = -k\vec{r}, which produces ellipses with the source at the centre (not a focus). Every other power-law gives orbits that precess — rosette shapes that never close.

This remarkable fact is called Bertrand's theorem, and it tells you that Kepler's first law (closed elliptical orbits) is a near-miraculous property of the 1/r^2 force specifically. Change the exponent even slightly — to 1/r^{2.001}, say — and Mercury's orbit would trace a slowly rotating ellipse instead of a closed one.

In fact, Mercury's orbit does precess, by 43 arcseconds per century beyond what Newtonian gravity predicts — an anomaly that Einstein's general relativity explained exactly in 1915. GR predicts that gravity is very slightly stronger than 1/r^2 in regions of deep gravitational potential, enough to nudge Mercury's ellipse around over millennia. For every other planet, the effect is too small to measure.

So Kepler's laws are exact for pure Newtonian 1/r^2 gravity, and they are approximately true for real gravity in our solar system. The small deviations — Mercury's perihelion precession, the bending of starlight near the Sun, the precession of binary pulsars — are the fingerprints of relativity.

A last word on conservation

The whole of Kepler's analysis rests on two conservation laws: energy (the first law is the shape of a conic-section; the energy decides which conic) and angular momentum (the second law is direct angular-momentum conservation). These two laws alone — applied to Newton's inverse-square force — are enough to reconstruct every orbital trajectory you will ever meet. Conservation laws are the reason the 1/r^2 force's behaviour is universal rather than messy: instead of tracking every moment of force and acceleration, you exploit the two invariants and the problem collapses. You have already seen this template in rolling motion (energy conservation collapsed the incline problem in three lines) and in collisions (momentum conservation gave you the elastic-collision formulas). The same conceptual move, applied to central forces, produces Kepler's laws and the entire edifice of celestial mechanics.

Where this leads next