Physics · Gravitation

Kepler’s third law for a two-body orbit

Q: How do you apply Kepler third law in practice?

For an unperturbed bound two-body orbit, $T^2=\frac{4\pi^2a^3}{G(M+m)}$. Here a is relative semimajor axis and M+m is total mass. With a in metres, masses in kilograms and SI G, T is in seconds. For a very light companion, replace M+m by M. Comparing two orbits with the same gravitational parameter gives $\frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}$. The shorthand $T^2=a^3$ without a constant requires a suitable normalisation; it is not a general equation for arbitrary units. On an ellipse, neither instantaneous separation nor altitude above a surface may replace a. State the model and unit choices before inserting numerical values.

Q: How do you calculate the altitude of a geostationary satellite?

In an idealised Earth model, set the period of a prograde circular orbit to one sidereal day, approximately 86164 s. Using $\mu{}\approx{}\frac{3.986\times{}10^{14} m^{3}}{s^{2}}$ gives $r=\sqrt[3]{\frac{\mu T^2}{4\pi^2}}$ ≈ 42164 km from Earth’s centre. For altitude above the equator, subtract the equatorial radius; approximately 6378 km gives about 35786 km. A mean Earth radius gives a slightly different altitude. Matching period alone does not make the orbit geostationary: it must also be equatorial, circular and prograde. An inclined or elliptical orbit with the same period is geosynchronous but does not remain apparently above a fixed ground point.

Q: How do you determine the mass of a celestial body with Kepler 3?

The general rearrangement first gives $M+m=\frac{4\pi^2a^3}{GT^2}$, the partners’ total mass. Use the relative orbit’s semimajor axis and their shared period. Only for negligible companion mass does the result approximately equal central mass alone. A binary with comparable masses cannot use that simplification. Separating masses needs additional information, such as the ratio of barycentric orbits. Geometry and inclination can also affect reconstruction of a from a limited observed orbital arc. The calculation requires positive values, matching units and a sufficiently accurate two-body approximation; measuring T alone cannot determine mass. Check which orbit the quoted semimajor axis actually describes.

Q: Why is $\frac{T^{2}}{a^{3}}$ the same for all planets?

In the simplified model of light planets around the same Sun, $\frac{T^2}{a^3}$ is approximately $\frac{4\pi^2}{GM}$. The common central mass explains the shared ratio. The general two-body form includes M+m, so substantial companion mass changes the constant. Real many-body perturbations are also absent from the simple equation. The circular derivation uses $\frac\mu{r^2}=\frac{v_{\mathrm{rel}}^2}r$ and circumference $2\pi r$. Ellipses additionally require constant areal rate and $h^2=\mu a(1-e^2)$, giving the same rule with semimajor axis. Agreement with observations supports the model but is not an experiment-free proof of the gravitational law. Other central bodies have different constants.

Q: What is the semi-major axis of an elliptical orbit?

The semimajor axis a is half the longest ellipse axis, measured from ellipse centre to edge. A bound relative Kepler orbit has $a=\frac{r_{\mathrm{closest}}+r_{\mathrm{farthest}}}2$. These separations are measured from the focus; current separation varies between them and does not generally equal a. For a circle, a is constant radius. It is also not generally time-averaged separation or mean altitude above a surface. At fixed gravitational parameter, period depends only on a rather than eccentricity. This statement belongs to the unperturbed two-body model, not an exact equality for arbitrary real planetary and comet orbits. Distinguish relative a from individual barycentric semimajor axes when masses are comparable.

An unperturbed bound two-body orbit satisfies $\frac{T^2}{a^3}=\frac{4\pi^2}{G(M+m)}$ . Semimajor axis is relative; for a light companion, total mass may be approximated by central mass.

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Formula

Calculate

\frac{T ^{2}}{a ^{3}} = \frac{4 π ^{2}}{G ( M + m )}

T in s · a in m · G in $\mathrm{m^3\,kg^{-1}\,s^{-2}}$ · M, m in kg

Variables & units

Variables & units
Symbol	Meaning	Unit
$T$	Shared orbital period of both bodies	s
$a$	Relative semimajor axis (sum of barycentric semimajor axes)	m
$G$	Gravitational constant	m³/(kg·s²)
M, m	Masses of the two gravitationally interacting bodies	kg

LaTeX

\frac{T^2}{a^3} = \frac{4\pi^2}{G(M+m)}

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Diagram: an (exaggerated) elliptical orbit with a planet, the central body M at one focus, the semi-major axis a is marked from the centre to the edge. — Ellipse of a light companion around M in the unperturbed model. a is the semimajor axis; for m ≪ M, T²/a³ approximates 4π²/(GM).

Orbital period calculator (Kepler’s third law)

Find orbital period, semi-major axis or gravitational parameter for an undisturbed orbit of two bodies.

Find

Semi-major axis a

Semi-major axis measured from the centre, e.g. 6771 km for the ISS: 6371 km Earth radius plus 400 km altitude.

Gravitational parameter μ = G(M + m)

μ = G · (M + m), 398,600 km³/s² for the Earth, about 1.327·10¹¹ km³/s² for the Sun.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for T, substitute, evaluate

1. Rearrange for T
$T=2\pi\sqrt{\frac{a^3}{\mu}}$
2. Substitute values (in SI units)
- a = 6771 km = 6 771 000 m
- μ = 398 600 km³/s² = 3.986 · 10¹⁴ m³/s²
$T=2\pi\sqrt{\frac{6771000^3}{398600000000000}}=5544.86$
3. Result
T = 92.41 min

Result: Orbital period T

T = 92.41 min

With a = 6771 km and μ = 398 600 km³/s², T = 92.41 min.

Quick tips

For the same μ, T₂ ÷ T₁ = (a₂ ÷ a₁)^(3/2): four times the semi-major axis, eight times the period.
a is measured from the centre, not the surface: for satellites add the Earth radius of 6371 km to the altitude first.

Assumptions and limits

Unperturbed bound Newtonian two-body orbit. a is relative-orbit semi-major axis and μ = G(M+m) the total parameter. Not altitude, instantaneous separation or one barycentric semi-axis; a, T and μ are positive.

Derivation & background

For a relative circle, $\frac\mu{r^2}=\frac{v_{\mathrm{rel}}^2}r$ and $v_{\mathrm{rel}}=\frac{2\pi r}T$ give the relation using $\mu=G(M+m)$ . For ellipses, constant areal rate and $h^2=\mu a(1-e^2)$ produce the same period rule. Here h denotes specific angular momentum rather than altitude. Different central masses or substantial companions generally produce different ratios.

Exam blueprint

Validity range

Unperturbed bound Newtonian two-body orbit $(0\le{}e<1)$ , point masses or suitable exterior spherical fields. a is the relative semimajor axis; $GM$ approximates m≪M.

Derivation steps

The circular orbit explains scaling; the general two-body ellipse follows from constant areal rate and its semimajor axis.

1Define $\mu=G(M+m)$ and relative semimajor axis a rather than an individual barycentric orbit.
2For a circle $a=r$ , use $\frac\mu{r^2}=\frac{v_{\mathrm{rel}}^2}r$ and $v_{\mathrm{rel}}=\frac{2\pi r}T$ , obtaining $T^2=\frac{4\pi^2r^3}\mu$ .
3The ellipse conserves $h=|\vec r\times\vec v_{\mathrm{rel}}|$ . Areal rate is $\frac h2$ and area $\pi ab$ .
4Hence $T=\frac{2\pi ab}h$ , with $b=a\sqrt{1-e^2}$ .
5The orbit $r=\frac{h^2}{\mu(1+e\cos\theta)}$ gives $a=\frac{h^2}{\mu(1-e^2)}$ from nearest and farthest separation.
6Insert $h^2=\mu a(1-e^2)$ into the period equation. Eccentricity cancels, leaving $T^2=\frac{4\pi^2a^3}\mu$ .
7Only for $m\ll M$ simplify to $\mu\approx GM$ ; different central or substantial companion masses change the ratio.

Rearrangements

Total mass

M+m=\frac{4\pi^2a^3}{GT^2}

Use relative semimajor axis. This approximates central mass only for a negligible companion.

Orbital period

T=2\pi\sqrt{\frac{a^3}{G(M+m)}}

For a bound two-body orbit. Eccentricity does not change period at fixed semimajor axis and total mass.

Relative semimajor axis

a=\sqrt[3]{\frac{G(M+m)T^2}{4\pi^2}}

Gives centre separation for a circle, not altitude. Subtract the appropriate Earth radius for geostationary altitude and check the other orbital conditions.

Task variant

Jupiter needs 11.86 years per solar orbit. What is its semi-major axis in AU?

For the same solar central mass and negligible companion masses, $\left(\frac a{1\,\mathrm{AU}}\right)^3\approx\left(\frac T{1\,\mathrm{year}}\right)^2$ .
Thus $a\approx\sqrt[3]{11.86^2}\,\mathrm{AU}\approx5.2\,\mathrm{AU}$ .

The Moon orbits Earth in 27.32 days with relative semimajor axis $a=3.844\cdot10^{8}\,\mathrm{m}$ . Estimate Earth’s mass by neglecting the Moon’s mass compared with it.

$T\approx2.36\cdot10^6\,\mathrm s$ .
Neglecting lunar mass gives $M\approx\frac{4\pi^2a^3}{GT^2}$ .
Substitution gives $M\approx\frac{39.48\cdot5.68\cdot10^{25}}{6.674\cdot10^{-11}\cdot5.57\cdot10^{12}}\,\mathrm{kg}\approx6.0\cdot10^{24}\,\mathrm{kg}$ .
Without that approximation, Kepler’s law determines the sum of Earth and Moon masses.

Common mistakes

Using a satellite altitude above ground as the orbital radius.

For a circle use $r=R_E+h$ ; for an ellipse obtain semimajor axis from centre separations. The mean Earth radius of about 6371 km differs from equatorial radius.

Substituting periods in days or years into the SI form.

The form with G and M needs T in seconds; AU-year units only for pure ratios in the solar system.

Comparing orbits around different central bodies.

The ratio $\frac{T^2}{a^3}=\frac{4\pi^2}{G(M+m)}$ agrees only for the same gravitational parameter. The same central body alone is insufficient unless companion masses are negligible or equal.

Exam context

Typical: computing the geostationary orbit altitude, determining central masses from moon or satellite data and checking planetary data via the ratio $\frac{T^2}{a^3}$ .

Formula cluster

Gravity and orbits

Circular motion explains the special case; angular momentum and constant areal rate extend the law to an ellipse.

F = G·m₁m₂/r²

Newton's Law of Universal Gravitation

Fz = m·v²/r

Centripetal Force (Circular Motion)

Worked example

Two light companions of the same source with $a_2=1.524a_1$ have $\frac{T_2}{T_1}=1.524^{3/2}\approx1.8814$ . This is a ratio calculation; $T^2=a^3$ without a constant is not a general SI equation. A geostationary circle has centre separation about 42 164 km. Besides matching period, it must be equatorial and prograde relative to Earth’s rotation.

Applications

Satellite orbits (GPS, geostationary), mass determination of stars and planets, exoplanet detection, spaceflight mission planning

How do you calculate with Kepler Third Law?

Here is how to work through a typical Kepler Third Law (T²/a³ = 4π²/[G(M+m)]) task step by step:

1
Task
Jupiter needs 11.86 years per solar orbit. What is its semi-major axis in AU?
Solution path
1. For the same solar central mass and negligible companion masses, $\left(\frac a{1\,\mathrm{AU}}\right)^3\approx\left(\frac T{1\,\mathrm{year}}\right)^2$ .
2. Thus $a\approx\sqrt[3]{11.86^2}\,\mathrm{AU}\approx5.2\,\mathrm{AU}$ .
2
Task
The Moon orbits Earth in 27.32 days with relative semimajor axis $a=3.844\cdot10^{8}\,\mathrm{m}$ . Estimate Earth’s mass by neglecting the Moon’s mass compared with it.
Solution path
1. $T\approx2.36\cdot10^6\,\mathrm s$ .
2. Neglecting lunar mass gives $M\approx\frac{4\pi^2a^3}{GT^2}$ .
3. Substitution gives $M\approx\frac{39.48\cdot5.68\cdot10^{25}}{6.674\cdot10^{-11}\cdot5.57\cdot10^{12}}\,\mathrm{kg}\approx6.0\cdot10^{24}\,\mathrm{kg}$ .
4. Without that approximation, Kepler’s law determines the sum of Earth and Moon masses.

Understand and apply the equation

Orbital period depends on orbital size and the total mass of a two-body system. For an ellipse, use its semimajor axis rather than instantaneous separation. This guide explains the usual approximation, connects circular and elliptical derivations, and provides original examples with explicit unit checks.

Which orbit size determines the period?

A more distant companion takes longer to orbit within the same gravitational system. For an ellipse, the relevant fixed geometric quantity is its semimajor axis: half the longest ellipse axis. Instantaneous separation instead changes along the orbit. Begin with the circle, then explore why the same rule extends to ellipses.[1]

Prior knowledge: radius, squared and cubed quantities, and period. A relative orbit describes one partner’s position relative to the other.

What is Kepler’s third law, and which mass counts?

\begin{aligned}&T^2=\frac{4\pi^2a^3}{G(M+m)}=\frac{4\pi^2a^3}{\mu}\\&\mu=G(M+m)\end{aligned}

a is the semimajor axis of the relative separation-vector orbit; $\mu$ is the gravitational parameter in $\mathrm{m^3\,s^{-2}}$ .[3]

Consider two point masses M and m interacting only through their mutual Newtonian attraction. Both orbit their common centre of mass. Their individual semimajor axes $a_M$ and am sum to relative semimajor axis a. Both have the same period. The usual $GM$ denominator neglects m compared with M.[3]

\begin{aligned}&a=a_M+a_m\\&a_M=\frac{m}{M+m}a\\&a_m=\frac{M}{M+m}a\end{aligned}

For comparable masses, distinguish the relative orbit from the two barycentric orbits.[3]

If $\frac mM$ is small, $\frac{T^2}{a^3}$ is approximately $\frac{4\pi^2}{GM}$ . Two light companions of the same source then share this ratio. It is not universally equal across different central bodies or non-negligible companion masses.[5]

View diagram: Fourfold semimajor axis means eightfold period

Fourfold semimajor axis means eightfold period

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At fixed gravitational parameter,

\frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}

. Four times the relative semimajor axis gives eight times the period. This applies to unperturbed bound circles and ellipses; semimajor axis is neither instantaneous separation nor altitude.[1]

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What is the semi-major axis of an ellipse?

\begin{aligned}&a=\frac{r_{\mathrm p}+r_{\mathrm a}}2\\&r_{\mathrm p}=a(1-e)\\&r_{\mathrm a}=a(1+e)\end{aligned}

rp and ra are minimum and maximum relative separations; 0 ≤ e < 1 describes a bound ellipse.[1]

The major axis passes through both foci and has length 2a. Its semimajor axis runs from the ellipse centre to the edge. The gravitational reference point is at a focus, not the ellipse centre. In the circular limit $e=0$ these points coincide and a is constant orbital radius.[1]

a is not mean altitude above a planet’s surface. Given closest and farthest altitudes, first add planet radius to each, then average the resulting separations. Time-averaged separation along an ellipse is generally not equal to a either. The third law uses the geometric semimajor axis.

How does Newton’s law give the circular-orbit relation?

Newton’s law of gravitation supplies mutual attraction in the two-body model used here. Let $\vec r$ point from M to m. Subtracting their accelerations gives $\ddot{\vec r}=-\frac{\mu\vec r}{r^3}$ . A relative circular orbit has radial acceleration $\frac{v_{\mathrm{rel}}^2}{r}$ . Here $v_{\mathrm{rel}}$ is their relative speed, not automatically either body’s speed about the centre of mass.[3]

\begin{aligned}&\frac{\mu}{r^2}=\frac{v_{\mathrm{rel}}^2}{r}\\&v_{\mathrm{rel}}=\frac{2\pi r}{T}\\&T^2=\frac{4\pi^2r^3}{\mu}\end{aligned}

Substitution gives the third law for $a=r$ . This calculation alone does not prove the elliptical case.[2]

Explore further: Why ellipses follow the same $a^3$ rule

For the Newtonian relative orbit, specific angular-momentum magnitude $h=|\vec r\times\vec v_{\mathrm{rel}}|$ is constant, in $\mathrm{m^2\,s^{-1}}$ . Central acceleration produces no torque, so swept area per time is $\frac h2$ . Over one period the swept area is $\pi ab$ , where $b=a\sqrt{1-e^2}$ .[1][4]

T=\frac{2\pi ab}{h}=\frac{2\pi a^2\sqrt{1-e^2}}h

This expression still contains h. The inverse-square force relates h to orbital shape.[4][3]

In polar coordinates, $r^2\dot\theta=h$ . Set $u=\frac1r$ and differentiate with respect to angle $\theta$ instead of time. The radial motion equation becomes $u^{\prime\prime}+u=\frac\mu{h^2}$ , where primes denote angle derivatives. Its solution is $u=\frac\mu{h^2}(1+e\cos\theta)$ when $\theta=0$ points toward closest approach.[3]

\begin{aligned}r&=\frac{\frac{h^2}{\mu}}{1+e\cos\theta}\\{} a&=\frac{h^2}{\mu(1-e^2)}\\{} h^2&=\mu a(1-e^2)\end{aligned}

The relation for a also follows by averaging separations at $\theta=0$ and $\theta$ = π.[4][3]

T^2=\frac{4\pi^2a^4(1-e^2)}{h^2}=\frac{4\pi^2a^3}{\mu}

Eccentricity cancels: bound ellipses with the same a and $\mu$ have the same period.

This derivation assumes Newtonian attraction and an unperturbed two-body orbit. It establishes a consequence within the model rather than the experimental origin of the gravitational law. For $e\ge1$ there is no bound complete orbital period of this kind.

Calculating period, semi-major axis and total mass

\begin{aligned}T&=2\pi\sqrt{\frac{a^3}{\mu}}\\{} a&=\left(\frac{\mu T^2}{4\pi^2}\right)^{\frac{1}{3}}\\{} M+m&=\frac{4\pi^2a^3}{GT^2}\end{aligned}

T, a and $\mu$ must be positive; G is positive. The mass rearrangement initially gives total mass.

With a in metres and $\mu$ in $\mathrm{m^3\,s^{-2}}$ , $\frac{a^3}\mu$ has units of seconds squared. Convert kilometres to metres before cubing; a factor of 1000 in a gives a factor of one billion in $a^3$ . Convert days to seconds by multiplying by 86400.

Writing $T^2=a^3$ without a constant uses a special normalisation, not a general SI equation. For two orbits with the same $\mu$ , periods and semimajor axes may each use matching arbitrary units: $\frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}$ . Different $\mu$ adds the factor $\sqrt{\frac{\mu_1}{\mu_2}}$ .

Calculating an ellipse and two comparable-mass partners

An elliptical orbit has varying speed during a revolution. Speed on a circular orbit is a special case with the appropriate constant radius; for comparable masses, relative motion must still be distinguished from motion about the centre of mass.[1][2]

Period of a specified ellipse

Given: Relative two-body model with $\mu=4.00\cdot10^{14}\,\mathrm{m^3\,s^{-2}}$ ; closest separation 7200 km, farthest separation 10800 km.
Find: Find semimajor axis and period.

Approach: For the relative ellipse use $a=\frac{r_{\min}+r_{\max}}2$ and $T=2\pi\sqrt{\frac{a^3}{\mu}}$ . Average centre separations and convert to metres.
$a=\frac{7200+10800}{2}=9000\,\mathrm{km}=9.00\times10^6\,\mathrm m$
Cube semimajor axis and divide by $\mu$ .
$\frac{a^3}{\mu}=\frac{7.29\times10^{20}}{4.00\times10^{14}}=1.8225\times10^6\,\mathrm{s^2}$
Use the positive root and factor 2π.
$T=2\pi\sqrt{1.8225\times10^6}=8482.30\,\mathrm s\approx141.37\,\mathrm{min}$

Result: a is 9000 km; model period is approximately 8482 s or 141.4 minutes.

Check: $e=\frac{r_{\max}-r_{\min}}{r_{\max}+r_{\min}}=\frac{10800-7200}{10800+7200}=0.20$ . A circle with the same a and $\mu$ has the same period; inserting 7200 km alone describes a different orbit.

Two companions of comparable mass

Given: $M=3.00\cdot10^{30}\,\mathrm{kg}$ , $m=1.00\cdot10^{30}\,\mathrm{kg}$ ; relative semimajor axis $a=2.00\cdot10^{11}\,\mathrm{m}$ ; $G=6.67430\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ . Unperturbed model.
Find: Calculate period and both barycentric semimajor axes.

Approach: Both masses contribute to $\mu=G(M+m)$ ; the relative semimajor axis determines $T=2\pi\sqrt{\frac{a^3}{G(M+m)}}$ . Use total mass to find $\mu$ .
$\mu=G(M+m)=2.66972\times10^{20}\,\mathrm{m^3/s^2}$
Calculate period from the relative orbit.
$\begin{aligned}T&=2\pi\sqrt{\frac{(2.00\times10^{11})^3}{2.66972\times10^{20}}}\,\mathrm s\\&\approx3.43947\times10^7\,\mathrm s\\&\approx398.09\,\mathrm d\end{aligned}$
Partition the semimajor axis inversely to mass.
$\begin{aligned}a_M&=5.00\times10^{10}\,\mathrm m\\{} a_m&=1.50\times10^{11}\,\mathrm m\end{aligned}$

Result: Both have a period of approximately 398.1 days. Their barycentric semimajor axes sum to 2.00×10¹¹ m.

Check: $Ma_M=ma_m$ checks the centre of mass. Neglecting m in the denominator would overestimate period by $\sqrt{\frac43}\approx1.1547$ .

What does a period measurement determine, and when is an orbit geostationary?

T and relative semimajor axis first determine $\mu$ , and G then gives $M+m$ . Separating masses requires additional information, such as the ratio of barycentric orbits. For a very light satellite, total mass is effectively planet mass. Measurement uncertainty in a and T determines accuracy. The vis-viva equation for elliptical orbits adds instantaneous separation to determine local relative-orbit speed; period alone is insufficient.[1][2]

A geostationary orbit needs more than the correct period: it must be circular, equatorial and rotate in the same direction as Earth. The same period on an inclined or elliptical orbit gives geosynchronous motion, not a fixed sky position. To obtain altitude, subtract the selected planet radius from circular orbital radius.[2][6]

Other bodies, oblateness, drag, thrust manoeuvres and relativistic effects may change a real orbit. The two-body formula then approximates the motion or describes instantaneous orbital elements. A positive calculated T does not establish collision avoidance or stability under these perturbations.[3]

Five problems with solutions

Calculate · Foundation

Two light companions of the same source have a₂ = 4a₁. Find $\frac{T_2}{T_1}$ .

Show hint

Raise the semimajor-axis ratio to the power $\frac32$ .

Learning goal: Calculate period scaling from a semimajor-axis ratio.

Show solution and calculation

Approach: At fixed $\mu$ , compare relative semimajor axes through $\frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}$ .
$\frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}=4^{3/2}=8$ .

T_2=8T_1

Calculate · Application

A relative ellipse has closest separation 3.00×10⁷ m and farthest separation 5.00×10⁷ m. For $\mu=1.00\cdot10^{15}\,\mathrm{m^3\,s^{-2}}$ , find a and T.

Show hint

First find a as half the sum.

Learning goal: Find ellipse semimajor axis from extreme separations and calculate period.

Show solution and calculation

Approach: First find $a=\frac{r_{\min}+r_{\max}}2$ , then use $T=2\pi\sqrt{\frac{a^3}{\mu}}$ .
$a=4.00\cdot10^7\,\mathrm m$ .
$T=2\pi\sqrt{\frac{6.40\cdot10^{22}}{10^{15}}\,\mathrm{s^2}}\approx5.0265\cdot10^4\,\mathrm s$ , approximately 13.96 hours.

T=16000\pi\,\mathrm s

Transfer · Application

$a=1.00\cdot10^{9}\,\mathrm{m}$ and T = 10.0 days. Use $G=6.67430\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ to find total mass.

Show hint

Convert days to seconds and use the mass rearrangement.

Learning goal: Infer total mass from orbit size and period converted to seconds.

Show solution and calculation

Approach: Convert days to seconds and use $M+m=\frac{4\pi^2a^3}{GT^2}$ .
With $T=864000\,\mathrm s$ , this gives $M+m\approx7.9237\cdot10^{26}\,\mathrm{kg}$ .
The measurement determines the sum rather than individual masses.

Check: substituting total mass into $T=2\pi\sqrt{\frac{a^3}{G(M+m)}}$ gives ten days again.

M+m\approx7.9237\times10^{26}\,\mathrm{kg}

Interpret · Application

Two negligibly light satellites orbit the same point mass in the Newtonian model. Orbit A is circular with radius 12 000 km; orbit B is elliptical with closest distance 6000 km and greatest distance 18 000 km. Compare their periods. Does equal period also mean both have the same speed throughout their orbits?

Show hint

For an ellipse, semimajor axis is the mean of closest and greatest distances. The third law gives period, not the local speed profile.

Learning goal: Recognize equal semimajor axes as the period criterion despite different orbit shapes.

Show solution and calculation

Approach: establish each semimajor axis and apply Kepler’s third law as a ratio.
Both have semimajor axis 12 000 km and therefore equal periods.
Distance and speed vary along the ellipse; equal period does not mean equal local speed.

Check: eccentricity is 0.50 for the ellipse and zero for the circle, but $a^3$ agrees.

\begin{aligned}a_B&=\frac{r_{\min}+r_{\max}}2=\frac{6000+18\,000}{2}\,\mathrm{km}=12\,000\,\mathrm{km}\\\frac{T_B}{T_A}&=\left(\frac{a_B}{a_A}\right)^{3/2}=1\\e_B&=\frac{r_{\max}-r_{\min}}{r_{\max}+r_{\min}}=0.50\end{aligned}

[1][3]

Error analysis · Challenge

Two equal-mass stars follow circular orbits around their common centre of mass. Each orbital radius is $b=2.0\cdot10^{9}\,\mathrm{m}$ and both have the same period T. A calculation uses $a=b$ in $T^2=\frac{4\pi^2a^3}{G(M+m)}$ . Which semimajor axis belongs in this equation, and by what factor does the wrong a underestimate total mass?

Show hint

The equation describes the relative orbit: separation of one star from the other. Equal masses lie on opposite sides of the centre of mass.

Learning goal: Separate the relative two-body orbit from barycentric orbits when inferring mass.

Show solution and calculation

Approach: add the two barycentric radii to obtain the relative circular orbit.
The required semimajor axis is 4.0·10⁹ m.
For fixed T, inferred total mass is proportional to its cube.
Using b therefore gives only one eighth of the correct mass.

Check: doubling a multiplies $a^3$ by eight while the measured period remains unchanged.

\begin{aligned}a_{\mathrm{rel}}&=b+b=2b=4.0\cdot10^9\,\mathrm m\\M+m&=\frac{4\pi^2a_{\mathrm{rel}}^3}{GT^2}\\\frac{(M+m)_{\mathrm{wrong}}}{(M+m)_{\mathrm{correct}}}&=\left(\frac{b}{2b}\right)^3=\frac18\end{aligned}

[3]

What to remember

Use the relative orbit’s semimajor axis and both partners’ total mass. A circle is the simple special case; an ellipse with the same semimajor axis has the same period at fixed gravitational parameter. Axis and time units must match the constant used. Matching period alone does not make a satellite geostationary.[1]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made period, relative semi-major axis, total mass, circular/elliptical derivations and geostationary conditions accessible through learning questions; retained comparable-mass calculations and linked gravitation, circular-orbit speed and vis-viva in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an individual introduction and conclusion, checked existing calculations, clarified notation and model references, and added an original learning figure.

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Frequently asked questions about Kepler Third Law

How do you apply Kepler third law in practice?+

For an unperturbed bound two-body orbit, $T^2=\frac{4\pi^2a^3}{G(M+m)}$ . Here a is relative semimajor axis and M+m is total mass. With a in metres, masses in kilograms and SI G, T is in seconds. For a very light companion, replace M+m by M. Comparing two orbits with the same gravitational parameter gives $\frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}$ . The shorthand $T^2=a^3$ without a constant requires a suitable normalisation; it is not a general equation for arbitrary units. On an ellipse, neither instantaneous separation nor altitude above a surface may replace a. State the model and unit choices before inserting numerical values.

How do you calculate the altitude of a geostationary satellite?+

In an idealised Earth model, set the period of a prograde circular orbit to one sidereal day, approximately 86164 s. Using $\mu{}\approx{}\frac{3.986\times{}10^{14} m^{3}}{s^{2}}$ gives $r=\sqrt[3]{\frac{\mu T^2}{4\pi^2}}$ ≈ 42164 km from Earth’s centre. For altitude above the equator, subtract the equatorial radius; approximately 6378 km gives about 35786 km. A mean Earth radius gives a slightly different altitude. Matching period alone does not make the orbit geostationary: it must also be equatorial, circular and prograde. An inclined or elliptical orbit with the same period is geosynchronous but does not remain apparently above a fixed ground point.

How do you determine the mass of a celestial body with Kepler 3?+

The general rearrangement first gives $M+m=\frac{4\pi^2a^3}{GT^2}$ , the partners’ total mass. Use the relative orbit’s semimajor axis and their shared period. Only for negligible companion mass does the result approximately equal central mass alone. A binary with comparable masses cannot use that simplification. Separating masses needs additional information, such as the ratio of barycentric orbits. Geometry and inclination can also affect reconstruction of a from a limited observed orbital arc. The calculation requires positive values, matching units and a sufficiently accurate two-body approximation; measuring T alone cannot determine mass. Check which orbit the quoted semimajor axis actually describes.

Why is

\frac{T^{2}}{a^{3}}

the same for all planets?+

In the simplified model of light planets around the same Sun, $\frac{T^2}{a^3}$ is approximately $\frac{4\pi^2}{GM}$ . The common central mass explains the shared ratio. The general two-body form includes M+m, so substantial companion mass changes the constant. Real many-body perturbations are also absent from the simple equation. The circular derivation uses $\frac\mu{r^2}=\frac{v_{\mathrm{rel}}^2}r$ and circumference $2\pi r$ . Ellipses additionally require constant areal rate and $h^2=\mu a(1-e^2)$ , giving the same rule with semimajor axis. Agreement with observations supports the model but is not an experiment-free proof of the gravitational law. Other central bodies have different constants.

What is the semi-major axis of an elliptical orbit?+

The semimajor axis a is half the longest ellipse axis, measured from ellipse centre to edge. A bound relative Kepler orbit has $a=\frac{r_{\mathrm{closest}}+r_{\mathrm{farthest}}}2$ . These separations are measured from the focus; current separation varies between them and does not generally equal a. For a circle, a is constant radius. It is also not generally time-averaged separation or mean altitude above a surface. At fixed gravitational parameter, period depends only on a rather than eccentricity. This statement belongs to the unperturbed two-body model, not an exact equality for arbitrary real planetary and comet orbits. Distinguish relative a from individual barycentric semimajor axes when masses are comparable.

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Subject references and further reading

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[1]OpenStax, University Physics Vol. 1, §13.5 Kepler’s Laws of Planetary Motion

Related formulas

\omega = 2\pi f = \frac{2\pi}{T} \qquad v = \omega \cdot r

Angular Velocity and Angular Frequency

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

Vis-Viva Equation for Elliptical Orbits

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

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Kepler’s third law for a two-body orbit

Variables & units

Orbital period calculator (Kepler’s third law)

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How do you calculate with Kepler Third Law?

Which orbit size determines the period?

What is Kepler’s third law, and which mass counts?

Fourfold semimajor axis means eightfold period

What is the semi-major axis of an ellipse?

How does Newton’s law give the circular-orbit relation?

Explore further: Why ellipses follow the same a3a^3a3 rule

Calculating period, semi-major axis and total mass

Calculating an ellipse and two comparable-mass partners

Period of a specified ellipse

Two companions of comparable mass

What does a period measurement determine, and when is an orbit geostationary?

Five problems with solutions

Two light companions of the same source have a2 = 4a1. Find T2T1\frac{T_2}{T_1}T1​T2​​.

A relative ellipse has closest separation 3.00×107 m and farthest separation 5.00×107 m. For μ=1.00⋅1015 m3 s−2\mu=1.00\cdot10^{15}\,\mathrm{m^3\,s^{-2}}μ=1.00⋅1015m3s−2, find a and T.

a=1.00⋅109 ma=1.00\cdot10^{9}\,\mathrm{m}a=1.00⋅109m and T = 10.0 days. Use G=6.67430⋅10−11 m3 kg−1 s−2G=6.67430\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}G=6.67430⋅10−11m3kg−1s−2 to find total mass.

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Explore further: Why ellipses follow the same $a^3$ rule

Two light companions of the same source have a₂ = 4a₁. Find $\frac{T_2}{T_1}$ .

A relative ellipse has closest separation 3.00×10⁷ m and farthest separation 5.00×10⁷ m. For $\mu=1.00\cdot10^{15}\,\mathrm{m^3\,s^{-2}}$ , find a and T.

$a=1.00\cdot10^{9}\,\mathrm{m}$ and T = 10.0 days. Use $G=6.67430\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ to find total mass.