Dynamical Masses from Kepler's Third Law: A Derivation

Dynamical Masses from Kepler’s Third Law: A Derivation

One of the most important reasons binary stars are astrophysical laboratories is that their masses can be measured dynamically.

The central relation is

(1)

where

• M₁ and M₂ are the stellar masses,
• P is the common orbital period,
• a is the semimajor axis of the relative orbit,
• G is the Newtonian gravitational constant.

Equation (1) is the Newtonian two-body form of Kepler’s third law.

It is the basis of the dynamical-mass discussion in BIN01.

The most important geometric point is

(2)

where a₁ and a₂ are the semimajor axes of the two stars about the center of mass.

The a appearing in Equation (1) is not the orbit size of either star individually.

1 Why this relation measures mass

For an isolated Newtonian binary, orbital motion is produced by the gravitational attraction of the two stars.

The orbital size and orbital period therefore encode the strength of The Gravitational Field.

A larger total mass produces stronger gravitational acceleration.

For a fixed orbital size, a larger total mass implies a shorter orbital period.

For a fixed period, a larger orbital size requires a larger total mass.

Equation (1) converts this statement into a quantitative measurement.

The mass is called dynamical because it is obtained from the motion caused by gravity rather than inferred only from luminosity, temperature, or a stellar-evolution model.

2 First derivation: circular orbit intuition

The clearest introductory derivation begins with circular orbits.

Let the two stars have masses

(1)

and

(2)

Let their distances from the center of mass be

(3)

and

(4)

Their separation is

(3)

Because both stars complete one revolution in the same orbital period P, they have the same angular speed

(4)

3 Force balance for star 1

The gravitational force magnitude is

(5)

For circular motion, star 1 requires centripetal force

(5)

Gravity supplies that force:

(6)

Cancel M₁:

(6)

4 Force balance for star 2

The same gravitational force acts on star 2.

Its circular-motion equation is

(7)

Cancel M₂:

(7)

5 Add the two equations

Add Equations (6) and (7):

(8)

Factor the left side:

(9)

Using

(10)

we obtain

(11)

Multiply by a²:

(8)

Solve for total mass:

(12)

Now insert

(13)

Then

M1 + M2	= 2	(14)
	= .	(15)

Thus

(9)

This is the desired relation.

6 What the circular derivation teaches

The circular derivation is especially useful because every step has a simple physical meaning.

It shows that:

1. gravity provides the centripetal acceleration,
2. both stars have the same angular speed,
3. both stars contribute to the orbital separation,
4. the total mass, not only one stellar mass, controls the relative orbital period.

However, Equation (9) is not restricted to circular orbits.

The same result holds for elliptical Keplerian binaries.

7 Second derivation: the general two-body problem

Let the inertial-frame positions be

(16)

and

(17)

Define the relative vector

(10)

Newton’s equations are

M1r1	= −Gr,	(18)
M2r2	= + Gr.	(19)

Divide the first by M₁ and the second by M₂, then subtract:

r	= r1 −r2	(20)
	= −Gr − Gr.	(21)

Therefore

(11)

The two-star problem has become one effective Kepler problem for the relative coordinate.

8 The gravitational parameter of the relative orbit

Equation (11) has the same form as the standard Kepler equation

(22)

with

(12)

The total stellar mass therefore plays the role that a central mass would play in a test-particle Kepler problem.

This is why the period of a binary depends on

(23)

9 Angular momentum and areal velocity

Because the force in Equation (11) is a central force,

(24)

Define specific angular momentum

(13)

Then

(14)

The motion is therefore planar.

The areal velocity is

(15)

This is Kepler’s second law.

10 Specific angular momentum of an ellipse

For a Kepler ellipse with semimajor axis a and eccentricity e,

(16)

Thus

(17)

The semiminor axis of the ellipse is

(18)

The area of the ellipse is therefore

A	= πab	(25)
	= πa2.	(19)

11 Use the time required to sweep the whole ellipse

The orbital period is the time required for the radius vector to sweep the entire orbital area.

Since

(26)

we have

(27)

Substitute Equations (15) and (19):

P	=	(28)
	= .	(29)

Now insert Equation (17):

(30)

The eccentricity factors cancel:

(31)

Simplify the powers of a:

(20)

Square:

(32)

Rearrange:

(21)

This is the same result obtained from the circular derivation.

12 Why eccentricity disappears

A striking feature of Equation (21) is that it contains no eccentricity.

At first this can seem surprising.

An eccentric binary moves much faster near periastron than near apastron.

However, the ellipse area contains the factor

(33)

and the angular momentum contains the same factor.

They cancel when the full orbital period is calculated.

Therefore

(34)

This does not mean that the instantaneous speed or separation is independent of eccentricity.

Only the period for a given a and total mass has this property.

13 Relative orbit versus barycentric orbits

The individual stellar semimajor axes satisfy

(22)

and

(23)

Adding them gives

(35)

so

(24)

This is the orbital length scale required in Equation (21).

Using only a₁ or a₂ in the total-mass formula would give the wrong result.

14 Astronomical-unit form

Equation (21) becomes especially simple in standard astronomical units.

For Earth’s orbit around the Sun,

(36)

to the usual Newtonian approximation.

Therefore, when

• a is in AU,
• P is in years,
• mass is in solar masses,

Equation (21) becomes

(25)

This is the form most often used for visual binaries.

15 Visual binaries and parallax

A visual-binary orbit usually yields an angular semimajor axis

(37)

Kepler’s law requires a physical semimajor axis.

If the annual parallax is

(38)

then

(26)

Substitute this into Equation (25):

(27)

This is the direct observational route from an angular visual orbit to total dynamical mass.

16 Numerical example

Suppose a visual binary has

a′′	= 0.50 arcsec,	(39)
ϖ	= 25 mas,	(40)
P	= 20 yr.	(41)

Convert parallax to arcseconds:

(42)

The physical semimajor axis is

a	=	(43)
	= 20 AU.	(44)

Then

	≈	(45)
	=	(46)
	= 20.	(47)

Therefore

(48)

This example also illustrates the strong cubic dependence on orbital size.

17 Sensitivity to measurement errors

From

(49)

take a logarithmic differential:

(50)

Thus

(28)

For a visual binary,

(51)

so

(52)

Therefore

(29)

For independent random uncertainties, the approximate variance relation is

(30)

If the physical semimajor axis is obtained from angular semimajor axis and parallax,

(31)

when correlations can be neglected.

This is why accurate parallaxes are so valuable for visual-binary mass measurements.

18 Total mass versus individual masses

Equation (21) gives

(53)

It does not by itself separate the two component masses.

A second observable is required.

For example, if the barycentric orbit sizes are known,

(32)

For a double-lined spectroscopic binary,

(33)

where K₁ and K₂ are the radial-velocity semiamplitudes.

Combining total mass with a mass ratio gives the individual masses.

19 Physical interpretation

Equation (21) can be written schematically as

(54)

The dimensional structure reflects Newtonian gravity.

Since

(55)

we have

(56)

with dimensions of mass.

The combination is therefore not arbitrary.

It is the natural mass scale formed from the orbital length, orbital time, and gravitational constant.

20 Common mistakes

1. Using the semimajor axis of one star instead of the relative semimajor axis.
2. Using the instantaneous separation instead of the semimajor axis.
3. Treating one star as fixed when both masses are significant.
4. Applying the circular centripetal-force argument directly at every point of an eccentric orbit.
5. Concluding from the circular derivation that Kepler’s third law is valid only for circular motion.
6. Forgetting that the general elliptical derivation uses the semimajor axis, not periastron or apastron distance.
7. Using angular semimajor axis directly in Equation (21) without converting it to physical units.
8. Forgetting to convert milliarcseconds to arcseconds before using the simple parallax conversion.
9. Assuming total dynamical mass automatically gives the individual stellar masses.
10. Ignoring the cubic sensitivity of dynamical mass to semimajor-axis uncertainty.

21 Practice exercises

1. Reproduce the circular derivation beginning with the force balance for each star.
2. Show explicitly why adding the two circular equations produces the total mass M₁ + M₂.
3. Derive the relative equation of motion by subtracting the two Newton equations.
4. Explain why the relative problem depends on M₁ + M₂ rather than on either mass alone.
5. Starting from constant areal velocity, derive the elliptical form of Kepler’s third law.
6. Show exactly where the factor cancels in the elliptical derivation.
7. A binary has a = 10 AU and P = 25 yr. Estimate its total mass.
8. A visual binary has a^′′ = 0.40 arcsec, parallax 20 MAS, and period 30 yr. Find the total mass.
9. If the orbital semimajor axis has a one-percent uncertainty and the period uncertainty is negligible, estimate the resulting fractional mass uncertainty.
10. Explain what additional measurement is required to separate a total dynamical mass into individual stellar masses.

22 Summary

For a binary star, the Newtonian two-body problem gives

(57)

The relative orbit therefore behaves as a Kepler orbit controlled by the total mass.

For an ellipse,

(58)

Thus

(59)

The same formula is obtained from the simpler circular force-balance derivation.

In astronomical units,

(60)

For a visual binary,

(61)

The combination of orbit size, distance, and period therefore provides a direct dynamical measurement of stellar mass.

References

References