Orbits and the Wider Universe

What you'll learn

How Kepler’s laws and Newton’s gravitation describe planets, satellites and binary stars.
How orbital data can be used to find the mass of stars, planets and galaxies.
How Doppler shifts and Hubble’s law connect spectra to motion and cosmic expansion.
Why dark matter and critical density matter in modern cosmology.

Starting point: gravity makes orbits happen

An orbit is the curved path of one object around another due to gravity. For a small satellite of mass $m$ orbiting a much larger central mass $M$ , the gravitational force acts towards the centre and provides the centripetal force needed for circular motion.

Circular orbit with gravity providing centripetal force

Definition

Newton’s law of gravitation

Two point masses, or two spherically symmetric masses treated as if their mass is at their centres, attract with force

G\frac{M_1M_2}{r^2}

where $G$ is the gravitational constant and $r$ is the separation of their centres.

For circular motion,

\frac{v^2}{r}

so the required centripetal force is $F = mv^2/r$ .

Kepler’s three laws

Kepler described planetary motion before Newton explained it.

First law: elliptical orbits

Planets move in ellipses, with the Sun at one focus. A circle is the special case where the two foci coincide.

Second law: equal areas in equal times

A line from the Sun to a planet sweeps out equal areas in equal time intervals. This means a planet moves faster when it is closer to the Sun.

Third law: period and orbital size

For objects orbiting the same central mass,

T2∝r3T^2 \propto r^3

for circular orbits, where $T$ is the orbital period and $r$ is the orbital radius.

Example

Comparing orbital periods

A planet orbits a star at four times Earth’s orbital radius around the same star. Estimate its orbital period in Earth years.

Use Kepler’s third law as a ratio:
$T22T12=r23r13\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$
Substitute $r_2 = 4r_1$ :
$T22T12=43=64\frac{T_2^2}{T_1^2} = 4^3 = 64$
Take the square root:
$T2T1=8\frac{T_2}{T_1} = 8$
So the orbital period is 8 Earth years.

Deriving Kepler’s third law for a circular orbit

For a satellite in a circular orbit, gravity provides the centripetal force:

GMmr2=mv2rG\frac{Mm}{r^2} = \frac{mv^2}{r}

The satellite mass $m$ cancels:

GMr2=v2rG\frac{M}{r^2} = \frac{v^2}{r}

v2=GMrv^2 = \frac{GM}{r}

For one complete orbit, the distance travelled is the circumference $2πr2\pi r$ , so

\frac{2\pi r}{T}

Substitute this into $v^2 = GM/r$ :

(2πrT)2=GMr\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}

Rearranging gives

T2=4π2r3GMT^2 = \frac{4\pi^2r^3}{GM}

This is Kepler’s third law for a circular orbit.

Key Idea

Mass from orbital data

If you know the orbital radius and period of a circular orbit, you can find the central mass using

\frac{4\pi^2r^3}{GT^2}

If you know orbital speed instead, use

\frac{v^2r}{G}

Example

Finding the mass of a planet from a satellite orbit

A satellite orbits a planet with orbital radius $4.22 \times 10^7 \text{ m}$ and period $8.62 \times 10^4 \text{ s}$ . Calculate the planet’s mass.

Choose the period form because $r$ and $T$ are given:
$\frac{4\pi^2r^3}{GT^2}$
Substitute the values:
$\frac{4\pi^2(4.22 \times 10^7 \text{ m})^3}{(6.67 \times 10^{-11} \text{ N m}^2\text{ kg}^{-2})(8.62 \times 10^4 \text{ s})^2}$
Calculate:
$\times 10^{24} \text{ kg}$
This is approximately the mass of Earth.

Common Mistake

Using the wrong radius

The orbital radius $r$ is measured from the centre of the central body to the orbiting object, not from the surface.

Spiral galaxies and dark matter

In a spiral galaxy, stars orbit the galactic centre. If most mass were in the bright visible region, then far from the centre the enclosed mass would be roughly constant, giving

\sqrt{\frac{GM}{r}}

so orbital speed should decrease as distance increases.

But observations show that the rotation curves of spiral galaxies stay approximately flat: stars far out move too fast to be held by visible matter alone.

Galaxy rotation curve showing evidence for dark matter

Definition

Dark matter

Dark matter is matter that does not emit, absorb or reflect enough electromagnetic radiation to be seen directly, but whose gravitational effects can be observed.

A flat rotation curve implies that the mass enclosed within radius $r$ keeps increasing with $r$ . This suggests a large, roughly spherical dark matter halo around the visible galaxy.

Example

Estimating unseen mass in a galaxy

At radius $5.0 \times 10^{20} \text{ m}$ from a galactic centre, stars orbit at $2.0 \times 10^5 \text{ m s}^{-1}$ . Estimate the enclosed mass.

Use the circular orbit result:
$\frac{v^2r}{G}$
Substitute:
$\frac{(2.0 \times 10^5 \text{ m s}^{-1})^2(5.0 \times 10^{20} \text{ m})}{6.67 \times 10^{-11} \text{ N m}^2\text{ kg}^{-2}}$
Calculate:
$\times 10^{41} \text{ kg}$
If the visible mass is much less than this, the difference is evidence for dark matter.

The Higgs boson and dark matter

The Higgs boson is a particle discovered at CERN in 2012. It is linked to the Higgs field, which is involved in giving many fundamental particles mass.

The Higgs boson itself is not normally considered to be dark matter because it is unstable and decays very quickly. However, some theories suggest dark matter particles may interact with ordinary matter through the Higgs field, sometimes called a “Higgs portal”. Evidence from particle physics can therefore constrain possible dark matter models.

Two-body orbits and the centre of mass

When two stars orbit each other, neither is fixed. Both orbit their common centre of mass, also called the barycentre.

Binary system centre of mass

For two spherically symmetric bodies separated by distance $r$ ,

r = r_1 + r_2

and the centre of mass condition is

m_1r_1 = m_2r_2

r1=m2m1+m2rr_1 = \frac{m_2}{m_1 + m_2}r

and

r2=m1m1+m2rr_2 = \frac{m_1}{m_1 + m_2}r

Their mutual orbital period for circular orbits is

2\pi\sqrt{\frac{r^3}{G(m_1 + m_2)}}

Example

Finding a binary centre of mass and period

Two stars have masses $2.0 \times 10^{30} \text{ kg}$ and $1.0 \times 10^{30} \text{ kg}$ . Their separation is $3.0 \times 10^{10} \text{ m}$ . Find the centre of mass distances and orbital period.

Find the distance of the first star from the centre of mass:
$mr_1 = \frac{1.0 \times 10^{30}}{3.0 \times 10^{30}}(3.0 \times 10^{10} \text{ m}) = 1.0 \times 10^{10} \text{ m}$
Find the second distance:
$mr_2 = 3.0 \times 10^{10} \text{ m} - 1.0 \times 10^{10} \text{ m} = 2.0 \times 10^{10} \text{ m}$
Calculate the period:
$2\pi\sqrt{\frac{(3.0 \times 10^{10} \text{ m})^3}{(6.67 \times 10^{-11})(3.0 \times 10^{30} \text{ kg})}}$ $\times 10^6 \text{ s}$
This is about 27 days.

Doppler shift of spectral lines

A spectral line is a narrow bright or dark line at a particular wavelength, produced by atoms absorbing or emitting specific photon energies.

If a star moves along the line joining it to Earth, its observed wavelengths shift. This line-of-sight component of velocity is called radial velocity.

For speeds much less than the speed of light,

Δλλ=vc\frac{\Delta\lambda}{\lambda} = \frac{v}{c}

where $Δλ\Delta\lambda$ is the change in wavelength, $λ\lambda$ is the rest wavelength, $v$ is radial velocity, and $c$ is the speed of light.

Tip

Redshift and blueshift

If the observed wavelength is larger than the rest wavelength, $Δλ\Delta\lambda$ is positive and the source is receding. If it is smaller, the source is approaching.

Example

Finding radial velocity from a spectral line

A spectral line with rest wavelength 656.3 nm is observed at 656.9 nm. Find the radial velocity.

Find the wavelength shift:
$nm\Delta\lambda = 656.9 \text{ nm} - 656.3 \text{ nm} = 0.6 \text{ nm}$
Substitute into the Doppler relationship:
$c\frac{\Delta\lambda}{\lambda}$ $\times 10^8 \text{ m s}^{-1})\frac{0.6}{656.3}$
Calculate and interpret the sign:
$\times 10^5 \text{ m s}^{-1}$
The wavelength increased, so the star is moving away from Earth.

Masses in an edge-on double system

For a circular binary system viewed edge-on, the maximum radial velocities are the true orbital speeds.

From radial velocity data, measure the velocity amplitudes $v_1$ and $v_2$ , and the period $T$ .

Then

r1=v1T2πr_1 = \frac{v_1T}{2\pi}

and

r2=v2T2πr_2 = \frac{v_2T}{2\pi}

\frac{(v_1 + v_2)T}{2\pi}

The total mass is

m1+m2=4π2r3GT2m_1 + m_2 = \frac{4\pi^2r^3}{GT^2}

Also, because both bodies orbit the same centre of mass,

m_1v_1 = m_2v_2

so the slower body is the more massive one.

Example

Finding masses from radial velocity data

An edge-on binary system has period $1.00 \times 10^7 \text{ s}$ . The radial velocity amplitudes are $3.0 \times 10^4 \text{ m s}^{-1}$ and $6.0 \times 10^4 \text{ m s}^{-1}$ . Find the two masses.

Find the separation:
$\frac{(3.0 \times 10^4 + 6.0 \times 10^4)\text{ m s}^{-1}(1.00 \times 10^7 \text{ s})}{2\pi}$ $\times 10^{11} \text{ m}$
Find the total mass:
$s)2m_1 + m_2 = \frac{4\pi^2(1.43 \times 10^{11} \text{ m})^3}{(6.67 \times 10^{-11})(1.00 \times 10^7 \text{ s})^2}$ $kgm_1 + m_2 = 1.7 \times 10^{31} \text{ kg}$
Use the mass ratio. Since $m_1v_1 = m_2v_2$ , the object with speed $3.0 \times 10^4 \text{ m s}^{-1}$ is twice as massive as the other:
$kgm_1 = 1.2 \times 10^{31} \text{ kg}$ $kgm_2 = 5.8 \times 10^{30} \text{ kg}$

Common Mistake

Inclination matters

If the orbit is not edge-on, the measured radial velocity is smaller than the true orbital speed. The simple mass method above only works directly for the edge-on case.

Hubble’s law and the age of the universe

For distant galaxies, Hubble’s law is

v = H_0D

where $v$ is the galaxy’s recessional velocity, $D$ is its distance, and $H_0$ is the Hubble constant.

The SI unit of $H_0$ is per second, s⁻¹, although astronomers often quote it in kilometres per second per megaparsec.

If a galaxy has moved at approximately constant speed since the Big Bang,

time≈Dv\text{time} \approx \frac{D}{v}

Using $v = H_0D$ ,

time≈DH0D=1H0\text{time} \approx \frac{D}{H_0D} = \frac{1}{H_0}

So $1/H_0$ gives an estimate of the age of the universe.

Example

Estimating the age of the universe from Hubble’s constant

Take $Mpc−1H_0 = 70 \text{ km s}^{-1}\text{ Mpc}^{-1}$ . Estimate the age of the universe.

Convert to SI units using 1 Mpc $\times 10^{22} \text{ m}$ :
$mH_0 = \frac{70\,000 \text{ m s}^{-1}}{3.09 \times 10^{22} \text{ m}}$ $s−1H_0 = 2.27 \times 10^{-18} \text{ s}^{-1}$
Take the reciprocal:
$s\frac{1}{H_0} = \frac{1}{2.27 \times 10^{-18} \text{ s}^{-1}} = 4.41 \times 10^{17} \text{ s}$
Convert to years:
$yr\frac{4.41 \times 10^{17} \text{ s}}{3.16 \times 10^7 \text{ s yr}^{-1}} = 1.4 \times 10^{10} \text{ yr}$
The estimate is about 14 billion years.

Common Mistake

Forgetting Hubble units

Before using $1/H_0$ , convert $H_0$ into s⁻¹. Do not take the reciprocal while it is still in kilometres per second per megaparsec.

Critical density of a flat universe

The critical density is the average density needed for a flat universe in this simple model.

Imagine a small galaxy of mass $m$ at the edge of a uniform spherical region of radius $R$ and density $ρ\rho$ . The mass inside the sphere is

\frac{4}{3}\pi R^3\rho

Use conservation of energy:

\frac{1}{2}mv^2 - G\frac{Mm}{R}

For the critical case, take total energy as zero:

12mv2=GMmR\frac{1}{2}mv^2 = G\frac{Mm}{R}

Substitute $v = H_0R$ and $\frac{4}{3}\pi R^3\rho$ :

12mH02R2=G(43πR3ρ)mR\frac{1}{2}mH_0^2R^2 = G\frac{\left(\frac{4}{3}\pi R^3\rho\right)m}{R}

Cancel $mR^2$ :

12H02=43πGρ\frac{1}{2}H_0^2 = \frac{4}{3}\pi G\rho

ρc=3H028πG\rho_c = \frac{3H_0^2}{8\pi G}

Example

Calculating critical density

Use $s−1H_0 = 2.27 \times 10^{-18} \text{ s}^{-1}$ to estimate the critical density.

Substitute into the equation:
$ρc=3H028πG\rho_c = \frac{3H_0^2}{8\pi G}$
Calculate:
$kg−2)\rho_c = \frac{3(2.27 \times 10^{-18} \text{ s}^{-1})^2}{8\pi(6.67 \times 10^{-11} \text{ N m}^2\text{ kg}^{-2})}$
Quote the result:
$m−3\rho_c = 9.2 \times 10^{-27} \text{ kg m}^{-3}$
This is extremely small by everyday standards.

Exam technique

In the exam

For orbit calculations, identify whether you are using $v$ , $T$ , or both, then choose between $M = v^2r/G$ and $4\pi^2r^3/(GT^2)$ .
In Doppler questions, use the rest wavelength for $λ\lambda$ and the shift for $Δλ\Delta\lambda$ ; the sign tells you approach or recession.
For Hubble questions, convert $H_0$ to s⁻¹ before estimating age or critical density.

Self review

Check yourself

Why does a flat galaxy rotation curve imply extra unseen mass?
In a binary star system, why does the more massive star have the smaller radial velocity amplitude?
Why is $1/H_0$ only an approximation to the age of the universe?

What you'll learn

How Kepler’s laws and Newton’s gravitation describe planets, satellites and binary stars.
How orbital data can be used to find the mass of stars, planets and galaxies.
How Doppler shifts and Hubble’s law connect spectra to motion and cosmic expansion.
Why dark matter and critical density matter in modern cosmology.

Starting point: gravity makes orbits happen

Circular orbit with gravity providing centripetal force

Definition

Newton’s law of gravitation

Two point masses, or two spherically symmetric masses treated as if their mass is at their centres, attract with force

G\frac{M_1M_2}{r^2}

where $G$ is the gravitational constant and $r$ is the separation of their centres.

For circular motion,

\frac{v^2}{r}

so the required centripetal force is $F = mv^2/r$ .