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Sample determination of the mass and orbit of a star’s companion
Data:
Mass of star: MS = 1.00  1030 kg [i.e. 0.5 MSun]
Orbital speed of star: vS = 1.00  103 m s1
Orbital period T = 1.00  107 s [~ 1/3 year]
Notes:
1. In an exam the some data are likely to be presented graphically, e.g. a graph of radial velocity
against time could be given. Alternatively a graph of the wavelength of a given spectral line
against time could be given from which, using the Doppler relationship, the speed can be inferred.
2. These data are made up. Real recent data are given at the end of the exercise.
What is the nature of the orbiting companion? Is it a planet, a star or a brown dwarf? The boundary
between a planet and a brown dwarf is normally taken as 13  mass of Jupiter [MJ = 1.899  1027 kg]. The
lower limit for a star is approximately 0.08 solar masses, i.e. 1.6  1029 kg. Below this mass a stellar-type
object never gets hot enough in its centre to initiate the p-p chain.
We assume, as usual, that we see the system edge on.
Structured questions:
(a) Determine the radius of the orbit of the star around the centre of mass of the system:
Using the relationships: distance = speed  time and circumference of a circle = 2r
2r = 1.00  103  1.00  107
So r = 1.59  109 m [= 1.59 million km]
This radius is greater than the radius of the star, so the centre of mass of the system is outside the star.
(b) Determine the radius of the orbit of the companion: - Assume that MS >> mP
The period of orbit of the pair of objects about their centre of mass is given by:
T  2
d3
G  M S  mP 
In this equation d is the distance between the objects.
If we are assuming that MS >> mP, then we can ignore mP in the equation.
Doing so, the equation becomes: T  2
d3
d3
or, squaring T 2  4 2
.
GM S
GM S
Rearranging and inserting the figures:
GMT 2 6.67  1011  1.00  1030  1.00  1014
d3 

 1.69  1032 m3
2
2
4
4
So d = 5.53  1010 m [= 55 million km]
This figure of 5.53  1010 m is the separation between the two objects, the star and its companion. It
is essentially the same as the radius of the companion’s orbit because we are assuming that MS >> mP,
which means that the centre of mass is very close to the centre of the star.
If we want to be a bit more picky: d = 5.53  1010 m and rS = 1.59  109 m
So rP = 5.53  1010  1.59  109 m = 5.37  1010 m.
(c) Determine the mass of the companion:
From these data, the easiest way of getting at the mass of the companion is to think of the Centre of
Mass as the balance point of the system and apply the Principle of Moments.
Using this idea: mPrP = MSrS.
M S rS 1.00  1030  1.59  109

 2.95  1028 kg – that’s using the less accurate value of rP. If
rP
5.53  1010
we use the other value of rP we get 2.87  1028 kg. This mass figure is very close to the boundary
between a planet and a brown dwarf.
So mP 
Now the real data:
The faint star, GJ 1046, was discovered to have an orbiting companion. The relevant data are:
Mass of star: MS = 0.398 MSun = 7.92 ± 0.14 1029 kg
Orbital speed of star: vS = 1830.7 ± 2.2 m s1
Orbital period T = 168.848 ± 0.030 days
[N.B. Ignore the uncertainty figures; they are only included to give an idea of the accuracies involved]
Use the same method to show that the mass of the companion is about 25 Jupiter masses and therefore is
likely to be a brown dwarf.
If you want to see the original paper from which these data come, type GJ1046 brown dwarf Kurster into
Google and choose the first link.
A couple of extra notes:
The method above assumes that we see the system edge on. If it is tilted, the value of m calculated is actually
m sin i , where i is the angle of inclination; i is 90 for an edge-on view and 0 for a face-on view.
For GJ1046, the paper referred to estimates the probability that m > 0.08 MS at 6.3%. In order to determine
the mass of the central star, the parallax of GJ1046 was determined at 71.11 mas [milli-arc seconds]. This
enabled the power output of the star to be determined and compared to models of stellar evolution.
Gareth Kelly
26 October 2008