Two-body problem

I was asked this by somebody doing a homework question, and I realized
that I don't know how to solve for the mass of the smaller body. Here's
the question:

"You observe a main sequence type B5 star that moves as if it is in a
binary system, but no companion is visible. If the period of the system
is 8.4 years and the semi-major axis is 8 AU, what is the mass of the
system? What is the mass of the companion?"

Using Kepler's 3rd Law, I was able to deduce that the star is about 7.25
solar masses, which conforms well to known masses of Class B5 stars (eg.
Rho Aurigae, http://is.gd/w85MlS). Here's the formula I used:

G · m · t² = 4 · π² · r³

Now, how do you solve for the mass of the companion? I assume it is a
planet, being much smaller than the star.

Yousuf Khan

Dear Yousuf Khan:

On Monday, March 10, 2014 6:09:50 AM UTC-7, Yousuf Khan wrote:
....
Now, how do you solve for the mass of the companion?

http://lasp.colorado.edu/education/o...lar_masses.pdf

David A. Smith

On Mon, 10 Mar 2014 09:09:50 -0400, Yousuf Khan wrote:


If the period of the system
is 8.4 years and the semi-major axis is 8 AU, what is the mass of the
system? What is the mass of the companion?"


http://en.wikipedia.org/wiki/Elliptic_orbit

Scroll down to "Orbital Period."

Get total mass from u, which should be G*m_1 + G*m_2

On 10/03/2014 10:06 AM, dlzc wrote:
Dear Yousuf Khan:

On Monday, March 10, 2014 6:09:50 AM UTC-7, Yousuf Khan wrote:
...
Now, how do you solve for the mass of the companion?

http://lasp.colorado.edu/education/o...lar_masses.pdf

I think the problem with the equation at this link is, that in order to
find the mass of the planet, we would need to know the velocity of the
star itself. The people who published this paper know the velocity of
the star through Doppler shift information. We're given no such velocity
information about this star in this problem. This is the equation that
I'm talking about from their paper:

m_planet = m_star * v_star * p_planet / ( 2 * pi * a_planet )

Where,
m_star = mass of star = 7.25 solar masses
v_star = velocity of star = unknown
p_planet = orbital period of planet = 8.4 years
a_planet = semi-major axis of orbit = 8 AU

Yousuf Khan

On 10/03/2014 10:17 AM, Frank Colessi wrote:
On Mon, 10 Mar 2014 09:09:50 -0400, Yousuf Khan wrote:


If the period of the system
is 8.4 years and the semi-major axis is 8 AU, what is the mass of the
system? What is the mass of the companion?"


http://en.wikipedia.org/wiki/Elliptic_orbit

Scroll down to "Orbital Period."

Get total mass from u, which should be G*m_1 + G*m_2

Unfortunately, it doesn't quite work that way, the u is estimated from
assuming that the star is much much bigger than the planet, not the
general case.

Yousuf Khan

In article ,
Yousuf Khan writes:
"You observe a main sequence type B5 star that moves as if it is in a
binary system, but no companion is visible. If the period of the system
is 8.4 years and the semi-major axis is 8 AU, what is the mass of the
system? What is the mass of the companion?"

This appears to be the case of a single-line spectroscopic binary
except that somehow we know the inclination is 90 degrees. I don't
think the problem can be solved from only the information given, but
it should be OK if we can look up the mass from the spectral type.

The first step is essentially Kepler's law, where M = R^3/T^2 in
solar system units. As the OP wrote, this gives a mass of 7.26 solar
masses. That would be the correct mass of the other object if the
visible star were massless, which is not the case.

From a quick glance at some reference sites, and without working
through all the math for myself, I _think_ the mass derived above
corresponds to m^3/(m+M)^2, where m is the mass of the unknown object
and M is the mass of the visible B star. _Allen's Astrophysical
Quantities_ gives the mass of a B5V star as 5.8 solar masses, so we
just plug that in to get 7.26 = m^3/(m+5.8)^2 and solve for m. This
is a cubic equation, and as far as I can tell, it has one real root
around 13.62 solar masses.

I won't swear this is correct; as I wrote, I haven't checked the
derivation properly. Something along these lines should be correct,
though. The Keplerian calculation gives some indicative system mass,
and you plug the mass of the known star into the right formula to
give the mass of the unknown star.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA

On 10/03/2014 5:48 PM, Steve Willner wrote:
This appears to be the case of a single-line spectroscopic binary
except that somehow we know the inclination is 90 degrees. I don't
think the problem can be solved from only the information given, but
it should be OK if we can look up the mass from the spectral type.

The first step is essentially Kepler's law, where M = R^3/T^2 in
solar system units. As the OP wrote, this gives a mass of 7.26 solar
masses. That would be the correct mass of the other object if the
visible star were massless, which is not the case.

From a quick glance at some reference sites, and without working
through all the math for myself, I _think_ the mass derived above
corresponds to m^3/(m+M)^2, where m is the mass of the unknown object
and M is the mass of the visible B star. _Allen's Astrophysical
Quantities_ gives the mass of a B5V star as 5.8 solar masses, so we
just plug that in to get 7.26 = m^3/(m+5.8)^2 and solve for m. This
is a cubic equation, and as far as I can tell, it has one real root
around 13.62 solar masses.

So you're taking the unknown star to be the B5 (or more specifically the
B5V) star? My interpretation of the question was that the known star was
the B5-class star. If the unknown companion were the B5 star, then it
would be pretty easy to see, it wouldn't be an "unseen" companion.

Yousuf Khan

In article ,
Yousuf Khan writes:
So you're taking the unknown star to be the B5

No, the known star was the B5. The unseen companion works out to be
more massive, presumably a black hole. That depends, of course, on
my having used the correct formula, which I still don't vouch for.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA

On Monday, March 10, 2014 6:09:50 AM UTC-7, Yousuf Khan wrote:
I was asked this by somebody doing a homework question, and I realized

that I don't know how to solve for the mass of the smaller body. Here's

the question:



"You observe a main sequence type B5 star that moves as if it is in a

binary system, but no companion is visible. If the period of the system

is 8.4 years and the semi-major axis is 8 AU, what is the mass of the

system? What is the mass of the companion?"



Using Kepler's 3rd Law, I was able to deduce that the star is about 7.25

solar masses, which conforms well to known masses of Class B5 stars (eg.

Rho Aurigae, http://is.gd/w85MlS). Here's the formula I used:



G · m · t² = 4 · π² · r³



Now, how do you solve for the mass of the companion? I assume it is a
planet, being much smaller than the star.

Yousuf Khan

Sadly, we have more than sufficient supercomputers to deal with three body simulations, but sadly those public-funded machines and of those we also pay to program and operate them are taboo or off-limits as for those of your outsider kind.

On 11/03/2014 12:24 PM, Steve Willner wrote:
In article ,
Yousuf Khan writes:
So you're taking the unknown star to be the B5

No, the known star was the B5. The unseen companion works out to be
more massive, presumably a black hole. That depends, of course, on
my having used the correct formula, which I still don't vouch for.

Oh, you're saying that the visible star, the B5-type star, is 5.8 solar
masses, while the invisible mass star is 7.25 solar masses? That's an
interesting way to turn the problem around.

Yousuf Khan