# Solving for the mass of the larger star in a binary star system. (Larger star is much larger, so orbit is essentially circular)

I'm a teaching assistant for a class on Newtonian mechanics, and was confronted with a way of solving for the larger mass in this two-star system that gets the right symbolic solution, but seems to forget the factor of $$\frac{1}{2}$$ for kinetic energy.

The problem gives a distance between the two stars r, a velocity v, and states that the larger has such a large mass that the orbit of the smaller star is nearly circular.

Here's their solution:

$$U_g = E_k$$

Gravitational and kinetic energy must be equal if the star is in orbit.

$$U_g = -G \frac{m_1m_2}{r}$$

$$E_k = mv^2$$

This is missing its factor of $$\frac{1}{2}$$ but we ignore this and set the two energies equal to each other and solve.

$$m_2v^2 = -G \frac{m_1m_2}{r}$$

We will ignore the minus sign on gravitational potential energy, and $$m_2$$ cancels out.

$$v^2 = G \frac{m_1}{r}$$

We solve for $$m_1$$ and get:

$$\frac{v^2r}{G} = m_1$$

Which happens to be the same symbolic solution that is in the answer key.

How is this correct when kinetic energy is equal to $$\frac{1}{2} mv^2$$ ?

• Why should the extra mass of the larger star make the orbit circular? Dec 15, 2020 at 23:49

Gravitational and kinetic energy must be equal if the star is in orbit.

No it need not be. Forces must balance, and total energy must constant but not zero.

So why do you get the result you do ? This comes down to the viral theorem. As the Wikipedia page shows, the kinetic energy (in a time averaged gravitational system) is equal to half the potential energy. As a circular orbit means we do not need to worry about time averaging we can just use :

$$E_{KE}=-\frac 1 2 U$$

I think you can probably already see where that factor of $$2$$ you are worried about is coming from.

For a circular orbit we balance forces :

$$m\frac {v^2} r = \frac {GMm} {r^2}$$

And we get :

$$v^2=\frac {GM} r$$

So KE is given by :

$$E_{KE}=\frac 1 2 m v^2 = \frac 1 2 m \frac {GM}r=-\frac 1 2 U$$

So you can see that the viral theorem does indeed work here.

The other misconception (in the original question) is :

the larger has such a large mass that the orbit of the smaller star is nearly circular

A gravitational closed orbit of any two objects (in Newtonian mechanics) can be any ellipse (which of course includes circles). We know of thousands of examples of asteroids and comets in our own solar system that have extremely elliptical orbits and clearly their masses are much, much smaller than the Sun's.

• Thanks! That clears up a lot, though I will have to wait until future physics classes to make total sense of the Virial Theorem that you mentioned. Dec 17, 2020 at 6:02

The first equation means that the system is at the dissociation limit as the total energy is zero. For the circular orbit that you probably want your students to study instead the virial theorem holds.

Your first equation is inaccurate. In a circular orbit, the kinetic energy is half the magnitude of the gravitational potential energy.

Assuming the secondary mass is small, Newton's second law can be written $$m \frac{v^2}{r} = \frac{GMm}{r^2}$$ Thus $$\frac{1}{2} mv^2 = \frac{1}{2}\left(\frac{GMm}{r}\right)$$