# Deducing the mass of the Sirius system - Feynman exercises 3.17

83 views I do not even know how to try to deduce the mass M of the Sirius system in terms of that of the sun, as I do not not how to interpret the given data.

Could you give me a hint?

EDITED AT THIS POINT

1) Estimating the distance to the Sirius binary system.

I calculated the hypotenuse of the triangle $S E_1 SS$ (Sun - position of the Earth located at the top of a vertical circular orbit - Sirius System):

$$\theta = 0.378^{\circ} arc \times \frac{1}{3600} \times \frac{2 \pi}{360} = 1.83 \times 10^{-6}arc sec$$

$$D = \frac{D'}{sin(\theta)} = 8.16\times 10^{16} m$$

Where:

D = the distance from the Earth to the Sirius binary system.

D' = Distance from the Sirius system to the Earth ($15 \times 10^{10} m$).

Here's an illustration of what I have done: 2) Measuring the semi-major axis of the ellipse and estimating the period of the orbit from the figure.

To compute the semi-major axis I have assumed the major axis is under the scale [0'',12''].

As Sammy Gerbil said :'To measure the semi-major axis from the figure, magnify the diagram to fill your screen, place the edge of a sheet of paper along the diagonal line marked on the ellipse, mark off the ends of this line on the paper, then transfer the paper edge to the scale and read off the length in arc seconds. I get 13.8" for the major axis length, so the semi-major axis is a=6.9". The calculation is then:'

$$a=8.16\times 10^{16}m\times \frac{6.9}{3600}\times \frac{2\pi}{360} = 2.73 \times 10^{12}m$$

Now let's estimate the period of the orbit.

Is this the right thought?

$$\frac{dA}{dt} = SM \times SM \times \frac{d \theta}{2dt}$$

It will not be accurate as 'the two sides of most of the triangles are different'.

$A_1$: On process...

asked Oct 20, 2018
edited Nov 16, 2018
1) I should have looked better as it is stated at a). Then in the semi-major distance calculation you just used the sine function based on the angle 6.9'' arc, I see.

2) Using Heron's Formula I got an $A_1$ (please see image above) of an order $10^{24}$. Does that make sense for you?
My advice above is to measure areas on the diagram in $mm^2$ or $\text{arc-seconds}^2$. There is no need to find the actual area of the orbit of Sirius B. What scale or units you use for area is irrelevant, because you will be calculating the ratio of two areas, which is the same whatever units you work in.

The check you should make is to calculate a value for the period of the orbit, and compare that with the official value found by internet research.
I have not forgotten about this problem. However, due to the fact that I have an exam of statistical mechanics soon I wanted to ask you about other questions before continuing  with this one. I hope you have no problem with it. As always, thank you for your help professor.
Sr, so now the next step is to calculate the period of the orbit. But for that we need to calculate the area of several triangles. The thing is that I am still a little bit confused about this. I am wondering whether it would be possible to calculate firstly the areas of just two triangles using Heron's formula, secondly the corresponding time and finally make an approximation having as a reference the ratio of those areas and the time taken to cover them. I think it could work as we have the scale.
I do not think that will work, but you are welcome to try it.

The reason it will not work is because it will give us no new information. We already know the times for these 2 triangles. The areas should be in the same ratio as the times. If our calculation confirms that, we have learnt nothing new. Everything in the calculation is known.

What we need to do is compare the area of one or more triangles which have known orbital time, with the area of the whole ellipse which has an unknown orbital time (=period). Then we can use Kepler's 2nd Law (equal areas swept out in equal times) to find unknown period.

So we need to meaure the area of the whole ellipse in some way. Not just a couple of triangles.

Please re-read previous comments : **the scale does not matter** in this calculation. We are using ratios, so it is irrelevant what units we use for area. We can use $mm^2$ on any convenient magnification.