0
$\begingroup$

I am modeling a two binary star system, and I am wondering if this is the case.

The way I have it right now is that I first figure out the mass center, and then the radius from each of the planet to the mass center.

I then figure out the acceleration for the first planet with:

$acceleration_1=\frac{\frac{G*m_1*m_2}{(r_1+r_2)^2}}{m_1}$ where $r_1$ and $r_2$, together make the full distance between the planets.

I then figure out the speed by $\frac{v^2}{r_1}=a_1$

The same could be said for the other planet:

$acceleration_2=\frac{\frac{G*m_1*m_2}{(r_1+r_2)^2}}{m_2}$ where $r_1$ and $r_2$, together make the full distance between the planets.

I then figure out the speed by $\frac{v^2_2}{r_2}=a_2$

The distance around the whole circle (The path they will they travel if they travel in a circle) is $r_1π2$ and $r_2π2$

I then figure out the time it takes to rotate one period by dividing distance by speed and then rotate the planets round the radius(from the planet to the mass center) in that time manner.

I now read that because their masses are unequal they are not supposed to rotate in a circle. Have I done this incorrectly?

This is how it looks at the moment:

https://gyazo.com/5c08ed6eec4f6af5758c83baaf7572be

$\endgroup$
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/61116/2451 , physics.stackexchange.com/q/229650/2451 , physics.stackexchange.com/q/288976/2451 and links therein. $\endgroup$ – Qmechanic Nov 19 '16 at 13:22
  • $\begingroup$ $\frac{v^2}{r}=a$ is a condition for a circular orbit, but orbits are rarely circular. The correct way to find velocity would be to use integration schemes. The simplest integration scheme is Euler integration, which goes as follows: $\vec x_{new}=\vec x_{old}+\vec v\cdot dt, \vec v_{new} = \vec v_{old} + \vec a\cdot dt$. Other important schemes are Runge-Kutta and leapfrog integration. $\endgroup$ – user3502079 Nov 19 '16 at 13:48
  • $\begingroup$ @user3502079 Hi user, so let's say I give both the two planets a mass and start velocity. 1. Calculate acceleration based on the formula given above based on the distance between the two planets. Then use the Euler above (or Runge kutta), then calculate the new position and the next velocity. Then find the new distance between them and find the new acceleration, and repeat. Is this the correct way of doing it? Do I even have to bother with the mass center at all? $\endgroup$ – David Lund Nov 19 '16 at 22:20
  • $\begingroup$ Are you asking about 2 planets (or stars) orbiting their common CM? Or 2 planets orbiting a much more massive binary-star? $\endgroup$ – sammy gerbil Nov 19 '16 at 23:57
  • $\begingroup$ @sammygerbil 2 planets orbiting each other or a common CM I guess. $\endgroup$ – David Lund Nov 20 '16 at 0:02
1
$\begingroup$

Revised Answer in response to your comments :

Your calculations are correct.

Generally the orbits will be ellipses, but the 2 planets can each orbit the CM in circles with constant speed around the circle. (The acceleration is a change in direction rather than a change in speed.) If the masses are equal they will orbit on the same circle; if they are different the circles will be concentric. Either way, they always keep on opposite sides of the CM and therefore have the same period.

Your animation gives the impression that the planets are the same size (and therefore mass). If so, their orbits are incorrect, because the CM rotates.

$\endgroup$
  • $\begingroup$ Hey, Sammy. The objects does not have equal mass but it is in fact still incorrect, because I am forcing the planets to go in a circle no matter what (and they wont ever go in an ellipse or other shapes). So what I think will be correct is to set the mass and velocity of planet 1 and planet 2. Calculate their acceleration in instance of time $a_i=\frac{\frac{G*m_i*m_{i+1}}{(d)^2}}{m_i}$ using this formula on both of the planets. Then use this acceleration to calculate the change in velocity for each of the planets, and their next position. Continue this over and over. $\endgroup$ – David Lund Nov 20 '16 at 18:40
  • $\begingroup$ What annoys me about this is that I will have no idea of what path they will take (ellipse or circle), and will just have to enter velocities and masses that seem reasonable. $\endgroup$ – David Lund Nov 20 '16 at 18:46
0
$\begingroup$

Force, in your case you have written down the equation for the force for Newtonian gravity, is not the same as acceleration. So either equate forces, in which case you would have to use the centripetal force and not acceleration, or equate accelerations, in which case you can use Newton's second law to find the acceleration from the force obtained using Newtonian gravity. Both methods will result in the same answer.

And in general binary stars will orbit each other in elliptical orbits, but a circular orbit is a special case of an elliptical orbit, namely with an eccentricity of zero. Since for elliptical orbits in general there is not closed form solution for the position as a function of time (it would require you to iteratively solve Kepler's equation), therefore in your case it might be easier to just stick to the circular orbits.

$\endgroup$
  • $\begingroup$ About the first one. I did divide the formula by the mass of the object, so I did use newtons second law. I forgot to include the m in the formula. I am definitely gonna attempt to do elliptical orbits. $\endgroup$ – David Lund Nov 19 '16 at 15:56
  • $\begingroup$ Would it be correct to do it the way I did it for circular orbits though? Even though those are not very often seen in the real world. $\endgroup$ – David Lund Nov 19 '16 at 16:04
  • $\begingroup$ @DavidLund If you indeed correctly equated accelerations or forces, then yes. But currently it is not clear in your question what $m$ represents in your equation for acceleration. Did you mean $a_i=\frac{G\,m_1\,m_2}{m_i\,(r_1+r_2)^2}$, such that $\frac{v_i^2}{r_i}=a_i$ for $i$ is equal to 1 or 2? $\endgroup$ – fibonatic Nov 19 '16 at 16:11
  • $\begingroup$ I changed it again. I think I am more clear this time. $\endgroup$ – David Lund Nov 19 '16 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.