Calculating Orbital Period of 3 Planets with Similar Mass

AI Thread Summary
The discussion focuses on calculating the orbital period of three equally spaced planets with similar mass, orbiting around a central planet. Participants emphasize the need to equate centripetal force with gravitational attraction to derive the correct equations. The gravitational attraction between the planets must consider the distances involved, particularly that the distance from the outer planets to the center planet is R, while the distance between the outer planets is 2R. Adjustments to the equations are suggested, including proper notation for distances squared. The final consensus indicates that the orbital period can be derived correctly by addressing these gravitational relationships.
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Homework Statement



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The three planets (v1, v2 and v3) in the diagram all have similar mass and are in a line equally spaced so that v1 and v3 are orbiting around v2 synchronously. If the mass of each of the planets are M and the radius of the orbit is R, what is the orbital period?

Homework Equations



T=\frac{2*pi*r}{v}

v=\sqrt{GM/r}

Fg=GMm/r^2

The Attempt at a Solution



I am pretty sure you just need to set two equations equal to each other and solve for the variable T, but I am unsure which two equations this is. It would be appreciated if somebody could explain all this to me. Thank you.
 
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One equation.
Equate the centripetal force of v1 to the gravitational pull exerted on v1 by v2 and v3.
 
rude man said:
One equation.
Equate the centripetal force of v1 to the gravitational pull exerted on v1 by v2 and v3.

\frac{Mv^2}{R}=\frac{2*G*M*M}{R^2}

v=\sqrt{2GM}

T=\frac{2*pi*R}{v}=\frac{2*pi*R}{sqrt(2GM)}
 
woaini said:
\frac{Mv^2}{R}=\frac{2*G*M*M}{R^2}

v=\sqrt{2GM}

T=\frac{2*pi*R}{v}=\frac{2*pi*R}{sqrt(2GM)}

You don't have the correct expression for the total gravitationl attraction of v2 and v3 on v1.

I also suggest changing mv^2/R to m(w^2)R.
 
rude man said:
You don't have the correct expression for the total gravitationl attraction of v2 and v3 on v1.

I also suggest changing mv^2/R to m(w^2)R.

How would I find the total gravitational attraction? Since they both have the same radius and masses, I assume that the gravitational forces would be the same and therefore have two times that.
 
woaini said:
How would I find the total gravitational attraction? Since they both have the same radius and masses, I assume that the gravitational forces would be the same and therefore have two times that.

Look at the picture. Is the distance from v1 to v2 the same as the distance of v1 to v3?
 
rude man said:
Look at the picture. Is the distance from v1 to v2 the same as the distance of v1 to v3?


\frac{Mv^2}{R}=\frac{G*M*M}{R^2}+\frac{G*M*M}{2R^2}

v=\sqrt{3GM}

T=\frac{2*pi*r}{\sqrt{3GM}}
 
Still not right. re-examine the second term on the right.
 
rude man said:
Still not right. re-examine the second term on the right.

\frac{Mv^2}{R}=\frac{G*M*M}{R^2}+\frac{G*M*M}{2R^2}

v=\sqrt{5/4*GM}

T=\frac{2*pi*r}{\sqrt{5/4*GM}}
 
  • #10
woaini said:
\frac{Mv^2}{R}=\frac{G*M*M}{R^2}+\frac{G*M*M}{2R^2}

v=\sqrt{5/4*GM}

T=\frac{2*pi*r}{\sqrt{5/4*GM}}

What is the distance between v1 and v3? Ergo, what is the grav. attraction between them?
 
  • #11
rude man said:
What is the distance between v1 and v3? Ergo, what is the grav. attraction between them?

Isn't the distance 2R?

Alright, I figured out I simplified to v wrong resulting in a wrong answer.
 
Last edited:
  • #12
woaini said:
Isn't the distance 2R?
Yes, but when you write it, you should write (2R)^2, not 2R^2.
 
  • #13
tms said:
Yes, but when you write it, you should write (2R)^2, not 2R^2.

Yes, so it's 4R^2
 
  • #14
woaini said:
Yes, so it's 4R^2

So it is.
 
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