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Three identical stars of mass M are located at the vertices of an equilateral triangle with side L. At what speed must they move if they all revolve under the influence of one another’s gravity in a circular orbit circumscribing, while still preserving, the equilateral triangle?

Three identical stars of mass M are located at the vertices of an equilateral triangle with side L. At what speed must they move if they all revolve under the influence of one another’s gravity in a circular orbit circumscribing, while still preserving, the equilateral triangle?

Grade:10

2 Answers

Jitender Pal
askIITians Faculty 365 Points
7 years ago
The position of the three stars which are located at the vertices of an equilateral triangle is shown below.
235-1483_im8.JPG
From the figure, the stars orbit about a point O inside the equilateral triangle. The distance from the center of its orbit is given by

235-1259_im9.JPG
Now, the necessary centripetal force is provided by the net force due to the three stars.
Thus,
235-1008_im10.JPG
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the attached solution to your problem
From the geometry we know, angle OBC and angle OCB = 30o
Hence, OBcos30o + OCcos30o = BC
or, 2rcos30o = L
or, r = L/2cos30o
 
The, gravitational force on each star due to the other,
Fg = GM2/L2
Since, each star experience same force from two different stars, the net resultant force needs to be calculated.
The net force will be directed towards the center of the circle as the other components cancel out.
Hence, Net force on each star, F = 2 * GM2/L2 * cos30o
or, F = 2GM2cos30o/L2 
 
This force will provide the centripetal force necessary to keep the stars in circular motion.
Hence, F = Fc
or, 2GM2cos30o/L2 = Mv2/r = 2Mv2cos30o/L (substitutinf value of r)
or, v = √GM/L
 
Hope it helps.
Thanks and regards,
Kushagra

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