Question

Three particles of identical mass m are acted on only by their mutual gravitational attraction. They are located at the vertices of an equilateral triangle with sides of length d. Consider the motion of any one of the particles about the system center of mass and use Newton’s second law to determine the angular velocity ω required for d to remain constant.

Answer 1

Answer:

Explanation:

The particles will be in constant angular motion . Let us consider forces on a single particle . It will experience three forces .

1 ) Attraction by other two masses

2 ) centripetal force necessary for circular motion.

Resultant of two like attractive forces F acting at 60 degree

= 2R Cos 30

= √3R

Centripetal force

= mω²r

so

mω²r  = √3R

R = GMm/ d² ( R is force of attraction on a particular mass by other masses )

r is radius of circle in which three particles rotate about their centre of mass .

Its value will be equal to

r = d / √3

Putting these values above we have

mω²d/√3  = √3 x GMm/ d²

ω = $\sqrt{\frac{3Gm}{d^3} }$