2020 A/L Physics Past Paper · Question 48 (English Medium)

Three stars at the corners of an equilateral triangle — period of revolution

Three stars, each of mass mm, are situated at the three corners of an equilateral triangle of side aa as shown in the figure. Assume that these three stars revolve on a circular path about the centroid of the triangle, maintaining the initial distance between them. If only the mutual gravitational forces between the stars act on them, the period of revolution of this system is given by

  1. (1) 2πa32GM2\pi\sqrt{\dfrac{a^3}{2GM}}
  2. (2) 2πa33GM2\pi\sqrt{\dfrac{a^3}{3GM}}
  3. (3) 2π3a3GM2\pi\sqrt{\dfrac{3a^3}{GM}}
  4. (4) 2π2a3GM2\pi\sqrt{\dfrac{2a^3}{GM}}
  5. (5) 2π3a32GM2\pi\sqrt{\dfrac{3a^3}{2GM}}
Answer and explanation

Answer: (2) 2πa33GM2\pi\sqrt{\dfrac{a^3}{3GM}}

What the question asks

We must find: the time TT for one complete revolution about the centroid — in terms of aa, GG, and the star's mass.

Relevant theory

The gravitational force of attraction between two particles of masses M and m situated at a distance r is given by,

Since a body is moving with an acceleration only under the action of an external force, a body performing circular motion is doing so under the action of an external force acting in the direction of the centripetal acceleration. This force is thus known as "centripetal force" and is given by.

F=mv2r=mrω2F = \frac{mv^2}{r} = mr\omega^2

Period of rotation is the time taken for one complete rotation.

T=2πωT = \frac{2\pi}{\omega}

Required equations

F1=Gm2a2F_1 = \frac{Gm^2}{a^2}

F=mrω2F = m r \omega^2

T=2πωT = \frac{2\pi}{\omega}

Reasoning and calculation

Symbols: mm — the mass of each star (written as MM in the options).

aa — the distance between stars.

rr — the distance of a star from the centroid (the radius of the orbit).

ω\omega — the angular velocity.

On one star, two pulls act, each Gm2/a2Gm^2/a^2, toward the other two stars — the angle between them is 60∘60^\circ.

By symmetry the components perpendicular to the centroid cancel.

The components toward the centroid (cos⁡30∘\cos 30^\circ) add.

F=2×Gm2a2cos⁡30∘=3 Gm2a2F = 2 \times \frac{Gm^2}{a^2}\cos 30^\circ = \frac{\sqrt3\,Gm^2}{a^2}

r=23×32a=a3r = \frac{2}{3} \times \frac{\sqrt3}{2}a = \frac{a}{\sqrt3}

ma3ω2=3 Gm2a2m\frac{a}{\sqrt3}\omega^2 = \frac{\sqrt3\,Gm^2}{a^2}

ω2=3Gma3\omega^2 = \frac{3Gm}{a^3}

T=2πω=2πa33GmT = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{a^3}{3Gm}}

Reading the options

This matches 2πa3/3Gm2\pi\sqrt{a^3/3Gm} — option (2) (the star's mass written there as MM).

The others come from adding the forces as plain numbers, or from a wrong radius.

Checks

Dimension check: a3Gm\dfrac{a^3}{Gm}: m3N m2kg−2×kg=m3m3s−2=s2\dfrac{\text{m}^3}{\text{N m}^2\text{kg}^{-2} \times \text{kg}} = \dfrac{\text{m}^3}{\text{m}^3\text{s}^{-2}} = \text{s}^2; the root is s ✓.

Boundary limit: as mm grows (stronger pull), TT falls.

As aa grows, T∝a3/2T \propto a^{3/2} — the shape of Kepler's third law ✓.

Conclusion

T=2πa33GmT = 2\pi\sqrt{\frac{a^3}{3Gm}}

The resultant 3 Gm2/a2\sqrt3\,Gm^2/a^2 supplies the centripetal force m(a/3)ω2m(a/\sqrt3)\omega^2: T=2πa3/3GmT = 2\pi\sqrt{a^3/3Gm} — the answer is (2).

Adding the two gravitational pulls as plain numbers (2Gm2/a22Gm^2/a^2).

They are vectors at 60∘60^\circ: only the components toward the centre add — 3 Gm2/a2\sqrt3\,Gm^2/a^2.

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