Gravitation: Diagram-Based Questions (3)

hard • 3 min read
Tags Gravitation

Question

Three identical point masses, each of mass mm, sit at the vertices of an equilateral triangle of side aa. Find the magnitude of the gravitational force on any one mass due to the other two, and the gravitational potential energy of the system.

Solution — Step by Step

Pick mass at vertex AA. The other two masses at BB and CC each pull it with force:

F=Gm2a2F = \frac{Gm^2}{a^2}

These two forces make an angle of 60°60° with each other (since the triangle is equilateral and both pull toward the opposite vertices).

Net force magnitude:

Fnet=F2+F2+2F⋅Fcos⁡60°=2F2+F2=F3F_{\text{net}} = \sqrt{F^2 + F^2 + 2F\cdot F \cos 60°} = \sqrt{2F^2 + F^2} = F\sqrt{3}

So Fnet=3 Gm2a2F_{\text{net}} = \dfrac{\sqrt{3}\, Gm^2}{a^2}, directed along the median from AA toward the centroid.

PE is computed per pair, and there are three pairs (AB,BC,CA)(AB, BC, CA):

U=−3×Gm2a=−3Gm2aU = -3 \times \frac{Gm^2}{a} = -\frac{3Gm^2}{a}

The negative sign means the system is bound — work must be done to separate the masses to infinity.

Why This Works

Gravitational force is a vector, but gravitational PE is a scalar property of pairs. That’s why we add force components but sum PE directly across all distinct pairs. Counting pairs is where students slip — for NN masses there are (N2)\binom{N}{2} pairs, not NN. With three masses we get three pairs.

The geometry trick — that two equal forces at 60°60° give a resultant of 3\sqrt{3} times each — is worth memorising. It also appears in electrostatics for charges at triangle vertices.

Alternative Method

For the force, you can resolve each pull into components along the median from AA and perpendicular to it. The perpendicular components cancel by symmetry, and the along-median components each equal Fcos⁡30°=F3/2F\cos 30° = F\sqrt{3}/2. Two of them add to F3F\sqrt{3}. Same answer, useful when you also need the direction.

Whenever a configuration has a rotation or reflection symmetry, exploit it. In this triangle, the net force on each mass points toward the centroid, and all three forces have equal magnitude 3 Gm2/a2\sqrt{3}\, Gm^2/a^2. Recognising symmetry saves you from doing three separate calculations.

Common Mistake

Students often double-count pairs in the PE calculation, writing U=−6⋅Gm2/aU = -6 \cdot Gm^2/a by treating ABAB and BABA as different. Each pair contributes once. The cleanest way is to list pairs explicitly: (A,B),(B,C),(C,A)(A,B), (B,C), (C,A) — three pairs, three terms, done.

The other mistake is using cos⁡30°\cos 30° when adding two forces at angle 60°60° — confusing the angle between the vectors with the angle each makes with the resultant. Always start with the parallelogram law Fnet2=F12+F22+2F1F2cos⁡θF_{\text{net}}^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta where θ\theta is the angle between the forces.

Final answer: Fnet=3 Gm2a2F_{\text{net}} = \dfrac{\sqrt{3}\, Gm^2}{a^2}, U=−3Gm2aU = -\dfrac{3Gm^2}{a}.

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