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Three identical point masses M are placed at the vertices of an equilateral triangle with side length d. They interact only through mutual gravitational attraction, moving so that the side length d remains constant. Given that sqrt(GM/d) = sqrt(3) m/s, what is the magnitude of the relative velocity (in m/s) of one particle with respect to another?
- 3
- 6
- 3*sqrt(3)
- sqrt(3)
Correct answer: 3
Solution
Each mass moves in a circle of radius r = d/sqrt(3) about the centroid. Net gravitational force on one mass from the other two: F_net = 2 * (GM²/d²) * cos(30 deg) = sqrt(3)*GM²/d² (along the line from vertex toward centroid). Centripetal condition: sqrt(3)*GM²/d² = M*v²/r = M*v²*sqrt(3)/d. So v² = GM/d. Thus v = sqrt(GM/d) = sqrt(3) m/s. The relative velocity between two particles: each moves with speed v = sqrt(3) m/s. The angle between their velocities at any instant is 60 deg (since they are at 120 deg apart on circle, tangent velocities differ by 60 deg). |v_rel| = 2v*sin(30 deg) = 2v*(1/2) = v = sqrt(3) m/s. Wait: angle between velocity vectors for masses at 120 deg separation on a circle: velocities are perpendicular to radii, so angle between velocities = angle between radii = 120 deg... Hmm. Let me recalculate: angle between velocity vectors = angle between the radii = 120 deg. |v_rel|² = v² + v² - 2v²*cos(120 deg) = 2v²(1 - cos120) = 2v²*(1 + 1/2) = 3v². |v_rel| = v*sqrt(3) = sqrt(3)*sqrt(3) = 3 m/s.
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