Final answer:
The speed of each planet in mutual orbit around the center of an equilateral triangle is found using the laws of gravitation and circular motion, resulting in √(3Gm/(2l)), which corresponds to option B.
Step-by-step explanation:
The student asked for the speed of each planet in a system where three planets of identical mass m form the vertices of an equilateral triangle, each rotating in a circular orbit around the center of the triangle due to their mutual gravitation. To find this speed, we cannot directly apply Kepler's laws as they are meant for planets orbiting the Sun. Instead, we consider the gravitational force between the planets and use Newton's law of universal gravitation and circular motion to derive the formula.
Each planet will experience a gravitational force from the other two planets. For a planet to maintain a circular orbit, the gravitational forces must provide the necessary centripetal force to keep the planet in orbit.
For a single planet, the gravitational force exerted by one of the other planets is calculated by Newton's law of universal gravitation, F = Gm2/l2. Since the forces from the two planets add vectorially to provide the centripetal force, and because they are perpendicular, we can use Pythagoras's theorem to find the resultant force: Fnet = √(2) * Gm2/l2, which is equal to the mass of the planet times its orbital velocity squared divided by the radius of the orbit, Fnet = mv2/r.
Since the side of the triangle is l and the distance from the center of the triangle to a vertex is (l/√(3)), we can substitute this into our equation for r and solve for v to get:
v2 = (Gm)/(√(3)/2l)
v = √(3Gm/(2l))
This gives us the option B. √(3GM/2L) as the correct answer.