In deep space, three masses are held in fixed positions (by rods light compared

Question

In deep space, three masses are held in fixed positions (by rods light compared to the masses) at three comers of a square of side length m as shown. A fourth mass is placed at point A, and released from rest. By symmetry it accelerates straight down and reaches point B (the midpoint between the two 3.0x109 kg masses). sin (36.87 degree) 0.6 cos(36.87 degree) 0.8 How fast is the fourth mass moving at point B? (Note: the acceleration is not constant.) If the fourth mass is 70 kg, what is the magnitude of the net force acting on it at point A?

Explanation / Answer

let us assume the mass of fourth particle be m

Now writing the potential energy of system when m is at A

Note by simple math AB = 4m

distance between A and 3*10^9 is 5m

and B and 5.64*10^9 is 3m

GmM/R = energy

Initial energy = {G*(3*10^9)*m/5}*2 + G*(5.64*10^9)*m/7

Note two time coz of two equal mases and third one for other mass.

Final energy = {G*(3*10^9)*m/3}*2 + G*(5.64*10^3)*m/9 - 1/2mv^2

equating and solving

v = 1.581*10^-5 m/s

net force on 70kg at A

{G*(3*10^9)*70/25}*2*cos(36.87) + G*(5.64*10^9)*70/49

net force = 1.434 N downward

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