Three uniform spheres are fixed at the positions shown in the diagram. Assume th

Question

Three uniform spheres are fixed at the positions shown in the diagram. Assume they are completely isolated and there are no other masses nearby. (a) What is the magnitude of the force on a 0.20 kg particle placed at the origin (Point P)? What is the direction of this force? (b) If the 0.20 kg particle is placed at (x,y) = (-500 m, 400 m) and released from rest, what will its speed be when it reaches the origin? (c) How much energy is required to separate the three masses so that they are very far apart?

Explanation / Answer

A)
Gravitational force = G m1 m2 / d^2

force on 0.20 kg

due to upper 1kg :

F1 = (6.67 x 10^-11 x 1 x 0.20 ) / (0.5^2) (j) = 5.336 x 10^-11 j N

due to 2 kg :

F2 =[(6.67 x 10^-11 x 2 x 0.20) /(0.5^2 + 0.5^2) ] ( cos45i + sin45j)

F2 = (3.773i + 3.773j) x 10^-11 N

due to 1 kg on x axis.

F3 = (6.67 x 10^-11 x 1 x 0.20 ) / (0.5^2) (i) = 5.336 x 10^-11 i N

Fnet = F1 + F2 + F3 = (9.109i + 9.109j) x 10^-11 N

magnitude = sqrt(9.109^2 + 9.109^2) x 10^-11 = 12.88 x 10^-11 N

direction = tan^-1(9.109/9.109) = 45 deg

b) Gravt. PE of two mass system = - G M1 m2 / d

PE at (-500, 400)

PEi = -[G(1)(0.20) /sqrt(500^2 + 399.5^2)] - [G(2)(0.20) /sqrt(500.5^2 + 399.5^2)] - [G(1)(0.20) /sqrt(500.5^2 + 400^2)]

PEi =- 8.333 x 10^-14 J

KEi = 0

when at origin,

PEf = -[G(1)(0.20) /0.50] - [G(2)(0.20) /sqrt(0.50^2 + 0.5^2)] - [G(1)(0.20) /0.5)]

= - 9.109 x 10^-11 J

KE = 0.20 v^2 / 2 = 0.10v^2

Using energy conservation,

PEi + KEi = PEf + KEf

- 8.333 x 10^-14 J + 0 = - 9.109 x 10^-11 J   + 0.1v^2

v = 3.02 x 10^-5 m/s

c) PE energy of system

PE = -[G(1)(2) /0.50] - [G(1)(1) /sqrt(0.50^2 + 0.5^2)] - [G(1)(2) /0.5)]

= - -6.279 x 10^-10 J

energy required to separate them = 6.279 x 10^-10 J

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