Question

Four identical particles of mass M are located at the corners of a square of side ‘a’. What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square?
(A) 1.35
√(GM/a)
(B) 1.16
√(GM/a)
(C) 1.41
√(GM/a)
(D) 1.21 √(GM/a)

Answer 1

Answer:

C. is the correct answers

Explanation:

gravitational energy = kinetic energy

Answer 2

Given -

4 particles are located at corners of a square of side 'a' each having mass M

To Find-

speed of the particle if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square.

Solution -

If the particles are moving in circular orbit then the force that will act on them will be centripetal force.

by geometry we can find distance of particle from centre of circle which is a/√2

Net force on particle towards centre of circle is

Fc=GM²/2a² + GM²/a²(√2)

=GM²/a²(1/2+√2)

also Fc=mv²/r

comparing both values of fc we get

mv²/r=GM²/a²(1/2+√2)

r=a/√2 (substituting in above equation)

v²=GM/a((1/2√2) +1)

v²=GM/a(1.35)

v=1.16√GM/a