Three particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is

Question

Three particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is (A) zero (B)  (3GM/L2)  (C)  (9GM/L2)  (D)  (12/√3)(GM/L2)

Tardigrade Admin · Accepted Answer

The net force acting on a unit mass placed at

O

due to three equal masses

M

at vertices

A, B

and

C

is the gravitational field intensity at point

O

. The gravitational force on the particle placed at the point of intersection of three medians.

F_{1} \to + F_{2} \to + F_{3} \to = 0

Since, the resultant of

F_{1} \to

and

F_{2} \to

is equal and opposite to

F_{3} \to

.

Q. Three particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is

zero

$\frac{3 GM}{L ^{2}}$

$\frac{9 GM}{L ^{2}}$

$\frac{12}{3} \frac{GM}{L ^{2}}$

Q. Three particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is

zero

L23GM​

L29GM​

3​12​L2GM​

$\frac{3 GM}{L ^{2}}$

$\frac{9 GM}{L ^{2}}$

$\frac{12}{3} \frac{GM}{L ^{2}}$