Question

Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational field intensity at the centroid of the triangle is

Answer 1

Answer:

Net gravitational field intensity at the centroid of the triangle will be ZERO

Explanation:

Since all three masses are placed at three corners of the equilateral triangle

so here we have gravitational intensity due to each mass at the centroid position is given as

$E = \frac{GM}{r^2}$

since distance of each vertices is same from the corner

So the gravitational field intensity will be same in magnitude due to each mass

Since gravitational field intensity is a vector quantity so here all three mass will have same field intensity and equally inclined to each other

So we have

$E_{net} = \vec E + \vec E + \vec E$

$E_{net} = 0$

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Answer 2

The resulatnt gravitational field intensity at the centroid of the triangle is zero.

Explanation:

Given that,

Mass of each particle = m

Side = a

We know that,

The gravitational field intensity is

$E=\dfrac{Gm}{r^2}$

Where, r = side of triangle

The distance from its vertex to the centroid is

$r=\dfrac{a}{\sqrt{3}}$

The gravitational field intensith for first partyicle

Put the value into the formula

$E_{1}=\dfrac{3Gm}{a^2}$

Similarly,

$E_{2}=\dfrac{3Gm}{a^2}$

$E_{3}=\dfrac{3Gm}{a^2}$

They are co planer and equal in magnitude and make an angle of 120°

All three forces being equal and 120° to each other.

Hence, The resulatnt gravitational field intensity at the centroid of the triangle is zero.

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