Question

Two bodies 'A' and 'B' having masses 'm' and '2m' respectively are kept at a distance 'd' apart.
A small particle is to be placed so that the net gravitational force on it, due to the bodies A and
B. is zero. Its distance from the mass A should be:-



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Answer 1

Answer:

Given:

Mass of A = m

Mass of B = 2m

Separation = d

To find:

Distance of a small particle from A such that gravitational force on it is zero.

Calculation:

Let mass of that particle be M;

and distance of M from A be x

Hence its distance from B be (d - x)

Since M is in gravitational equilibrium

∴ F1 = F2

$= > \frac{GMm}{ {x}^{2} } = \frac{G(2m)M}{ {(d - x)}^{2} }$

Cancelling G, M and m on both sides :

$= > \frac{1}{ {x}^{2} } = \frac{2}{ {(d - x)}^{2} }$

$= > \frac{1}{x} = \frac{ \sqrt{2} }{(d - x)}$

$= > d - x = \sqrt{2} x$

$= > \sqrt{2} x + x = d$

$= > ( \sqrt{2} + 1)x = d$

$= > x = \frac{d}{( \sqrt{2} + 1)}$

Hence the distance of mass M from A has been obtained.

Answer 2

Answer:

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