Question

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Two particles having masses M and 9M are separated by distance d. Find the position where third particle can be kept so that net gravitational force on that particle is zero.

Answer 1

2 particles of mass M and 9M are separated by a distance d.

To find:

Position where third particle can be kept such that the net gravitational force on the particle is zero.

Calculation:

Let that neutral position be x from mass M and hence (d - x) from mass 9M.

Net gravitational force will be zero , because the thought particle will experience same gravitational force from both sides. Hence will be in equilibrium.

$= > \dfrac{GMm}{ {x}^{2} } = \dfrac{G(9M)m}{ {(d - x)}^{2} }$

Cancelling the similar terms :

$= > \dfrac{1}{ {x}^{2} } = \dfrac{9}{ {(d - x)}^{2} }$

Taking square root in both sides :

$= > \dfrac{1}{x} = \dfrac{3}{(d - x)}$

$= > 3x = d - x$

$= > 4x = d$

$= > x = \dfrac{d}{4}$

So that position is d/4 from mass M and 3d/4 from mass 9M.

Answer 2

ANSWER:

★ DIAGRAM: ↓

←------x------→ ←-----d - x------→

Ⓜ ------ ⓜ ------- 9Ⓜ

←-------------- d ----------------→

____________________________

SOLUTION:

Let the neutral position be x from mass M and hence (d - x) from mass 9M

Gravitational force will be 0 because the thought particle will experience same gravitational force from both sides . They will be in equilibrium

→ GMm/x² = G(9M)m/(d-x)²

→ 1/x² = 9/(d - x)²

→ (1/x)² = (3/d - x)²

→ √(1/x)² = 3/d - x

→ 1/x = 3/d - x

Cross Multiply

→ 3x = d - x

→ 4x = d

→ x = d/4

Hence , position is d/4 for particle of having mass M & 3d/4 for particle of having mass 9M