Question 248: Gravitation

Question

Three identical spheres of mass m , are placed at the vertices of an equilateral triangle of length $a$. When released, they interact only through gravitational force and collide after a time $\mathrm{T}=4$ seconds. If the sides of the triangle are increased to length $2 a$ and also the masses of the spheres are made 2 m , then they will collide after__________seconds.

Accepted Answer

Dimensional Analysis
:
$    \mathrm{T} \propto \mathrm{m}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{a}^{\mathrm{z}}    $
Using dimensional analysis for gravitational interactions, we have:
$    \mathrm{T} \propto \mathrm{M}^{\mathrm{x}} G^{\mathrm{y}} a^{\mathrm{z}}    $
Where $ G $ has dimensions $\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}ight]$.
Solving for Exponents
:
$    \mathrm{T} \propto \mathrm{M}^{\mathrm{x-y}} \mathrm{L}^{3y+z} \mathrm{T}^{-2y}    $
Equating dimensions, we solve:
$    \mathrm{x}-\mathrm{y}=0 \Rightarrow \mathrm{x}=\mathrm{y}    $
$    -2\mathrm{y}=1 \Rightarrow \mathrm{y}=-\frac{1}{2}, \mathrm{x}=-\frac{1}{2}    $
$    3\mathrm{y}+\mathrm{z}=0 \implies \mathrm{z}=-3\mathrm{y}=\frac{3}{2}    $
Time Proportionality
:
$    \mathrm{T} \propto \mathrm{~m}^{-1/2} \mathrm{G}^{-1/2} \mathrm{a}^{3/2}    $
Which simplifies to:
$    \mathrm{T} \propto \left(\frac{a^3}{m}ight)^{1/2}    $
Applying New Conditions
:
When the side length becomes $2a$ and mass becomes $2m$:
$    \mathrm{T} = 4 	imes \left(\frac{(2a)^3}{2m}ight)^{1/2}    $
Simplifying:
$    \mathrm{T} = 4 	imes \left(\frac{8a^3}{2m}ight)^{1/2} = 4 	imes (4)^{1/2} = 8 	ext{ seconds}    $
Thus, the spheres will collide after 8 seconds under the new conditions.

Gravitation

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