Question

Three particles, two with masses $ m $ and one with mass $ M $ , might be arranged in any of the four configurations shown below. Rank the configurations according to the magnitude of the gravitational force on $ M $ , least to greatest

• A. (i), (ii), (iii), (iv)
• B. (ii), (i), (iii), (iv)
• C. (ii), (i), (iv), (iii)
• D. (ii), (iii), (iv), (i)

Answer 1

Answer: (B)

For configuration (i), gravitational force on $M$ due to $m$ and $m$

$F_{1}=\frac{GMm}{d^{2}}+\frac{GMm}{\left(2d\right)^{2}}=\frac{GMm}{d^{2}}\left[1+\frac{1}{4}\right]$
$=\frac{GMm}{d^{2}}\cdot\frac{5}{4}\ldots\left(i\right)$
For configuration $\left(ii\right)$, gravitational force on $M$ due to $m$ and $m$

$F_{2}=\frac{GMm}{d^{2}}-\frac{GMm}{d^{2}}=0 \dots (iii)$
For configuration (iii), gravitational force on $M$

$F_{3}=\sqrt{\left(F'\right)^{2}+\left(F''\right)^{2}}$
$(\because$ angle between $F'$ and $F''$ is $90^{\circ}$ )
$=\sqrt{\left(\frac{GMm}{d^{2}}\right)^{2}+\left(\frac{GMm}{d^{2}}\right)^{2}}=\frac{GMm}{d^{2}} \sqrt{2} \cdots\left(iii\right)$
For configuration $\left(iv\right)$, gravitational force on $M$ , where $0 < \,\theta <\,90^{\circ}$

Gravitational force on $M$,
$F_{4}=\sqrt{\left(F'\right)^{2}+\left(F''\right)^{2}+2F'F'' cos\,\theta}$
$=\sqrt{\left(\frac{GMm}{d^{2}}\right)^{2}+\left(\frac{GMm}{d^{2}}\right)^{2}}+2\frac{GMm}{d^{2}}\cdot\frac{GMm}{d^{2}} cos\,\theta$
$ = \frac{GMm}{d^2} \sqrt{ 1 + 1 + 2\cos \theta}$
$=\frac{GMm}{d^{2}}\sqrt{2\left(1+cos\,\theta\right)}$
$=\left(\frac{GMm}{d^{2}}\sqrt{2}\right)\sqrt{1+cos\,\theta} \ldots\left(iv\right)$
($\because$ for $0<\,\theta<\,90^{\circ}, \sqrt{1+cos\,\theta\,>})$
From Eqs. $\left(i\right)$, $\left(ii\right)$, $\left(iii\right)$ and $\left(iv\right)$, we get
$F_{2}<\,F_{1}<\,F_{3}<\,F_{4}$
Thus, the correct option is (b)