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 $m$

$F_1=\frac{GMm}{d^2}+\frac{GMm}{{\left(2d\right)}^2}=\frac{GMm}{d^2}\left[1+\frac{1}{4}\right]$
$=\frac{GMm}{d^2}\cdot \frac{5}{4}\dots \left(i\right)$
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 $M$

$F_3=\sqrt{{\left(F^{\prime }\right)}^2+{\left(F^{\prime \prime }\right)}^2}$
$(\because$ angle between $F^{\prime }$ and $F^{\prime \prime }$ 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 $M$,
$F_4=\sqrt{{\left(F^{\prime }\right)}^2+{\left(F^{\prime \prime }\right)}^2+2F^{\prime }{F}^{\prime \prime }cos\theta }$
$=\sqrt{{\left(\frac{GMm}{d^2}\right)}^2+{\left(\frac{GMm}{d^2}\right)}^2}+2\frac{GMm}{d^2}\cdot \frac{GMm}{d^2}cos\theta$
$=\frac{GMm}{d^2}\sqrt{1+1+2\cos \theta }$
$=\frac{GMm}{d^2}\sqrt{2\left(1+cos\theta \right)}$
$=\left(\frac{GMm}{d^2}\sqrt{2}\right)\sqrt{1+cos\theta }\dots \left(iv\right)$
($\because$ 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)