Question 63: Gravitation

Question

Three masses $200 \mathrm{~kg}, 300 \mathrm{~kg}$ and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m . They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process $\_\_\_\_$ J. (Gravitational constant $\mathrm{G}=6.7 	imes 10^{-11} \mathrm{~N} \mathrm{~m}^2 / \mathrm{kg}^2$ )

Accepted Answer

The work done ( W ) by an external agent to rearrange a system of masses against gravitational force is equal to the change in the Gravitational Potential Energy $(U)$ of the system.
$$ \mathrm{W}=\Delta \mathrm{U}=\mathrm{U}_{	ext {final }}-\mathrm{U}_{	ext {initial }} $$
For a system of three masses $m_1, m_2, m_3$ separated by distances $r_{12}, r_{23}, r_{31}$, the total potential energy is the sum of the potential energies of all pairs :
$$ \mathrm{U}=\mathrm{U}_{12}+\mathrm{U}_{23}+\mathrm{U}_{31}=-\mathrm{G}\left[\frac{\mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{r}_{12}}+\frac{\mathrm{m}_2 \mathrm{~m}_3}{\mathrm{r}_{23}}+\frac{\mathrm{m}_3 \mathrm{~m}_1}{\mathrm{r}_{31}}ight] $$
Since the masses are at the vertices of an equilateral triangle, all distances are equal ( $r_{12}=r_{23}=r_{31}=L$ ).
$$ \mathrm{U}=-\frac{\mathrm{G}}{\mathrm{~L}}\left(\mathrm{~m}_1 \mathrm{~m}_2+\mathrm{m}_2 \mathrm{~m}_3+\mathrm{m}_3 \mathrm{~m}_1ight) $$
Given masses are,
1. $\mathrm{m}_1=200 \mathrm{~kg}$
2. $\mathrm{m}_2=300 \mathrm{~kg}$
3. $\mathrm{m}_3=400 \mathrm{~kg}$
$$ U=-\frac{G}{L}(200 	imes 300+300 	imes 400+400 	imes 200)=-\frac{G\left(2.6 	imes 10^5ight)}{L} $$
So, the work done by external agent is,
$$ \mathrm{W}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} $$
$$\Rightarrow $$ $\mathrm{W}=\left[-\frac{\mathrm{G}\left(2.6 	imes 10^5ight)}{\mathrm{L}_{\mathrm{f}}}ight]-\left[-\frac{\mathrm{G}\left(2.6 	imes 10^5ight)}{\mathrm{L}_{\mathrm{i}}}ight]$
$$ \mathrm{W}=\mathrm{G}\left(2 . 	imes 10^5ight)\left(\frac{1}{\mathrm{~L}_{\mathrm{i}}}-\frac{1}{\mathrm{~L}_{\mathrm{f}}}ight) $$
The initial side is $=\mathrm{L}_{\mathrm{i}}=20 \mathrm{~m}$
The final side is $\mathrm{L}_{\mathrm{f}}=25 \mathrm{~m}$
Putting the values,
$$ W=\left(6.7 	imes 10^{-11}ight) 	imes\left(2.6 	imes 10^5ight) 	imes\left(\frac{1}{20}-\frac{1}{25}ight) $$
$$\Rightarrow $$ $W=\left(6.7 	imes 10^{-11}ight) 	imes\left(2.6 	imes 10^5ight) 	imes\left(10^{-2}ight)$
$$\Rightarrow $$ $$ \mathrm{W}=1.742 	imes 10^{-7} \mathrm{~J} $$
Therefore, the work done in this process is $1.74 	imes 10^{-7} \mathrm{~J}$.
Hence, the correct option is (B).

Answer

$4.77 	imes 10^{-7}$

Answer

$1.74 	imes 10^{-7}$

Answer

$9.86 	imes 10^{-6}$

Answer

$2.85 	imes 10^{-7}$

Gravitation

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