Question 66: Gravitation

Question

Net gravitational force at the center of a square is found to be $F_1$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_2$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_1}{F_2}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

Newton's Law of Universal Gravitation states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of this force is,
$$ \mathrm{F}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2} $$
where G is the gravitational constant, $\mathrm{m}_1$ and $\mathrm{m}_2$ are the masses, and r is the distance between their centers.
Let's consider square of side length a and the distance from any corner to the centre be r .
Let's place a test mass $\mathrm{m}_0$ at the center. The gravitational force exerted by a mass m at a corner on this test mass will be directed towards that corner with a magnitude:
$$ \mathrm{F}=\frac{\mathrm{Gm} \cdot \mathrm{~m}_0}{\mathrm{r}^2} $$
Let the force on $\mathrm{m}_0$ by another mass M is $\mathrm{F}_0=\frac{\mathrm{GM} \cdot \mathrm{m}_0}{\mathrm{r}^2}$
So, a mass of 2 M will exert a force of $2 \mathrm{~F}_0, 3 \mathrm{M}$ will exert $3 \mathrm{~F}_0$, and 4 M will exert $4 \mathrm{~F}_0$.
Case 1 : Initial arrangement
The magnitudes of forces on the mass kept at centre of square are;
$$ \mathrm{F}_1=4 \mathrm{~F}_0, \mathrm{~F}_2=3 \mathrm{~F}_0, \mathrm{~F}_3=2 \mathrm{~F}_0 	ext { and } \mathrm{F}_4=\mathrm{F}_0 $$
Since $F_2$ and $F_4$ pull in opposite directions, the net force along this diagonal is $3 F_0-F_0=2 F_0$ directed along $\mathrm{F}_2$.
The net force of $F_1$ and $F_3$ is along $F_1$ is $4 F_0-2 F_0=2 F_0$.
Now two net forces, each of magnitude $2 \mathrm{~F}_0$, acting at a $90^{\circ}$ angle to each other.
Then the resultant magnitude $(F)_1$ :
$$ (F)_1=\sqrt{\left(2 F_0ight)^2+\left(2 F_0ight)^2}=\sqrt{8 F_0^2} $$
$$\Rightarrow $$ $$ (\mathrm{F})_1=2 \sqrt{2} \mathrm{~F}_0 $$
Case 2 : Positions of 3 M and 4 M interchanged
The magnitudes of forces on the mass kept at centre of square are;
$\mathrm{F}_1=4 \mathrm{~F}_0, \mathrm{~F}_2=3 \mathrm{~F}_0, \mathrm{~F}_3=2 \mathrm{~F}_0$ and $\mathrm{F}_4=\mathrm{F}_0$
Since $\mathrm{F}_2$ and $\mathrm{F}_3$ pull in opposite directions, the net force along this diagonal is $3 \mathrm{~F}_0-2 \mathrm{~F}_0=\mathrm{F}_0$ directed along $F_2$.
The net force of $\mathrm{F}_1$ and $\mathrm{F}_3$ is along $\mathrm{F}_1$ is $4 \mathrm{~F}_0-\mathrm{F}_0=3 \mathrm{~F}_0$.
Now two net forces are acting at a $90^{\circ}$ angle to each other.
Then the resultant magnitude $(F)_2$ :
$$ (F)_2=\sqrt{\left(F_0ight)^2+\left(3 F_0ight)^2}=\sqrt{10 F_0^2} $$
$$\Rightarrow $$ $$ (\mathrm{F})_2=\sqrt{10} \mathrm{~F}_0 $$
The ratio $\frac{(\mathrm{F})_1}{(\mathrm{~F})_2}$ is :
$$ \frac{(\mathrm{F})_1}{(\mathrm{~F})_2}=\frac{2 \sqrt{2} \mathrm{~F}_0}{\sqrt{10} \mathrm{~F}_0}=\frac{2}{\sqrt{5}} $$
$$\Rightarrow $$ $$ \Rightarrow \frac{2}{\sqrt{5}}=\frac{\alpha}{\sqrt{5}} \Rightarrow \alpha=2 $$
Therefore, the value of $\alpha$ is 2 .
Hence, the correct option is (A).

Answer

2

Answer

$2 \sqrt{5}$

Answer

1

Answer

3

Gravitation

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