Four identical particles of mass m are kept a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the side length of the square be $a$. Consider one of the masses at a corner. It experiences gravitational forces from the other three masses.

Vedclass · Accepted Answer

$4 L$

Vedclass · Answer

(B) Let the side length of the square be $a$. Consider one of the masses at a corner. It experiences gravitational forces from the other three masses.Two masses are at a distance $a$ (adjacent corners),and one mass is at a distance $\sqrt{2}a$ (opposite corner).The forces from the two adjacent masses are $F = \frac{Gm^2}{a^2}$ each,acting at an angle of $90^\circ$. Their resultant is $\sqrt{F^2 + F^2} = \sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$.The force from the opposite mass is $F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$.Both these forces act along the diagonal of the square,so the net force is:$F_{	ext{net}} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left( \sqrt{2} + \frac{1}{2} ight) = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} ight)$.Given that $F_{	ext{net}} = \left( \frac{2\sqrt{2} + 1}{32} ight) \frac{Gm^2}{L^2}$,we equate the two expressions:$\frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} ight) = \left( \frac{2\sqrt{2} + 1}{32} ight) \frac{Gm^2}{L^2}$.$\frac{1}{2a^2} = \frac{1}{32L^2} \implies a^2 = 16L^2 \implies a = 4L$.

Four identical particles of mass $m$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{Gm^2}{L^2}$,the length of the sides of the square is

Similar Questions

Which of the following statements about the gravitational constant $G$ is true?

Four identical masses of $m$ are kept at the corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{G m^2}{L^2}$,then the length of the side of the square is

The mass of the moon is about $1.2\%$ of the mass of the earth. Compared to the gravitational force the earth exerts on the moon,the gravitational force the moon exerts on the earth:

Two spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F$. The space around the masses is now filled with a liquid of specific gravity $3$. The gravitational force between the bodies will now be

The value of $G$ on Earth is $6.67 \times 10^{-11} \ Nm^2/kg^2$. What is the value of $G$ on the Moon?

Vedclass Test Series

Exam Paper Generator

Online Exam Module