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  4. Three identical particle A, B and C of mass 100 kg each are placed in a straight line with AB = BC =13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately :

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Q. Three identical particle $A, B$ and$ C$ of mass $100 \,kg$ each are placed in a straight line with $AB = BC =13 \,m$. The gravitational force on a fourth particle $P$ of the same mass is $F$, when placed at a distance $13\, m$ from the particle B on the perpendicular bisector of the line $AC$. The value of $F$ will be approximately :

JEE MainJEE Main 2022Gravitation

Solution:

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$ F =\frac{ GMM }{ r ^2}+\sqrt{2} \frac{ GMM }{(\sqrt{2} r )^2}$
$ =\frac{ GMM }{ r ^2}\left(1+\frac{1}{\sqrt{2}}\right) $
$ =\frac{ G \times 10^4}{13^2}\left(1+\frac{1}{\sqrt{2}}\right) $
$ F \simeq 100 G $