Question

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

• A. 21 G
• B. 100 G
• C. 59 G
• D. 42 G

Answer 1

Answer: (B)

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