Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.4k views
in Physics by (91.6k points)
closed by
Three particles, each of mass `m` are placed at the vertices of an equilateral triangle of side `a`. What are the gravitation field and gravitational potential at the centroid of the triangle.

1 Answer

0 votes
by (91.1k points)
selected by
 
Best answer
Refer to Fig. `O` is the centriod of triangle `ABC`, where
`OA = (2)/(3) AD = (2)/(3) (AB sin 60^(@))`
`= (2)/(3) xx a xx (sqrt(3))/(2) = (a)/(sqrt(3))`
Thus, `OA = OB = OC = (a)/(sqrt(3))`
image
The gravitational intensity at `O` due to mass `m`
at `A` is, `I_(A) = (Gm)/((OA)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OA`.
Similarly the gravitational intensity at `O` due to mass `m` at `B` is,
`I_(B) = (Gm)/((OB)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OB`.
and gravitational intensity at `O` due to mass `m`
at `C` is,
`I_(C) = (Gm)/((OC)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OC`.
As `I_(A),I_(B)` and `I_(C)` are equal in magnitude and equally inclined to each other, the resultant gravitational intensity at `O` is zero.
gravitational potential at `O` due to masses at
`A,B` and `C` is
`V = - (Gm)/(OA) + (-(Gm)/(OB)) + (-(Gm)/(OC))`
`= - (3 Gm)/(OA) = (-3 Gm)/(a//sqrt(3)) = (-3 sqrt(3) Gm)/(a)`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...