Question

The bodies of equal masses (m) are placed at vertices of equilateral triangle, the gravitational feild intensity at the centre of the equilateral triangle is​

Answer 1

R is the resultant of the two forces which is equal to F

The gravitational field intensity at the centre of the triangle will be zero due to the symmetry and equal forces of attraction by all three masses

Hope this is helpful to you

Pls mark as brainliest

Follow me and I'll follow you back

Answer 2

Answer:

is the centriod of triangle ABCABC, where

OA=23AD=23(ABsin60∘)OA=23AD=23(ABsin60∘)

=23×a×3–√2=a3–√=23×a×32=a3

Thus, OA=OB=OC=a3–√OA=OB=OC=a3

The gravitational intensity at OO due to mass mm

at AA is, IA=Gm(OA)2=Gm(a/3–√)2IA=Gm(OA)2=Gm(a/3)2 along OAOA.

Similarly the gravitational intensity at OO due to mass mm at BB is,

IB=Gm(OB)2=Gm(a/3–√)2IB=Gm(OB)2=Gm(a/3)2 along OBOB.

and gravitational intensity at OO due to mass mm

at CC is,

IC=Gm(OC)2=Gm(a/3–√)2IC=Gm(OC)2=Gm(a/3)2 along OCOC.

As IA,IBIA,IB and ICIC are equal in magnitude and equally inclined to each other, the resultant gravitational intensity at OO is zero.

gravitational potential at OO due to masses at

A,BA,B and CC is

V=−GmOA+(−GmOB)+(−GmOC)V=-GmOA+(-GmOB)+(-GmOC)

=−3GmOA=−3Gma/3–√=−33–√Gma