three balls ,a,b,c are kept in a straight line the sepration between a c is 1 metre , and - Science - Gravitation

Question

three balls ,a,b,c are kept in a straight line the sepration
between a c is 1 metre , and b is kept at the midpoint between them
the masses of a,b,c are 100 g ,200g,300g,respectively find the net
gravitation force on a,b,c

Jyoti Pant · Accepted Answer

We have three balls a,b and c placed in straight line
Given,
ma=100 g=0 1 kgmb=200g=0 2 kgmc=300g=0 3 kgab=0 5 m=bcac=1mNow
using relation,F=Gmm'r2Force on a=Force on a due to b +Force on a
due to c=Gmambab2+Gmamcac2=6 67×10-11[0 1×0 20 52+0
1×0 312]=6 67×10-11[0 020 25+0 031]=7
34×10-10NSimilarly force on b,Force on b due to c -Force on b
due to a(Forces are in opposite direction)=Gmbmcbc2-Gmbmaba2=6
67×10-11[0 2×0 30 52-0 2×0 10 52]=6
67×10-11[0 060 25-0 020 25]=1 06×10-11NForce on c=Force
on c due to a +Force on c due to b=Gmcmaac2+Gmcmbcb2=6
67×10-11[0 3×0 112+0 3×0 20 52]=6
67×10-11[0 03+0 060 25]=1 8009×10-11N

three balls ,a,b,c are kept in a straight line. the sepration between a c is 1 metre , and b is kept at the midpoint between them. the masses of a,b,c are 100 g ,200g,300g,respectively . find the net gravitation force on a,b,c .