Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
mass M is divided into two parts xm and (1-xm). for a given separtion the value of x for which gravitationl attraction between two piceces of mass becomes maximum
Fnet = G(xm) (1-xm)/x^2For maxima we need to differentiate it,Hence,Fnet = Gm(1-xm)/xGm[-m*x + 1.(1-xm)]/x^2 = 0(1-xm) = mx2mx = 1x = 1/2m meter.where m is the mass.Put x in Fnet we get,Fnet = G/4x^2 N.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -