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A very large number of particles of same mass m are kept at horizontal distances of 1m, 2m, 4m, 8mand so on from (0, 0) point The total gravitational potential at this point (0, 0) is(A)- 8G m8-3.(B)-3G m(C)-4G m(D)- 2G m |
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Answer» Gravitational potential at pointris given by, V = - G*m/r According to the problem the gravitational potential by infinite masses at the origin is, V = -G*m (1 + 1/2 + 1/4 + 1/8+...) This is a geometric series with ratio 1/2 So, V = -G*m*2. As(sum of infinite terms of G.P. = 1/(1-1/2 ) = 2 = - 2*G*m |
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