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Mathematics : Summation of 1 - 1 + 1 - 1

Consider, S = 1 - 1 + 1 - 1 + . . . 1 - S = 1 - (1 - 1 + 1 - 1 + . . .) 1 - S = 1 - 1 + 1 - 1 + 1 - . . . 1 - S = S Here left S and right S are different. If we take odd number of terms then left S is 1 and right S is 0. If we take even number of terms then left S is 0 and right S is 1. Thank you. Praveen Kumar Sirohiwal

Mathematics : if one divisor of a dividend is greater than the square root of the dividend, then the quotient is smaller than the square root of the dividend.

We have ij = n, i > 0, j > 0 if i > sqrt(n) ij > sqrt(n) j n > sqrt(n) j sqrt(n) > j Hence we have to check the primality of a number by trying to divide it only upto the square root of the number. Thank you, Praveen Kumar Sirohiwal.

Gravitational equation of motion for an object falling on a planet.

Suppose mass of planet is M and mass of object is m. Object is r meters away from planet which is R+H where R is radius of planet and H is height. At any instance, object is x meters away from centre of mass of planet. According to Sir Issac Newton Gravitational force on object = ma = - GMm/x 2 a = - GM/x 2 dv/dt = - GM/x 2 dv/dt dx/dx = - GM/x 2 dx/dt dv = - GM/x 2 dx v dv = - GM/x 2 dx Integrating both sides v 2 /2 = GM/x + C At x = r, v = u u 2 /2 = GM/r + C C = u 2 /2 - GM/r  = (ru 2 - 2GM)/2r Now, v 2 = 2(GM/x + C) v = - sqrt(2) sqrt((GM + xC)/x) dx/dt = - sqrt(2) sqrt((GM + xC)/x) dt = - 1/sqrt(2) sqrt(x/(GM + xC)) dx Let xC = GM tan 2 c x = GM/C tan 2 c dx = GM/C 2 tan(c) sec 2 c dc   = 2GM/C tan(c) sec 2 c dc tan 2 c = xC/GM tan(c) = sqrt(xC/GM) sec 2 c = xc/GM + 1 sec(c) = sqrt((xC + GM)/GM) Putting in equation dt = - sqrt(GM/2C) sqrt(tan 2 c/(GM + GM tan 2 c)) 2GM/C tan(c) sec 2 c dc   = - sqrt(2

Two masses each of one kg, one meter apart. After how much time will they collide?

Suppose two masses are at rest and we release them. Centre of mass is at rest. At any instance, Mass is x meters away from centre of mass. Force = 1.a = -G.1.1/(2x)^2     (since mass is 1 kg) a = -G/4x^2 dv/dt = -G/4x^2 dv/dt dx/dx = -G/4x^2 dx/dt dv = -G/4x^2  dx v dv = -G/4x^2  dx Integrating both sides v^2 / 2 = G/4x + C At x = 1/2, we have v = 0 0 = G/2 + C C = -G/2 Putting in equation v^2 / 2 = G/4x - G/2 v^2 = G/2x - G v^2 = G (1 - 2x)/2x v = - sqrt(G) sqrt((1 - 2x)/2x) dx/dt = - sqrt(G) sqrt((1 - 2x)/2x) sqrt(G) dt = - sqrt(2x/(1 - 2x)) dx Let 2x = (sin(c))^2 Deriving, 2 dx = 2 sin(c) cos(c) dc dx = sin(c) cos(c) dc Putting in equation sqrt(G) dt = - sqrt(((sin(c))^2)/(1 - (sin(c)^2)) sin(c) cos(c) dc sqrt(G) dt = - sin(c)/cos(c)  sin(c) cos(c) dc sqrt(G) dt = - (sin(c))^2 dc sqrt(G) dt = - (1 - cos(2c))/2  dc sqrt(G) dt = - 1/2 dc + cos(2c)/2 dc Integrating both sides sqrt(G) t = - c/2 + sin(2c)/4 + C sqrt(G) t = - arcsin(sqrt(2x))/2 + sin(2(arcsin(sqrt(2x))/4 + C At