Science

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

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