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 t = 0, x = 1/2

0 = - arcsin(1)/2 + sin(2(arcsin(1)))/4 + C

0 = - PI/4 + sin(PI)/4 + C

0 = - PI/4 + 0 + C

C = PI/4

Putting in equation

sqrt(G) t = - arcsin(sqrt(2x))/2 + sin(2(arcsin(sqrt(2x))))/4 + PI/4

At collision x = r which is radius of object

sqrt(G) t = - arcsin(sqrt(2r))/2 + sin(2(arcsin(sqrt(2r))))/4 + PI/4

Thank you,
Praveen Kumar Sirohiwal