Hello, I do not know why the assumption that x/r = sin^2(theta) was made. Could

Question

Hello, I do not know why the assumption that x/r = sin^2(theta) was made. Could someone help make it clear why sin is squared? I am talking about a question in Analytical Mechanics 7th by Cassiday, it is problem 2.14.
The problem is: 2.14 A particle of mass    m is released from rest a distance    b from a fixed orgin of force that attracts    the particle according to the inverse square law:             F(x) = k/x^2         
   Show that the time required for the particle to reach the orgin is          Pi*m*b^3 /(8 k)

The problem is: 2.14 A particle of mass    m is released from rest a distance    b from a fixed orgin of force that attracts    the particle according to the inverse square law:             F(x) = k/x^2         
   Show that the time required for the particle to reach the orgin is          Pi*m*b^3 /(8 k)
2.14 A particle of mass    m is released from rest a distance    b from a fixed orgin of force that attracts    the particle according to the inverse square law:             F(x) = k/x^2         
   Show that the time required for the particle to reach the orgin is          Pi*m*b^3 /(8 k)

Explanation / Answer

F = -k/x^2, where negative means the force is attractive dv/dt = F/m = -(k/m)*1/x^2 dv/dx * dx/dt = -(k/m)*1/x^2 v dv/dx = -(k/m)*1/x^2 v dv = (k/m)*(-1/x^2) dx integrate, v^2/2 = (k/m)*(1/x) + C use x = b, v = 0, so C = -(k/m)*(1/b) v^2/2 = (k/m)*(1/x - 1/b) v = sqrt(2k/mb)*sqrt[(b - x)/x) dx/dt = sqrt(2k/mb)*sqrt[(b - x)/x)] sqrt[x/(b - x)] dx = sqrt(2k/mb) dt integrate, b*arctan{sqrt[x/(b - x)]} - sqrt[(x(b - x)] = sqrt(2k/mb) * t + C' use t = 0, x = b C' = -pi*b/2 find t when x = 0 t = pi*b/2 * sqrt[mb/(2k)] = pi*sqrt[mb^3/(8k)]

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