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## Main Question or Discussion Point

Consider two stationary masses, each 1 kg, placed a distance ##d_0## apart. Find the position ##x(t)## of each mass in terms of time ##t##. Consider only gravitation without relativistic effects.

Let the distance between the two masses at time ##t## be ##d(t)##. If we set the origin of our coordinate system at the centre of mass of the system, then ##d(t)=2\,x(t)## since the system is symmetrical about the origin.

Gain in kinetic energy = loss in gravitational potential energy

$$\frac{v^2}{2}=\frac{G}{d}-\frac{G}{d_0}$$

Substituting ##d=2x## and ##d_0=2x_0##,

$$v^2=\frac{G}{x}-\frac{G}{x_0}$$, where ##x_0=x(0)## is the initial ##x##.

$$\int\frac{1}{\sqrt{\frac{G}{x}-\frac{G}{x_0}}}dx=\int dt$$

Q1. We expect ##x<x_0## for ##t>0##. But when ##x<x_0##, LHS is complex while RHS is real. What's wrong?

Integration on LHS: http://www.wolframalpha.com/input/?i=integrate+(G/x-G/u)^(-1/2)dx (where ##u## represents ##x_0##)

Q2. When I solve the question using forces, I get a different expression for ##v^2## (shown below). What's wrong?

$$\ddot{x}=-\frac{G}{d^2}$$

Substituting ##d=2x##,

$$\ddot{x}=-\frac{G}{4x^2}$$

$$\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}=-\frac{G}{4x^2}$$

$$\int vdv=\int-\frac{G}{4x^2}dx$$

$$\frac{v^2}{2}=\frac{G}{4x}-\frac{G}{4x_0}$$

$$v^2=\frac{G}{2x}-\frac{G}{2x_0}$$, which differs from the above equation by a factor of ##\frac{1}{2}##.

Let the distance between the two masses at time ##t## be ##d(t)##. If we set the origin of our coordinate system at the centre of mass of the system, then ##d(t)=2\,x(t)## since the system is symmetrical about the origin.

Gain in kinetic energy = loss in gravitational potential energy

$$\frac{v^2}{2}=\frac{G}{d}-\frac{G}{d_0}$$

Substituting ##d=2x## and ##d_0=2x_0##,

$$v^2=\frac{G}{x}-\frac{G}{x_0}$$, where ##x_0=x(0)## is the initial ##x##.

$$\int\frac{1}{\sqrt{\frac{G}{x}-\frac{G}{x_0}}}dx=\int dt$$

Q1. We expect ##x<x_0## for ##t>0##. But when ##x<x_0##, LHS is complex while RHS is real. What's wrong?

Integration on LHS: http://www.wolframalpha.com/input/?i=integrate+(G/x-G/u)^(-1/2)dx (where ##u## represents ##x_0##)

Q2. When I solve the question using forces, I get a different expression for ##v^2## (shown below). What's wrong?

$$\ddot{x}=-\frac{G}{d^2}$$

Substituting ##d=2x##,

$$\ddot{x}=-\frac{G}{4x^2}$$

$$\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}=-\frac{G}{4x^2}$$

$$\int vdv=\int-\frac{G}{4x^2}dx$$

$$\frac{v^2}{2}=\frac{G}{4x}-\frac{G}{4x_0}$$

$$v^2=\frac{G}{2x}-\frac{G}{2x_0}$$, which differs from the above equation by a factor of ##\frac{1}{2}##.

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