Question

The distance between two objects with mass m1 and m2 are in the intergalactic space is equal to l. With no other forces acting except gravity, how soon will they collide?

Answer 1

Assuming that object of mass `m_2` is orbiting object of mass `m_1`.

We make use of Kepler's Third Law which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis `a` of its orbit.

`T^2propa^3`

It is given that the distance between the two objects is `l`. `:.` semi major axis `a=l/2`. To find the force equation we use circular orbit which is a special case of ellipse.

1. Using the concept of reduced mass `bbmu` which is a quantity which allows the two-body problem to be solved as if it were a one-body problem. We get

   `bbmu=1/(1/m_1+1/m_2)=(m_1m_2)/(m_1+m_2)` ....(1)
   where the force on this mass is given by the force between the two objects.

2. For a circular orbit of radius `r` equating centripetal force with the gravitational force we have

   `bbmuromega^2=G(m_1m_2)/r^2` .....(2)

Rewriting `ω=(2π)/T`, where `T` is the time period of circular orbit. Taking `r=l/2`, using (1) and solving (2) for `T` we get

`(m_1m_2)/(m_1+m_2)l/2((2pi)/T)^2=G(m_1m_2)/(l/2)^2`
`=>(1)/(m_1+m_2)(l/2)^3(4pi)/T^2=G`
`=>T=pisqrt(l^3/(2G(m_1+m_2)))`
3. In the limiting case, the two objects are moving in a straight line towards each other. If `T_c` is the time of collision, then

`T_c=T/2=pi/2sqrt(l^3/(2G(m_1+m_2)))`

Answer 2

Time to collision is:

• `pi sqrt( (L^3 )/( 8 G(m_1+ m_2)))`

Explanation:

KEY IDEA : Because there are no external forces on the system, the centre of mass (CoM) of the system will remain the same throughout, and this is also where any collision will occur.

Makes sense to place the origin of a co-ordinate system at the CoM with the objects on a number line such that, at `t = 0`:

`m_1 x_o + m_2 y_o = 0`

• `implies {(m_1: qquad x(0) = x_o = - m_2/m_1 y_o),(m_2: qquad y(0) = y_o):}`

Throughout, that same relationship, `- m_1 x = m_2 y`, will hold as there are no external forces acting on the system of objects

Considering only `m_2`, the point being that the co-ordinate system reduces this immediately to a problem in a single variable:

• Newton's Second Law

`sum bb F = m bb a`

`implies m_2 ddot y = - (Gm_1m_2)/(color(blue)(y - x))^2`

`ddot y = - (Gm_1)/(color(blue)(y(1 + m_2/m_1)))^2`

`ddot y = - alpha/y^2, qquad alpha = (Gm_1 )/(1 + m_2/m_1)^2`

From here, it's mostly just processing calculus.

Using integrating factor, `dot y`, and integrating wrt `t`:

`dot y ddot y = - alpha/y^2 dot y qquad implies dot y^2/2 = alpha/y + C`

IV's : `qquad dot y(0) = 0, qquad y(0) = y_o implies C = - alpha/y_o`

`dot y^2/2 = alpha(1/y - 1/y_o) = alpha/y_o((y_o - y)/(y))`

Bit of physics : Take the negative square root as `m_2` is moving back to the origin:

`dot y = - sqrt( ( 2alpha)/y_o((y_o - y)/(y))) = - sqrt( (2 alpha)/y_o) * sqrt(( y_o - y)/( y))`

That is separable . And if the objects collide at `t = tau`, then:

`int_0^tau \ dt = tau = - sqrt( y_o/( 2alpha))color(red)( int_(y_o)^0 \ dy \ sqrt(( y)/(y_o - y)))`

In terms of the indefinite version of the integral in red, this intermediate step is all over the internet, and it works really well:

`z = sqrt(( y)/(y_o - y)) implies y = (y_o z^2)/(1 + z^2) = y_o - y_o/(1 + z^2)`

`implies dy = (2 y_o z)/(1+z^2)^2 dz`

`tau = - sqrt( (2y_o^3)/( alpha)) int_(oo)^0 \ dz qquad (z^2 )/((1+z^2)^2) qquad square`

That intermediate step sets it up very nicely for IBP. Noting that:

`-1/2 d/dz( (1 )/(1+z^2) )= (z )/(1+z^2)^2`

Then by IBP :

`int \ dz qquad (z^2 )/((1+z^2)^2) = int \ dz qquad z * (-1/2 d/dz( (1 )/(1+z^2) ))`

`= - 1/2 ( (z )/(1+z^2) ) + 1/2 underbrace(int \ dz qquad (1 )/(1+z^2))_(= tan^(-1) z) + C`

ie:

`square = tau = - sqrt( (y_o^3)/( 2 alpha)) \ [ tan^(-1) z - ( (z )/(1+z^2) ) ]_(oo)^(0)`

`= pi sqrt( (y_o^3)/( 8 alpha)) = pi sqrt( (y_o^3 (1 + m_2/m_1)^2)/( 8 Gm_1))`

The stated initial separation of the objects, `L`, is:

• `L = y - x = y_o(1 + m_2/m_1)`

`implies tau = pi sqrt( (L^2 color(blue)(y_o) )/( 8 G color(blue)(m_1)))`

Not quite finished. Note that:

`L = y_o(1 + m_2/m_1) = y_o( (m_1 + m_2)/m_1) implies color(blue)(y_o/m_1) = L/(m_1 + m_2)`

`implies tau = pi sqrt( (L^3 )/( 8 G(m_1+ m_2)))`