Like any of you ever thought my posts were on topic ...

(crossposted, trim) I'm sure this will interest the s.e.d regulars and
several can solve it, and s.s.science & rec.models.rockets are my best
guess for where it actually is on topic. Any other groups likely to be
appropriate? This is related to my post about eMachineShop in other
groups this week.

Given two spheres of mass \r and density \rho, each has a mass of \rho V = 4/3 \pi \rho r^3.

Separate them, center to center, by 3r. That is, surface to surface separation of r, then the gravitational force between them is

F = G m m / (9r^2)

Using F = ma and waiving hands wildly while students sleep ...

a = -G m / (9r^2) a = -4/27 \pi \rho G r

Using iridium (for small spheres), \rho = 22,160 kg/m^3 G=6.673E-11 m^3 / (kg s^2) a = -688E-9 m/s^2 r

Notice, a is reletive to center of mass, not sphere to sphere, explaining a factor of 2 that likes to hide.

Solving s=1/2at^2 and bounding t by the initial value of a:

t = 1,205s or 20 min. Notice, it doesn't matter what r is. ..... Now, call the original position \s_0 at time t=0 and place the origin at the center of mass. At time t, the position is s, and the separation between centers will be 2s. There is surely more elegant notation.

a = 4/3 \pi G \rho r^3 / s^2

s(t) = \int \int (4/3 \pi G r^3) / (2s)^2 dt dt

Maybe I'm slow this morning, but I don't see the solution to this. Can someone with untarnished calculus or mathematica give a solution to this, for arbitrary s_0? Is there some aerospace engineering text where this is treated as an example or homework problem?

(crossposted, trim) I'm sure this will interest the s.e.d regulars and

Given two spheres of mass \r and density \rho, each has a mass of \rho V = 4/3 \pi \rho r^3.

Separate them, center to center, by 3r. That is, surface to surface separation of r, then the gravitational force between them is

F = G m m / (9r^2)

Using F = ma and waiving hands wildly while students sleep ...

a = -G m / (9r^2) a = -4/27 \pi \rho G r

Using iridium (for small spheres), \rho = 22,160 kg/m^3 G=6.673E-11 m^3 / (kg s^2) a = -688E-9 m/s^2 r

Notice, a is reletive to center of mass, not sphere to sphere, explaining a factor of 2 that likes to hide.

Solving s=1/2at^2 and bounding t by the initial value of a:

t = 1,205s or 20 min. Notice, it doesn't matter what r is. ..... Now, call the original position \s_0 at time t=0 and place the origin at the center of mass. At time t, the position is s, and the separation between centers will be 2s. There is surely more elegant notation.

a = 4/3 \pi G \rho r^3 / s^2

s(t) = \int \int (4/3 \pi G r^3) / (2s)^2 dt dt

Maybe I'm slow this morning, but I don't see the solution to this. Can someone with untarnished calculus or mathematica give a solution to this, for arbitrary s_0? Is there some aerospace engineering text where this is treated as an example or homework problem?