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rcgldr
#8
Dec18-09, 02:43 AM
HW Helper
P: 7,126
Quote Quote by diazona View Post
The trick I've usually seen is to multiply both sides by [itex]\dot{r}[/itex]
v= dr/dt

a = dv/dt

multiply by dv/dt by dr/dr:

a = (dr dv)/(dt dr) = v dv/dr

This gets you to the first step:

v dv/dr = -G (m1 + m2) / r2

v dv = -G (m1 + m2) dr / r2

for v= 0 at r0 you get:

[tex]1/2 v^2 = G (m_1 + m_2) / r - G (m_1 + m_2) / r_0 [/tex]

[tex]v = \sqrt{ 2 G (m_1 + m_2) / r - 2 G (m_1 + m_2) / r_0} [/tex]

[tex]v = \frac{dr}{dt} = \sqrt{ \frac{2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)} {r\ r_0} [/tex]

[tex]\frac{\sqrt{r_0 \ r} \ dr} {\sqrt{ 2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)}} = dt[/tex]

[tex]{\sqrt{ \frac{r_0}{2 G (m_1 + m_2)}}} \ \ \sqrt{\frac{r}{r_0 - r}} \ dr = dt[/tex]

Using arildno's method from this thread
http://www.physicsforums.com/showthread.php?t=306442

[tex]u = \sqrt{\frac{r}{r_0-r}}[/tex]

[tex]u^2 = {\frac{r}{r_0-r}}[/tex]
[tex]r = r_0 u^2 - ru^2[/tex]
[tex]r + r u^2 = r_0[/tex]
[tex]r(1 + u^2) = r_0[/tex]
[tex]r = \frac{r_0 u^2}{1 + u^2} = \frac{r_0 + r_0 u^2 - r_0}{1 + u^2} = \frac{ r_0(1 + u^2) - r_0}{1 + u^2} = r_0 - \frac{r_0}{1 + u^2}[/tex]

[tex]dr = \frac{2 r_0 u}{(1 + u^2)^2}[/tex]

at r = r0, u = ∞, at r = 0, u = 0.

[tex]t = \frac{2 {r_0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \ \int_0^\infty \frac{u^{2}}{(1+u^{2})^{2}}du= \frac{2 {r_0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \left [ \frac{1}{2} \left (\tan^{-1}(u)-\frac{u}{1+u^{2}} \right ) \right ] _0^\infty = \frac{\pi\ {r_0}^{3/2}}{2 \sqrt{2 G (m1 + m2)}} [/tex]

This matches arildno's method, except I kept m1 + m2 as variables, while he combined the two 1kg masses. I also checked this doing crude numerical integration via a spreadsheet.

To answer the original post, for an initial distance of 1 meter, t ~= 96136 seconds.
To check my math, I also tested with 2 meters, where t ~= 271915 seconds