Changes between Version 18 and Version 19 of SelfGravityDevel

Timestamp:

03/05/12 21:35:27 (13 years ago)

Author:

Baowei Liu

Comment:

—

SelfGravityDevel

@@ -62 +62 @@
 }}}
 
+=== Free-fall Time ===
+
+The free-fall time is the characteristic time that would take a body to collapse under its own gravitational attraction, if no other forces existed to oppose the collapse.
+Using the Gauss's theorem, the gravitional acceleration $g_{r}$ is given by
+{{{
+#!latex
+\begin{equation}\label{eq:graAcce}\tag{5}
+g_{r}=-\frac{GM_{r}}{r^{2}}
+\end{equation}
+}}}
+where [[latex($M_{r}$)]] is the mass inside [[latex($r$)]]
+{{{
+#!latex
+\begin{equation}\label{eq:insideMass}\tag{6}
+M_{r}=\int_{0}^{r}\rho4\pi r^{2}dr
+\end{equation}
+}}}
+So the equation of motion for a gas molecule under a control of the self-gravity can be written as
+{{{
+#!latex
+\begin{equation}\label{eq:motion}\tag{7}
+\frac{d^{2}r}{dt^{2}}=-\frac{GM_{r}}{r^{2}}
+\end{equation}
+}}}
+Equation (7) can be written as the form of the conservation of mechanical energy 
+{{{
+#!latex
+\begin{equation}\label{eq:conserE}\tag{8}
+\frac{1}{2}\left(\frac{dr}{dt}\right)^{2}-\frac{GM_{r}}{r}=E
+\end{equation}
+}}}
+Suppose the gas molecule is initially located at $r=R$. So the total energy
+{{{
+#!latex
+\begin{equation}\label{eq:totalE}\tag{9}
+E=-\frac{GM_{r}(R)}{R}
+\end{equation}
+}}}
+Now assume [[latex($M_{r}\approx M_{r}(R)$)]]. Equation (8) gives
+{{{
+#!latex
+\begin{equation}\label{eq:approx1}\tag{10}
+\frac{dr}{dt}=-[2GM_{r}(R)]^{\frac{1}{2}}\left(\frac{1}{r}-\frac{1}{R} \right)^{\frac{1}{2}}
+\end{equation}
+}}}
+Let [[latex($r=R\cos^{2}\eta$)]]. From the initial state we have [[latex($\eta(t=0)=0$)]]. So Equation (10) reduces to
+{{{
+#!latex
+\begin{equation}\label{eq:reduced}\tag{11}
+\cos^{2}\eta\frac{d\eta}{dt}=\left(\frac{GM_{r}(R)}{2R^{3}}\right)^{\frac{1}{2}}
+\end{equation}
+}}}
+Solve Equation (11) we have
+{{{
+#!latex
+\begin{equation}\label{eq:sol}\tag{12}
+t=\left(\frac{R^{3}}{2GM_{r}(R)} \right)^{\frac{1}{2}}\left(\eta+\frac{\sin2\eta}{2} \right)
+\end{equation}
+}}}
+Let [[latex($r=0$)]] or [[latex($\eta=\frac{\pi}{2}$)]]. we have the free-fall time
+{{{
+#!latex
+\begin{equation}\ref{eq:freeFall}\tag{13} 
+\begin{aligned}
+ t_{ff}&=\left(\frac{R^{3}}{2GM_{r}(R)}\right)^{\frac{1}{2}}\frac{\pi}{2}\\
+       &=\left(\frac{3\pi}{32G\bar{\rho}}\right)^{\frac{1}{2}}
+\end{aligned}
+\end{equation}
+}}}
+where
+{{{
+#!latex
+\begin{equation}\label{rhoBar}\tag{14}
+\bar{\rho}=\frac{M_{r}(R)}{\frac{4\pi R^{3}}{3}}
+\end{equation}
+}}}
+is the average density of the gas.
+
+
 == Implementation ==