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PiratePhysicist
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1. Problem: A particle falls to Earth starting from rest at a great height (may times Earth's radius ). Neglect air resistance and show that the particle requires approximately 9/11 of the total time of fall to traverse the first half of the distance.
F=\frac{-G M_e m}{r^2}
F=\frac{-G M_e m}{r^2}
a=\frac{-G M_e}{r^2}
\frac{\partial v}{\partial r} v = \frac{-G M_e}{r^2}
seperate and integrate
<br /> \frac{1}{2}v^2=\frac{G M_e}{r}+K
v^2 = \frac{2G M_e}{r}+K
v^2(r_0)=\frac{2G M_e}{r_0}+K=0
v^2=2G M_e \( \frac{1}{r}-\frac{1}{r_0}\)
Then take square root, replace v with the time derivative of r, separate and integrate and you get this mess:
\int{\sqrt{\frac{r_0 r}{r_0-r}}\partial r}=\sqrt{G M_e}t
Which comes out to having a tangent in it, and in general very messy. Also when you plug the two values r_0 and 0, you wind up with an infinity. So any ideas where I went wrong?
Homework Equations
F=\frac{-G M_e m}{r^2}
The Attempt at a Solution
F=\frac{-G M_e m}{r^2}
a=\frac{-G M_e}{r^2}
\frac{\partial v}{\partial r} v = \frac{-G M_e}{r^2}
seperate and integrate
<br /> \frac{1}{2}v^2=\frac{G M_e}{r}+K
v^2 = \frac{2G M_e}{r}+K
v^2(r_0)=\frac{2G M_e}{r_0}+K=0
v^2=2G M_e \( \frac{1}{r}-\frac{1}{r_0}\)
Then take square root, replace v with the time derivative of r, separate and integrate and you get this mess:
\int{\sqrt{\frac{r_0 r}{r_0-r}}\partial r}=\sqrt{G M_e}t
Which comes out to having a tangent in it, and in general very messy. Also when you plug the two values r_0 and 0, you wind up with an infinity. So any ideas where I went wrong?
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