Math not working right!

  • #1
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.

Homework Equations


[tex]F=\frac{-G M_e m}{r^2}[/tex]

The Attempt at a Solution


[tex]F=\frac{-G M_e m}{r^2}[/tex]
[tex]a=\frac{-G M_e}{r^2}[/tex]
[tex]\frac{\partial v}{\partial r} v = \frac{-G M_e}{r^2}[/tex]
seperate and integrate
[tex]
\frac{1}{2}v^2=\frac{G M_e}{r}+K[/tex]
[tex]v^2 = \frac{2G M_e}{r}+K[/tex]
[tex]v^2(r_0)=\frac{2G M_e}{r_0}+K=0[/tex]
[tex]v^2=2G M_e \( \frac{1}{r}-\frac{1}{r_0}\)[/tex]
Then take square root, replace v with the time derivative of r, seperate and integrate and you get this mess:
[tex]\int{\sqrt{\frac{r_0 r}{r_0-r}}\partial r}=\sqrt{G M_e}t[/tex]
Which comes out to having a tangent in it, and in general very messy. Also when you plug the two values [tex]r_0[/tex] and 0, you wind up with an infinity. So any ideas where I went wrong?
 
Last edited:

Answers and Replies

  • #2
dlgoff
Science Advisor
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Are you sure a=(dv/dt)v?
 
  • #3
It's not a = (dv/dt)v, it's a=(dv/dr)v
[tex] a = \frac{\partial v}{\partial r}*\frac{\partial r}{\partial t} = \frac{dv}{dt}[/tex]
Chain rule!
 
  • #4
dlgoff
Science Advisor
Gold Member
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Okay, sorry. Couldn't make out the r on this old computer monitor.
 

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