MHB Problem using integral form of Work-Energy Theorem

[skate_nerd]

The problem is that the Earth has lost all velocity and begins plummeting toward the sun. I need to find the time it takes for it to hit the sun.

Note: Primes indicate "dummy variables"

This solution begins with the Work K.E. Theorem:
$$\frac{1}{2}mv(x)^2-\frac{1}{2}mv_{o}^2=\int_{x_o}^{x}F(x')dx'$$
Where \(v_{o}=0\) and $$F(x')=F(r)=\frac{-GMm}{r^2}$$
Plugging it all in gives
$$\frac{1}{2}mv(r)^2-\frac{1}{2}m(0)^2=\int_{r_{au}}^{r(t)}-\frac{GMm}{r'^2}dr'$$
$$\frac{1}{2}mv(r)^2=-GMm\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'$$
Earth's mass cancels out, as expected, and then we want to solve to get the function \(v(r)\):
$$v(r)=\sqrt{-2GM\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'}$$
$$=\sqrt{-2GM(-\frac{1}{r(t)}+\frac{1}{r_{au}})}$$
$$=\sqrt{2GM(\frac{1}{r(t)}-\frac{1}{r_{au}})}$$
Next, we use the following formula to find time as a function of position:
$$t(r)=\int_{r_{au}}^{r_{sun}}\frac{1}{v(r')}dr'$$
$$=\frac{1}{\sqrt{2GM}}{\int_{r_{au}}^{r_{sun}}} \frac{1}{\sqrt{\frac{1}{r(t)}-\frac{1}{r_{au}}}}dr'$$
The above integral gets a very long and gross answer via wolframalpha, and when I try plugging in the bounds I end up with an imaginary number.
Anybody know where I went wrong?

 

[topsquark]

Re: problem using integral form of Work K.E. Thm

$$\frac{1}{2}mv(r)^2-\frac{1}{2}m(0)^2=\int_{r_{au}}^{r(t)}-\frac{GMm}{r'^2}dr'$$
$$\frac{1}{2}mv(r)^2=-GMm\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'$$
Earth's mass cancels out, as expected, and then we want to solve to get the function \(v(r)\):
$$v(r)=\sqrt{-2GM\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'}$$
$$=\sqrt{-2GM(-\frac{1}{r(t)}+\frac{1}{r_{au}})}$$
$$=\sqrt{2GM(\frac{1}{r(t)}-\frac{1}{r_{au}})}$$

Starting from here I would use v = dr/dt, giving
[math] T = \int \frac{dr}{\sqrt{2GM \left ( \frac{1}{r} - \frac{1}{r_{au}} \right ) }} [/math]

There may be some substitutions that will shortcut this, but I prefer to work one substitution at a time. The method isn't that bad, but you really need to see it before you can do it on your own.

Let u = 1/r. Then the integral becomes
[math]T = -\int \frac{1}{\sqrt{Au - B}}\frac{1}{u^2}du[/math]
(where A and B are the appropriate constants.)

Now let y = Au - B. Then
[math]T = -A \int \frac{1}{\sqrt{y}} \frac{1}{(y + B)^2} dy[/math]

Let [math]z = \sqrt{y}[/math]. Then
[math]T = -2A \int \frac{1}{(z^2 + B)^2} dz[/math]

Let [math]z = \sqrt{B}~tan( \theta )[/math]. Then
[math]T = -\frac{2A}{B^{3/2}} \int cos^2( \theta )~d \theta [/math]

You can take it from here.

-Dan

 

[skate_nerd]

Re: problem using integral form of Work K.E. Thm

Wow, yeah I don't think I would have ever come up with that on my own haha. Thanks for the help.

 

[topsquark]

Re: problem using integral form of Work K.E. Thm

Wow, yeah I don't think I would have ever come up with that on my own haha. Thanks for the help.

It highlights something I've noted in the more advanced Mathematics...keep redefining the problem until you get something familiar! (Cool)

-Dan