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#### skatenerd

##### Active member

- Oct 3, 2012

- 114

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?