Analysis of the physics in Interstellar

[phi1123]

So I recently watched the new movie Interstellar, and I've been inspired to do some more general relativity. At one point in the movie they mention that 1 hour on a planet orbiting a black hole is 7 years back on Earth, and so I decided my first project would be to figure out exactly how close they would have to be. So, taking the geodesic equations off of Wikipedia (because I'm too lazy to figure that out)
$0=\frac{d^2\phi}{dq^2}+\frac{2}{r}\frac{d\theta}{dq}\frac{dr}{dq}$
$0=\frac{d^2t}{dq^2}+w^{-1}\frac{r_s}{r^2}\frac{dr}{dq}\frac{dt}{dq}$
$0=\frac{d^2r}{dq^2}+\frac{w}{2}\frac{d(w^{-1})}{dr}(\frac{dr}{dq})^2-rw(\frac{d\phi}{dq})^2+\frac{c^2}{2}w\frac{dw}{dr}(\frac{dt}{dq})^2$
Where $w=(1-\frac{r_s}{r})$ and $r_s$ is the Schwarzschild radius. Using these three equations, setting $q=\tau$, and assuming circular orbits ($\frac{dr}{dq}=0$), I ended up with 4 equations and 4 unknowns. Solving for $\frac{dt}{d\tau}$, I end up with
$\frac{dt}{d\tau}=\sqrt{\frac{1}{1-\frac{3}{2}\frac{r_s}{r}}}$

SO my questions: Is this process valid, or am I missing something? I was expecting the time dilation to go to infinity at the event horizon, $r_s$, however it seems to become undefined at $\frac{3}{2}r_s$ instead. What is special about this distance?

 

[Danger]

You might want to check out the seemingly endless discussion of it in the Science Fiction section of PF Lounge entitled "Interstellar—Spectacularly Stupid Movie".

 

[Higgs Boson]

Danger meant to be friendly and provide you with a direct link to the thread he referenced in his post above ... but your math probably angered him into his rudeness.

https://www.physicsforums.com/threads/interstellar-spectacularly-stupid-movie.780388/

 

[Danger]

Danger meant to be friendly and provide you with a direct link to the thread he referenced in his post above ... but your math probably angered him into his rudeness.

I'm not sure who put you in charge of my social skills, but I can understand your comment in light of having read your "occupation" in your user profile.
My answer was in no way rude. I provided an exact location to which I thought that s/he should be directed in case that thread dealt with whatever was on his/her mind. For you to assume that s/he couldn't find it without a direct link was an insult by you, not me.

 

[Matterwave]

So I recently watched the new movie Interstellar, and I've been inspired to do some more general relativity. At one point in the movie they mention that 1 hour on a planet orbiting a black hole is 7 years back on Earth, and so I decided my first project would be to figure out exactly how close they would have to be. So, taking the geodesic equations off of Wikipedia (because I'm too lazy to figure that out)
$0=\frac{d^2\phi}{dq^2}+\frac{2}{r}\frac{d\theta}{dq}\frac{dr}{dq}$
$0=\frac{d^2t}{dq^2}+w^{-1}\frac{r_s}{r^2}\frac{dr}{dq}\frac{dt}{dq}$
$0=\frac{d^2r}{dq^2}+\frac{w}{2}\frac{d(w^{-1})}{dr}(\frac{dr}{dq})^2-rw(\frac{d\phi}{dq})^2+\frac{c^2}{2}w\frac{dw}{dr}(\frac{dt}{dq})^2$
Where $w=(1-\frac{r_s}{r})$ and $r_s$ is the Schwarzschild radius. Using these three equations, setting $q=\tau$, and assuming circular orbits ($\frac{dr}{dq}=0$), I ended up with 4 equations and 4 unknowns. Solving for $\frac{dt}{d\tau}$, I end up with
$\frac{dt}{d\tau}=\sqrt{\frac{1}{1-\frac{3}{2}\frac{r_s}{r}}}$

SO my questions: Is this process valid, or am I missing something? I was expecting the time dilation to go to infinity at the event horizon, $r_s$, however it seems to become undefined at $\frac{3}{2}r_s$ instead. What is special about this distance?

You're going to have to be more specific. You "took the geodesic equations off of wikipedia", which ones? For what metric? What is $w$ and what is $q$? I can't recall a standard metric or coordinate system which uses these symbols for coordinates.