Twin paradox with black hole (based on Interstellar) I'm an Undergrad student working on a summer project. I'm learning about Differential Geometry, Schwarzschild's Solution, General and Special Relativity. I want to include the twin paradox as well. In my project one of the twins makes a journey into space in a high-speed rocket and goes near a black hole and then returns to Earth. How much energy is needed for such a journey? What will be the age difference between the twins? What's the difference between using rotating and nonrotating black hole? Let's say they are 21 years old at the beginning of the journey, the rocket can travel 50% of the speed of light and the black hole is 5 light-years away from us. I'm not really sure about the mass of the black hole and how close the rocket gets to it. (A typical stellar-class of black hole has a mass between about 3 and 10 solar masses.)
 A: Kip Thorne (Nobel Prize of Physics 2017 and one of the authors of the famous and fat book "Gravitation" if you're looking for literature) was an advisor to make "Interstellar" as scientifically accurate as possible. He wrote a book and gave a talk called "The Science of Interstellar". At around 18:00 there is an animation about how passing a wormhole looks like and at 30:00 he talks about how he managed to achieve the enormous time dilatation on Miller's planet.
It depends not only on the distance to the black hole, but also if it's rotating or not. The innermost stable circular orbit (ISCO) for a non-rotating (and non-charged) black hole described by the Schwarzschild metric is $3r_\mathrm{S}$, where $r_\mathrm{S}=\frac{2GM}{c^2}$ is the Schwarzschild radius calculated with the mass $M$ of the black hole. Kip Thorne found out, that a planet that far away can't possibly experience as much time dilatation as Christopher Nolan requested.
That's why Kip Thorne decided to make the black hole rotating at more than 99% of the speed of light. For a rotating (and non-charged) black hole described by the Kerr metric the event horizon and ISCO change and you also get an ergosphere outside the event horizon that can be entered and left again. The formulas are a bit more complicated, a lot of good animations by Yukterez can be found here. Since the rotation of the black hole causes whirls in spacetime outside, you can not only extract energy by entering this ergosphere, what is called the Penrose process and is also used in "Interstellar" later during the swing-by at the black hole, but also have the planet be a lot closer to the black hole achieving the time dilatation requested.
Your questions can't be answered without more data: The acceleration of the spacecraft or its orbit around the black hole cause a change in time dilatation and you will need integration. The properties of the black hole (mass, rotation and charge) also have an influence as they dictate the shape of the orbit, which has to be calculated with the geodesic equation, which is also not trivial. General Relativity is pretty difficult.
If it's for a project, I don't think you will need precise calculations anyway. Maybe you can use the talk of Kip Thorne or the animations I have linked as inspiration or material, in this case, you can also present your project to a larger audience.
Gravitational time dilatation is given by:
$$t=\frac{\tau}{\sqrt{g_{00}}},$$
where $t$ is the time, that has passed far away from the black hole, $\tau$ the time that has passed in the spaceship close to the black hole and $g_{00}$ is the $00$-entry of the metric tensor. In particular for the Schwarzschild metric, we have:
$$g_{00}(r)
=1-\frac{r_\mathrm{S}}{r}
=1-\frac{2GM}{c^2r}.$$
For example for the ISCO of the Schwarzschild metric, you get $g_{00}(3r_\mathrm{S})=2/3$ and therefore:
$$t=\tau\sqrt{\frac{3}{2}}
\approx 1.22 t,$$
which is far from Miller's planet.
