# Wouldn't Millers planet be ripped to shreds by extreme tidal forces?

If you can survive lava-like temperatures there is another issue for the well-being of Miller's planet from Interstellar. Time passes on the planet 61000 times more slowly than for an observer far away from the black hole.

From this analysis:

Einstein's laws dictate that, as seen from afar, for example, from Mann's planet, Miller's planet travels around Gargantua's billion-kilometer circumference orbit once each 1.7 hours. This is roughly half the speed of light! Because of time's slowing, the Ranger's crew measures an orbital period sixty thousand times smaller than this: a tenth of a second. Ten trips around Gargantua per second. That's really fast! Isn't it far faster than light? No, because of the space whirl induced by Gargantua's fast spin. Relative to the whirling space at the planet's location, and using time as measured there, the planet is moving slower than light, and that's what counts. That's the sense in which the speed limit is enforced.

For circular orbits in Newtonian gravity the tidal force only depends on the orbital period, regardless of the mass being orbited. Consider two rocks placed side-by-side orbiting the Earth and moving as soldiers marching abreast. The rocks have the same initial velocity and are on two separate orbits. If each rock is in free-fall, they will be drawn together by the tidal force in a quarter of an orbital period since these two orbits intersect.

What if we don't orbit a body but instead fall into a point mass from rest at infinity? The tidal forces increase as we approach the object, with the F~$$t^{-2}$$ where $$t$$ is the proper time to impact. This scaling law does not depend on the mass of the object and General relativity gives the same answer as Newtonian gravity! After-all, nothing special happens when you enter a black hole if you don't try to leave...

A 1.7 hour orbit is just enough tidal force to tear apart iron planets (on large scales, solid iron and fluid iron behave the same way). However, it is not that far away from planetary survival and is less than the international space station (1.5 hour orbit) so lets forgive this. However, the tidal forces in a 0.1s orbit are so extreme that the two rocks would collide in 1/40 of a second. This is lethally high for humans even though it is a supermassive black hole and any planet would be pulverized into meter-sized chunks.

But an orbit around a near-extremal spinning black hole is a different situation than a free-fall into an ideal non-spinning mass. Do these order-of-magnitude estimates derived from Newtonian gravity still hold?

• How big is the tidal force? The difference between the force on both sides? I always wondered about the spike waterwaves. WTF...? Truly? It is said everything is scientifically well established. But the frying temperature already disproves this. Good question! May 25 at 19:17
• Try modified newtonian gravity, with $a(r) = \frac{GM}{r^2\sqrt{1-\frac{3R_s}{2r}}}$ where $R_s = \frac{2GM}{c^2}$, $r$ is the orbital radius in Scwarzchild coordinates, $a$ is the radial coordinate acceleration of the orbiting body
– g s
May 25 at 19:36
• I'm not sure if it's a safe approximation, but you might vary $r$ by the difference between centers of mass of the close and far hemisphere (about 2/3 planet diameter for earth) and use $F= m\Delta a$ for tidal force, where m is planet mass.
– g s
May 25 at 19:48
• Well, I guess the answer is no, since Kip Thorne designed it to be not ripped apart by tidal forces and set the mass of the black hole appropriately. But I guess you want some proof of that. May 25 at 20:11
• This is covered to an extent in Thorne's book "The Science of Interstellar". He writes"Gargantua's tidal forces almost tear the planet apart. Almost but not quite. Instead, they simply deform the planet. Deform it greatly. It bulges strongly toward and away from Gargantua". May 26 at 8:06

According to Fishbone, 1973, ApJ, 185, 43 the minimum density defining the Roche limit for something in the innermost stable, equatorial, circular orbit around a Kerr black hole is the same as that for an orbiting object in Newtonian physics$$^1$$.

Using Fig.2 in that paper, for a maximally spinning Kerr black hole, the limiting Roche density is $$\sim 3 \times 10^{18} (M/M_\odot)^{-2}$$ g/cm$$^3$$. The black hole that Miller's planet orbits has $$M \sim 10^8M_\odot$$ and so Miller's planet needs to have a density of $$>300$$ g/cm$$^3$$ in order to survive.

Double-checking these figures, a simple use of the Newtonian Roche limit is that $$\rho > (2.44)^3 \left(\frac{3}{4\pi}\right)\left(\frac{M}{r^3}\right)\ ,$$ where $$M$$ is the black hole mass, $$r \simeq 0.5 r_s$$ is the orbital "radius" and the $$(2.44)^3$$ coefficient accounts for the planet being a deformable fluid ellipsoid (see here). For $$M = 10^8M_\odot$$, then $$r = 1.48\times 10^{11}$$ m, and this gives gives $$\rho > 200$$ g/cm$$^3$$, in fair agreement.

Theis density looks 1-2 orders of magnitude too high for any kind of sensible planet. I'm quite unhappy with this answer, since Kip Thorne designed the world/black hole system so that it would survive tidal forces...??

If you make the planet a rigid, undeformable sphere, then the $$(2.44)^3$$ gets replaced by $$(1.26)^3$$, which reduces the threshold density to $$\sim 30$$ g/cm$$^3$$. This is still too high for a terrestrial-type rocky planet.

You really need the black hole to have had a mass $$>2\times 10^8 M_\odot$$ to give a sensibly dense planet a chance of surviving break-up. Apparently Kip Thorne didn't do that because he wanted to make the black hole mass a nice round number. In a footnote in Chapter 6 of his book The Science of Interstellar, where the black hole mass is discussed, Thorne says

A more reasonable value might be 200 million times the Sun’s mass, but I want to keep the numbers simple and there’s a lot of slop in this one, so I chose 100 million.

I guess that isn't the only (or even nearly the biggest) scientific inaccuracy in the film (see also Wouldn't Miller's planet be fried by blueshifted radiation?).

$$^1$$ Note that is by no means obvious that this should be the case. In general, the tidal force on an orbiting object is not given by the simple Newtonian formula. For example, the expression for the tidal forces in the radial direction for an object in a stable circular orbit around a Schwarzschild black hole are 1.5 times larger than the Newtonian value at the ISCO at $$r=3r_s$$ and would blow up to infinity at the unstable circular orbit at $$r=3r_s/2$$ (e.g. see chapter 9 of Exploring Black Holes by Taylor, Wheeler & Bertschinger).

Nevertheless if you consult chapter 19 of the same book, it gives general expressions for the tides for orbits around a Kerr black hole. If you work through these expressions, then when $$a$$ is maximal, and the ISCO is at $$r = r_s/2$$ then the tidal acceleration in the radial direction is given by $$g_{\rm tidal} \simeq \frac{2GM}{r^3}\Delta r\ ,$$ as in Newtonian physics. I think this is because at this radius, the orbiting object is essentially co-rotating with the dragged spacetime.

• Hmmm. Does the massive time dliation make a difference? I guess Fishbone wouldn't neglect to account for that... May 25 at 22:44
• Would the sky above change several times per second but you would not feel like you were spinning several times per second? That would be a weird effect indeed. Stationary observers feel a Coriolis effect so would that get mostly cancelled out by the spinning? May 26 at 5:02
• From "The Science of Interstellar" it appears that Thorne did not consider the planet to be a fluid sphere. This reduces the density by about a factor of 7, giving \rho ~50 g/cc, which he rounded to 10 g/cc. Still high, but not totally unreasonable. May 26 at 8:54
• @ClaraDiazSanchez The time dilation at the ISCO is independent of the mass of the black hole; it depends on the spin parameter and the multiple of the Schwarzschild radius you put the planet in orbit at. The possible tensioning constraint would be that for a 3 times bigger Schwarzschild radius, the solid angle of the black hole, as viewed from Miller's planet would be 9 times smaller and less cinematic...? But actually, I think even that isn't true - it will still fill nearly half the sky. May 26 at 9:20
• @ClaraDiazSanchez The amount of (extreme) time dilation only depends on $a$ (and $R/R_s$, but which has to be $\simeq 0.5$). Whereas the threshold Roche limit density at the ISCO depends only on $M^{-2}$, since $R \simeq 0.5R_s$ is fixed to get extreme time dilation. Therefore why not choose an actually plausible $\rho < 10$ g/cc (50 g/cc is much higher than any planet ever observed, what would it be made of?) and $M > 4\times 10^{38}$ kg? May 26 at 13:48

In answer to the question, the scientific advisor to Interstellar, Kip Thorne, considered the effects of tidal forces on Miller's planet in quite some detail, and in fact the mass of Gargantua was set by the requirement of having the planet not be ripped to shreds.

(Note: I gave a lot of this material in comments to ProfRob's answer, but given the ephemeral nature of comments I think it is better to give it in an answer of its own).

The first requirement is that the orbital radius of Miller's planet is just a little larger than Gargantua's horizon, $$R = G M/c^2$$, where $$M$$ is the mass of Gargantua. Following the Technical Notes in The Science of Interstellar, the tidal force on Miller's world is given by:

$$g_{\mathrm{tidal}} = \frac{2 G M}{R^3} r$$,

where $$r$$ is the radius of Miller's planet. The gravitational acceleration on Miller's planet is given by:

$$g = \frac{G m}{r^2}$$,

where $$m$$ is the mass of Miller's planet. For the planet not to be torn apart, $$g_{\mathrm{tidal}} < g$$, and from combining these two expressions we can deduce that the limiting density of Miller's planet is $$\rho = \frac{3}{2 \pi} \frac{M}{R^3}$$. Note that this assumes that the planet is a spherical rigid body; a more sophisticated argument used in ProfRob's answer assumes that the planet is a deformable fluid ellipsoid gives a similar result, but with a larger prefactor.

The value for $$\rho$$ thus sets the value of $$M / R^3$$, and combining this with the horizon equation gives a unique solution for the mass and radius of Gargantua. Using a value of $$\rho = 10,000$$ kg/m$$^3$$ gives $$M = 3.5 x 10^{38}$$ kg and $$R=2.5x 10^{11}$$ m. These values were then rounded off to $$2 x 10^{38}$$ kg (about 100 million suns), and an orbital radius of $$1.48 x 10^8$$ km.

This value of $$\rho$$ is pretty high. Most rocks have densities of about 2,000 - 3000 kg/m$$^3$$. A more reasonable value would be obtained by increasing the value of $$M$$ since $$\rho \propto 1 / M^2$$. However, as Thorne explained in The Science of Interstellar:

I want to keep the numbers simple and there's a lot of slop in this one, so I chose [Gargantua's mass to be] 100 million [solar masses].