The reasoning by which I deduce Gargantua’s mass is this: The mass m of Miller’s planet exerts an inward gravitational acceleration g on the planet’s surface given by Newton’s inverse square law: g = Gm/r2, where r is the planet’s radius. On the faces of the planet farthest from Gargantua and nearest it, Gargantua’s tidal gravity exerts a stretching acceleration (difference of Gargantua’s gravity between the planet’s surface and its center a distance r away) given by gtidal = (2GM/R3)r. Here R is the radius of the planet’s orbit around Gargantua, which is very nearly the same as the radius of Gargantua’s horizon. The planet will be torn apart if this stretching acceleration on its surface exceeds the planet’s own inward gravitational acceleration, so gtidal must be less than g: gtidal < g. Inserting the formulas above for g, gtidal, and R, and expressing the planet’s mass in terms of its density ρ as m = (4π/3)r3ρ, and performing some algebra, we obtain 
. I estimate the density of Miller’s planet to be ρ = 10,000 kilograms/meter3 (about that of compressed rock), from which I obtain M < 3.4 × 1038 kilograms for Gargantua’s mass, which is about the same as 200 million suns—which in turn I approximate as 100 million suns.
Using Einstein’s relativistic equations, I have deduced a formula that connects the slowing of time on Miller’s planet, S = one hour/(seven years) = 1.63 × 10–5 to the fraction α by which Gargantua’s spin rate is less than its maximum possible spin: 
. This formula is correct only for very fast spins. Inserting the value of S, we obtain α = 1.3 × 10–14; that is, Gargantua’s actual spin is less than its maximum possible spin by about one part in a hundred trillion.
The equations that I gave to Oliver James at Double Negative, for the orbital motion of light rays around Gargantua, are a variant of those in Appendix A of Levin and Perez-Giz (2008). Our equations for the evolution of bundles of rays are a variant of those in Pineult and Roeder (1977a) and Pineult and Roder (1977b). In several papers that we’ll make available at http://arxiv.org/find/gr-qc, Paul Franklin’s team and I give the specific forms of our equations and discuss details of their implementation and the simulations that resulted.
Here are the calculations that underlie my statements in Chapter 13. They are a nice example of how a scientist makes estimates. These numbers are very approximate; I quote them accurate to only one digit.
The mass of the Earth’s atmosphere is 5 × 1018 kilograms, of which about 80 percent is nitrogen and 20 percent is molecular oxygen, O2—that is, 1 × 1018 kilograms of O2. The amount of carbon in undecayed plant life (called “organic carbon” by geophysicists) is about 3 × 1015 kilograms, with roughly half in the oceans’ surface layers and half on land (Table 1 of Hedges and Keil [1995]). Both forms get oxidized (converted to CO2) in about thirty years on average. Since CO2 has two oxygen atoms (that come from the atmosphere) and just one carbon atom, and the mass of each oxygen atom is 16/12 that of a carbon atom, the oxidization of all this carbon, after all plants die, would eat up 2 × 16/12 × (3 × 1015 kilograms) = 1 × 1016 kilograms of O2, which is 1 percent of the atmosphere’s oxygen.
For evidence of sudden overturns of the Earth’s oceans and the theory of how they might be produced, see Adkins, Ingersoll, and Pasquero (2005). The standard estimate of the amount of organic carbon in sediments on the ocean bottoms that might be brought to the surface by such an overturn focuses on an upper sedimentary layer that is mixed by ocean currents and animal activity. This mixed layer’s carbon content is the product of an estimated rate of deposit of carbon into the sediments (about 1011 kilograms per year) and the average time it takes for its carbon to be oxidized by oxygen from ocean water (1000 years), giving 1.5 × 1014 kilograms, one-twentieth of that on land and in ocean surface layers (Emerson and Hedges 1988, Hedges and Keil 1995). However: (i) The estimated deposition rate could be wrong by a huge amount; for example, Baumgart et al. (2009), relying on extensive measurements, estimate a deposition rate in the Indian Ocean off Java and Sumatra that is uncertain by a factor of fifty and, extrapolated to the whole ocean could give as much as 3 × 1015 kilograms of carbon in the mixed layer (the same as on land and in the ocean’s surface layers). (ii) A substantial fraction of the deposited carbon could sink into a lower layer of sediment that does not get mixed into contact with seawater and oxidized except possibly during sudden ocean overturns. The last overturn is thought to have been during the most recent ice age, about 20,000 years ago—twenty times longer than the oxidation time in the mixed layer. So the unmixed layer could have twenty times more organic carbon than the mixed layer, and as much as twenty times that on land and in the ocean’s surface. If brought to the ocean surface by a new overturn and there oxidized, this is nearly enough to leave everyone gasping for oxygen and dying of CO2 poisoning; see the end of Chapter 12. Thus such a scenario is conceivable, though highly unlikely.
Christopher Nolan chose several kilometers for the diameter of Interstellar’s wormhole. The wormhole’s angular diameter as seen from Earth, in radians, is this diameter divided by its distance from Earth, which is about 9 astronomical units or 1.4 × 109 kilometers (the radius of Saturn’s orbit). Therefore, the wormhole’s angular diameter is about (2 kilometers)/(1.4 × 109 kilometers) = 1.4 × 10–9 radians, which is 0.0003 arc-seconds. Radio telescopes routinely achieve this angular resolution using transworld interferometry. Optical telescopes on the ground using a technique called “adaptive optics,” and the Hubble space telescope in space, achieve angular resolutions a hundred times worse than this in 2014. Interferometry between twin Keck telescopes in Hawaii in 2014 can achieve a resolution ten times worse than the wormhole’s angular diameter, and it is very plausible that in the era of Interstellar optical interferometry between more widely spaced optical telescopes will make possible resolutions better than the wormhole’s 0.0003 arc-seconds.
If you are familiar with Newton’s gravitational laws in mathematical form, then you may find it interesting to explore a modification of them by the astrophysicists Bohdan Paczynski and Paul Wiita (Paczynski and Wiita 1980). In this modification, the gravitational acceleration of a nonspinning black hole is changed from Newton’s inverse square law, g = GM/r2 to g = GM/(r – rh)2. Here M is the hole’s mass, r is the radius outside the hole at which the acceleration g is felt, and rh = 2GM/c2 is the radius of the nonspinning hole’s horizon. This modification is a surprisingly good approximation to the gravitational acceleration predicted by general relativity.[60] Using this modified gravity, can you give a quantitative version of Figure 17.2[61] and deduce the radius of the orbit of Miller’s planet? Your result will be only roughly correct, because the Paczynski-Wiita description of Gargantua’s gravity fails to take account of the dragging of space into a whirling motion by the black hole’s spin.
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This Paczynski-Wiita modification of gravity was used in developing the black hole’s influence on spacecraft orbits for a gravitational-slingshot video game associated with