Calculating tidal forces (Interstellar)

[KAL 9000]

Surface gravity depends not only on the mass of the planet, but the radius of the planet. If it depended only on mass, then you would be correct, but to determine the mass of Miller you need to know the radius of the planet.

surface gravity = -(G*M_planet)/R_planet^2

The negative sign is there indicating that the acceleration is always towards the center of the planet. Surface gravity would be (1.3*-9.81m/s^2) = -10.1m/s^2 (to one significant figure). So in the equation you would have:

-10.1 = -(6.67E-11 * M_planet)/R_planet^2

Rearranging, the negative signs will cancel, and we will multiply both sides by R_planet^2 giving:

10.1*R_planet^2 = 6.67E-11*M_planet

Then divide both sides by the Universal Gravitational Constant (6.67E-11)

(10.1*R_planet^2)/6.67E-11 = M_planet

So, we are missing two pieces of information. If you know the radius of the planet, then you can easily find the mass of the planet. Having not seen the film, I'm not sure what information they gave, so you may not be able to determine the mass of the planet.

Well, it seems reasonable to assume that Miller's Planet has (roughly) the same density as Earth. So, to find the radius of the planet, we would find the square root of (1.3 X Earth's volume) divided by 4/3pi. You could calculate the mass from that and the surface gravity.

Edited October 20, 2015 by KAL 9000