Consider Europa, the smallest moon of Saturn. It is thought that an ocean of liquid water exists under an outer crust. The ocean may be assumed static and incompressible, and the crust surrounding Europa will be neglected (i. e., assume Europa is only a spherical ocean). How will the ocean pressure vary as a function of radial position within the planet? The gravitational acceleration at the surface of a planet (9.81 m/s2 on earth) is defined as 2 , GM g R = where G is the universal gravitational constant, M is the mass of the planet, and R is the radius of the planet. Newton’s shell theorem states that the gravitational force experienced at a given radial position within a planet will only be dependent on the mass of the planet below the radial position considered such that we can write 2 ( ) , GM r g r = where M(r) is the mass of the planet within radial position r. Finally, the pressure-"height" relationship in spherical coordinates is given by . dp g dr = − Provide a relationship between pressure and radial position within the ocean planet (i. e., if two radial positions are selected, find a relationship describing the pressure at each of the positions).

A = - (ρ G M / 2R)      and        P = A (1 + r² / R²)

Explanation:

Let's solve this problem in parts, the variation of pressure with height is

         dP / dy = rho g         (1)

Therefore we must know the variation of the acceleration of gravity, for this we write Newton's second law where the force is the universal force of attraction

        F = m a

        G M m / r² = m a

        a = G M / r²

this acceleration called acceleration of gravity (g)

where the mass of the planet is the mass that is inside the surface formed by the point of interest; that is, the mass of the outer spherical shell does not affect the attraction of gravity, to find this mass we use the concept of density

         ρ = M / V

         M = ρ 4/3 π r³

where r is the radius of the satellite to the point where the acceleration is being calculated

we substitute

          a = g = G ρ 4/3 π r³ / r²

          g = G ρ 4/3 π  r

now let's use the density

          ρ= M / V = M / (4/3 pi R3)

we substitute  

          g = G M r / R³

we substitute this equation into the equation   1

         dP / dr = (ρ G M / R³) r

we integrate

          ∫ dP = (ρ G M / R³)    ∫ r dr

          P = (ρ G M / R³)    r² / 2

we evaluate between the lower limit (r, P)  and the upper limit r = R,  P = 0

          0 -P = (ρ G M / R³) /2      (R² - r²)

          P = - (ρ G M /2R)    (1  - r² / R²)

let's call

          A = - (ρ G M / 2R)

          P = A (1 + r² / R²)

the correct answer is b