## Gravitational mass

Posted by Chris on May 29, 2012 – 5:08 pm

Estimate the mass of the gravitational field of the Earth. What would the size of a cube of water with the same mass?

Use the following info (the real info is complex, but the following is close enough):

Relative density of inner core = 13, radius = 400 km

Relative density of outer core = 11, radius = 3400 km

Relative density of outer layers = 4.5, radius = 6400 km

June 2nd, 2012 at 6:38 am

Are you saying that the “cube of water” has similiar densities like the example?

June 2nd, 2012 at 9:15 am

Hi Ragknot. No – the water has relative density 1. I only put it in for the sense of wonder the result should give.

June 4th, 2012 at 6:31 am

I’ll give it a shot … The cube of water would have an edge of length 18,183km.

June 4th, 2012 at 6:44 am

inner core volume = (4/3)pi(400^3)

outer core volume = (4/3)pi(3400^3 – 400^3)

inner core volume = (4/3)pi(6400^3 – 3400^3)

density is mass per volume.

volume * density = mass

inner core mass = 13 * (4/3)pi(400^3)

outer core mass = 11 * (4/3)pi(3400^3 – 400^3)

inner core mass = 4.5 * (4/3)pi(6400^3 – 3400^3)

Total mass ~= 6,011,969,519,000

Water density is 1, therefore, …

total volume = ~= 6,011,969,519,000

Cube root of total volume gives us the edge of the cube ~= 18,183 km

June 4th, 2012 at 7:07 am

Hi cazayoux. You’ve misunderstood the question. That cube corresponds to the total mass of the Earth. The question is about the mass of the Earth’s gravitational field.

June 4th, 2012 at 6:15 pm

Here’s the starting point. I’ll post more in a day or so.

To do this problem, we need to show that a gravitational field stores energy. In fact the energy density u (J/m³) in a region where the field strength is g, is u = -g²/(8πG). Due to E = mc², we can define a mass density μ = u/c². Note that both u and μ are negative.

To see why, we’ll find the field associated with a large plate of areal mass density σ. Away from the edges of the plate, the g-field is perpendicular to the plate. Direct calculation using Newton’s law of graviation is tedious. A fairly easy way is to use the well known results for a thin spherical shell of of radius r, areal mass density σ and total mass M = 4πr²σ. Just outside the sphere, g = GM/r² = 4πGr²σ/r² = 4πGσ and just inside the sphere, g = 0. Consider a small section of the surface of the sphere. In isolation it must be making a field that points inwards and outwards equally. So the rest of the sphere must be adding a field that exactly cancels the inwardly pointing field, and it follows that it must be doubling the outwardly pointing field. We conclude that for a plate, g = 2πGσ.

Now split the plate into two equal plates. Note that for the pair of plates, the field strength is g. Let the area of each plate be A. Each plate has areal mass density σ/2 and produces a field g/2. Keep these two plates close to each other. Consider one of these plates to be fixed. The force acting on the second plate (due to the first) is (g/2)(σA/2) = gσA/4. If the plates are further separated by a small distance d, the work required to do that is gσAd/4. The volume of the universe outside the region defined by the plates is reduced by Ad. Where has that work gone to? It must have been used to remove the g-field in the region whose volume is Ad, and so the energy density of the g-field must be u = -(gσAd/4)/(Ad) = -gσ/4. But g = 2πGσ => σ = g/(2πG) and substituting that into the previous equation gives u = -g²/(8πG). From now on, I’ll drop the – sign.

The mass density of the g-field is μ = g²/(8πGc²).

G = 6.673*10

^{-11}, c² = 8.988*10^{16}For amusement, if g = 9.81, then μ = 6.385*10

^{-7}, => μ = 638.5 kg/(km)³NB the 8πG factor also appears in Einstein’s field equations of general relativity: G

^{μν}= 8πGT^{μν}. G^{μν}is the Einstein (curvature) tensor, and T^{μν}is the stress-energy tensor. Note that it is quite common to choose physical units such that G = c = 1, and so it is common to see G^{μν}= 8πT^{μν}.June 4th, 2012 at 6:49 pm

Wow, Chris, that’s a lot of info, but I have not read it yet. The relative density of water is 1.00, if the temperature is 4 deg C. One milliliter of it has a mass of exactly one gram.

June 19th, 2012 at 7:12 pm

For a spherically uniform distribution of mass, Newton’s law of gravitation has some rather nice consequences. First, for points outside the sphere, the g-field is identical to one the would be produced as if the entire mass of the sphere were concentrated at a point at the centre of the sphere. Second, the g-field inside a sphere at a distance r from the centre, is entirely due to the mass inside inside the sphere of radius r: the matter outside that radius doesn’t contribute to the internal g-field. Third, g-fields sum linearly.

In my last post I argued that μ = g

^{2}/(8πGc^{2}) where μ is the mass density of the g-field. Hence the total mass of the g-field is M = ∫μ dV, where the integral is talken over all space and dV is a volume element. For a sphere, it is natural to use dV = 4πr^{2}dr. So M = ∫_{0}^{∞}4πr^{2}g^{2}/(8πGc^{2}) dr = 1/(2Gc^{2})∫_{0}^{∞}g^{2}r^{2}drWe now need to find g everywhere. For a sphere of radius r and uniform density ρ, the mass is 4πρr

^{3}/3. For typing convenience, I’ll use k = 4π/3 ≈ 4.1888. So, the mass of the sphere is kρr^{3}. The mass of a hollow sphere of inner radius r_{i}and outer radius r_{o}is kρ(r_{o}^{3}– r_{i}^{3}).In this problem we have four regions to consider: the inner core (radius r

_{i}, density ρ_{i}), the middle layer (aka outer core, radius r_{m}, density ρ_{m}), the outer layer (radius r_{o}, density ρ_{o}) and the region outside the Earth (radius ∞, density 0).Let m

_{i}= kρ_{i}r_{i}^{3}, m_{m}= kρ_{m}(r_{m}^{3}– r_{i}^{3}) and m_{o}= kρ_{o}(r_{o}^{3}– r_{m}^{3})For r

_{i}> r > 0: g = kGρ_{i}r^{3}/r^{2}= kGρ_{i}rFor r

_{m}> r > r_{i}: g = G(m_{i}– kρ_{m}r_{i}^{3})/r^{2}+ kGρ_{m}rFor r

_{o}> r > r_{m}: g = G(m_{i}+ m_{m}– kρ_{o}r_{m}^{3})/r^{2}+ kGρ_{o}rFor r > r

_{o}: g = G(m_{i}+ m_{m}+m_{o})/r^{2}Using those results for g, we now find g

^{2}r^{2}.For r

_{i}> r > 0: g^{2}r^{2}= k^{2}G^{2}ρ_{i}^{2}r^{4}For r

_{m}> r > r_{i}:g

^{2}r^{2}= G^{2}(m_{i}– kρ_{m}r_{i}^{3})^{2}/r^{2}+ 2kG^{2}ρ_{m}(m_{i}– kρ_{m}r_{i}^{3})r^{2}+ k^{2}G^{2}ρ_{m}^{2}r^{4}For r

_{o}> r > r_{m}:g

^{2}r^{2}= G^{2}(m_{i}+ m_{m}– kρ_{o}r_{m}^{3})^{2}/r^{2}+ 2kG^{2}ρ_{o}(m_{i}+ m_{m}– kρ_{o}r_{m}^{3})r^{2}+ k^{2}G^{2}ρ_{o}^{2}r^{4}For r > r

_{o}: g^{2}r^{2}= G^{2}(m_{i}+ m_{m}+ m_{o})^{2}/r^{2}To be continued…

June 20th, 2012 at 3:54 pm

…continued.

After some pretty boring integrations etc., we end up with the total mass of the g-field is around 3*10

^{15}kg. As the density of water is 1000 kg m^{-3}that’d be a volume of 3*10^{12}m^{3}and so a cube of side around 14.4 km.