This is not my calc,but a friend of mine did this recently.so I wanna share this here
http://www.narutoforums.com/xfa-blog-entry/narutoverse-moon-density-calculation.36011/
NARUTOVERSE MOON DENSITY I have decided to ascertain how dense Narutoverse Moon is, and this is due to its unorthodox gravitational field strength. Evidently, it's GA is analogous to ours as shown on the MOVIE, hence the Narutoverse's Lunar GA equates 9.81 N/kg
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Step 1, I'll determine the mass of Narutoverse Moon due to its gravitational field strength. The formula g = -GM/r^2 will serve as the equation to determine the mass of the NV's
g = -GM/r2
Where;
g = 9.18 N/kg
G = -6.67 × 10^-11 N m^2 Kg-2 (Universal Gravitation C)
M = ?
r = (1.7 × 10^6m)^2
Solution; M = 9.81 × (1.7 × 10^6)^2/6.67 × 10^11 M = 4.2 x 10^23 Therefore, the Mass of Narutoverse Moon with G.A of 9.81 is 4.2 × 10^23kg. Step 2, the mean density of Narutoverse Moon. The Moon is a spherical mass, hence its volume can be calculated using 4/3πr^2
Density = p = M/V Where; M = 4.2 × 10^23 kg V = 4/3 × π × (1.7 × 10^6)^3
Solution; p = 4.2 × 10^23/4/3 × π × (1.7 × 10^6)^3 p = 20 487 ≈ 20 500 Kg/m^3 Hence, Narutoverse Moon is 3.7 times more denser than the Earth and 6.1 times more denser the our moon.
ENERGY NEEDED TO DISBAND NV'S LUNAR Now for the sake of future purposes, I'll also calculate for the Gravitational binding energy required for Narutoverse's lunar to cease being in a gravitationally bound state, as well as the gravitational potential energy exerted to disjoint Narutoverse lunar. Note that Narutoverse lunar had cavities around its segments, which wouldn't really have a differentiated aggregate, (if calculated) from the gravitational potential energy of a normal lunar.
Gravitational binding energy is the minimum energy that must be added to a system for the system to cease being in a gravitationally bound state https://en.m.wikipedia.org/wiki/Gravitational_binding_energy
Gravitational binding energy = GBE = 3GM^2/5r Where; G = 6.67 × 10^-11 M = 4.2 × 10^23 (The mass of NV's Moon) r = 1.7 × 10^6 U = ?
Solution; U = 3 × 6.67 × 10^-11 × (4.2 × 10^23)^2/8.5 × 10^6 U = 4.15 x 10^30j
ENERGY EXERTED TO DISJOINT THE MOON This is the energy a body possesses due to its position along a gravitational field or the potential energy needed to bring a mass to a specific point. In this scenario, its the gravitational potential neergy needed to disjoint one half of the lunar from another by 50 km. http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html
Gravitational potential energy of the separated half of the Moon = E = GMm/r
Where;
G = 6.67 × 10^-11
M = m = 2.1 × 10^23
r = 1.7 × 10^6
E = ?
Solution; E = 6.67 × 10^-11 × 2.1 × 10^23 × 2.1 × 10^23 / 2 × (3/8) × 1.7 × 10^6 E = 2.25e30j
Gravitational potential energy of the second separated part in reference to a split moon = E = GMm/r2 where; G = 6.67 × 10^-11 M = m = 2.1 × 10^23 r2 = 1.7 × 10^6 Moon split width = 50,000m E = ?
Solution; E = 6.67 × 10^-11 × 2.1 × 10^23 × 2.1 × 10^23 / 2 × (3/8) × 1.7 × 10^6 + 50 000 E = 2.17e30j In relation to Gravitation potential energy, the energy needed to disjoint the Moon equates the difference between the above aggregates, which is; 2.25e30-2.17e30 = 8.0e28 ≈ 20 exatons.*