The Physics
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Opus in profectus

# Gravity of Extended Bodies

## Discussion

### tidal forces

The tides, tidal forces, prolate spheroid, Roche limit

Derive the tidal force formula. Despite being called a force, it's more common to care about the field — that is, the gravitational field strength, also known as the acceleration due to gravity.

 g = Gm r2

Consider the tidal force of the Moon acting on the Earth. Let…

 r = separation between the center of the Earth and the Moon Δr = radius of the Earth, the object experiencing the tide m = mass of the Moon, the object causing the tide

The tidal force on the Earth is the difference between the gravitational field on its front (the point nearest the Moon) and the gravitational field at its center (the place we normally think of as the location of the Earth).

 Δg = gfront − gcenter Δg = Gm − Gm (r − Δr)2 r2

Work that algebra. Work it!

Δg =
 Gm ⎛⎜⎝ r2 − (r − Δr)2 ⎞⎟⎠ r2(r − Δr)2
Δg =
 Gm ⎛⎜⎝ r2 − (r2 − 2rΔr + Δr2) ⎞⎟⎠ r2(r2 − 2rΔr + Δr2)
Δg =
 Gm ⎛⎜⎝ 2rΔr − Δr2 ⎞⎟⎠ r4 − 2r3Δr + r2Δr2

Now the part that feels like a cheat. Eliminate the "small" terms — those with the most Δr in them. That gives us an approximate solution that's good enough for most purposes.

 Δg ≈ Gm ⎛⎜⎝ 2rΔr ⎞⎟⎠ r4

Simplify.

 Δg ≈ 2GmΔr r3

Let's derive the tidal force equation again, but this time let's do it the easy way — using calculus.

 Δg ≈ dg Δr dr
 Δg ≈ d ⎛⎜⎝ Gm ⎞⎟⎠ Δr dr r2
 Δg ≈ 2GmΔr r3

Wow.

### Roche limit

Good, now derive the Roche limit equation.

gtidal  =  gsurface

 2Gmplanetrmoon rroche3
=
 Gmmoon rmoon2

 2Gmplanet mmoon
=
 rroche3 rmoon3

 rroche = rmoon ∛ 2mplanet mmoon

For the Earth and the Moon…

 rroche = (1,737,000 m) ∛ 2(5.9723 × 1024 kg) 7.364 × 1022 kg
 rroche = 9470 km ≈ 1.5 Earth radii ≈ 140 Earth‑Moon distance

### flattening

Oblate spheroid

Equatorial radius a, polar radius b. The flattening factor (also called oblateness) is…

 ε = a − b a

### gravity inside & outside

Two ways to solve problems. In general…

 g(r) = − G ⌠⌠⌠⌡⌡⌡ r̂ dm r2

where…

 g(r) = gravitational field vector at any location in space G = gravitational constant dm = infinitesimal mass r = vector pointing out from infinitesimal mass to any location in space r̂ = direction of r r = magnitude of r

Since

 r Vg(r) = − ⌠⌡ g(r) · dr ∞

We get

 Vg(r) = − G ⌠⌠⌠⌡⌡⌡ dm r

For systems with spherical, cylindrical, or planar symmetry…

∯ g · dA = −4πGm

For spherically symmetric mass distributions…

 r g(r) = − G ⌠⎮⌡ ρ(r) 4πr2 dr r̂ r2 0