The tides, tidal forces, prolate spheroid, Roche limit
| [magnify] | |
Let …
| r = | separation between planet and moon |
| a, b = | radius of planet and moon, respectively |
| ma, mb = | mass of planet and moon, respectively |
Derive the tidal force formula.
| gtidal = | gfront | − | gback |
| gtidal = | Gmb | − | Gmb |
| (r − a)2 | (r + a)2 |
Work that algebra. Worlk it!
| gtidal = Gmb | ⎛ ⎝ |
(r + a)2 − (r − a)2 | ⎞ ⎠ |
= Gmb | ⎛ ⎝ |
(r2 + 2ra + a2) − (r2 − 2ra + a2) | ⎞ ⎠ |
| (r − a)2(r + a)2 | r4 − a4 |
Simplify.
| gtidal = Gmb | ⎛ ⎝ |
4ra | ⎞ ⎠ |
| r4 − a4 |
Super-simplify
| gtidal ≈ | 4Gmba |
| r3 |
Good, now derive the Roche limit.
| gtidal | ≈ | gsurface |
| 4Gmab | ≈ | Gmb |
| r3 | b2 |
| r ≈ b | ∛ | 4ma |
| mb | ||
oblate spheroid
Polar radius a, equatorial radius c. The flattening factor (also called oblateness) is …
| ℰ = | a − c |
| a |
Two ways to solve problems …
| g(r) = − G | ⌠⌠⌠ ⌡⌡⌡ |
r̂ dm | & | Vg = − G | ⌠⌠⌠ ⌡⌡⌡ |
dm |
| r2 | r2 | |||||
| all space | all space | |||||
| r | r | ||||||
| g(r) = − | G | ⌠ ⌡ |
ρ(r) 4πr2 dr r̂ | & | Vg(r) = − | ⌠ ⌡ |
g(r) · dr |
| r2 | |||||||
| 0 | ∞ | ||||||
Prove that the gravitational field inside a uniform mass distribution is 0. This is the key.
| The Physics Hypertextbook © 1998–2013 Glenn Elert |
No condition is permanent.