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# Re: [Phys-L] Ex: Re: Roche limit

Here's a nice way to explain:
https://www.thingiverse.com/thing:3379138

These 3D models are equipotential surfaces around the pairs of massive
objects. If the material of the satellite expands too far, it will spill
over the edge and be attracted to the other object. For orbits around the
Earth-Sun system there are several "Lagrange-Points" where the forces go to
zero. Some are saddle points, some are peaks. An object inserted into
these locations carefully will "orbit" the Lagrange Point and remain stable
relative to both the earth and sun.

For high-school level explanation, print out one of these, plug the bottom,
and fill the hole slowly with water. When the level gets high enough, it
will spill out over the edge. I'd suggest the 4/1 model as being possibly
the most interesting for this exercise.

Paul

On Thu, Jan 30, 2020 at 5:17 PM Albert J. Mallinckrodt <ajm@cpp.edu> wrote:

One more note on the Roche limit:

As mentioned by John Denker, the “simplest" derivation of the Roche limit
is obtained by setting the gravitational field of the satellite at its
surface (the "hold it together" part) equal to the radial gravitational
tidal force of the primary body between the center of the satellite and the
nearest or most distant point on its surface (the "rip it apart” part).

It yields the result

S = [2 (rho_primary / rho_satellite ) ]^(1/3) R = 1.26 (rho_primary /
rho_satellite )^(1/3) R

where R is the radius of the primary (orbited) body … AND is flatly wrong
because it neglects the tidal component of the centrifugal field.
Including the centrifugal component of the tidal force leads to

S = [3 (rho_primary / rho_satellite ) ]^(1/3) R = 1.44 (rho_primary /
rho_satellite )^(1/3) R.

But this result is still at VERY significant odds with the original
calculation of Roche in 1849

S = 2.44 (rho_primary / rho_satellite )^(1/3) R

which assumes a fluid satellite in hydrostatic equilibrium … and makes the
calculation significantly less straightforward!

It’s easy to understand why a fluid satellite will be torn apart at much
larger distances from the primary because of the fact that the tidal forces
themselves will deform the satellite into a radially elongated ellipsoidal
shape which, in turn, further increases the tidal forces.

See https://en.wikipedia.org/wiki/Roche_limit

John Mallinckrodt
Cal Poly Pomona

On Jan 27, 2020, at 12:23 PM, John Denker via Phys-l <
phys-l@mail.phys-l.org<mailto:phys-l@mail.phys-l.org>> wrote:

The Roche derivation goes like this

The basic gravitational field of the primary goes like
1/S^2 where S is the separation between the two bodies,
i.e. orbital radius. This is just Newton's law of
universal gravitation.

The tide-producing field is the *difference* you get by
subtracting the basic field's value at the satellite
surface from its value at the satellite center. This
is easy to visualize in the /accelerated/ frame that
follows the satellite center. Using Einstein's principle
of equivalence we can zero out the acceleration at the
center, and then the surface tidal acceleration is just
the aforementioned difference. Conceptually, this is
not complicated at all.

Plugging in, the Roche limit occurs when:
m/M ∝ (r/S)^3

or if you want to get fancy:
m/M = 2(r/S)^3

That's it.

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