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Derivation of the Roche Limit Formula#

Assuming that the satellite and the planet it orbits are both spherical and that there are no forces other than their gravitational forces.
Let $u$ be the surface mass of the satellite (i.e., the point where the line connecting the centers of the two spheres intersects the satellite's surface). There are two forces acting on this surface mass $u$:

The gravitational force exerted by the satellite and the gravitational force exerted by the planet (since the satellite freely falls within the planet's gravitational field, the tidal force is equivalent to the planet's gravitational force).

Let $F_G$ be the gravitational force exerted on $u$ by the satellite. According to Newton's law of universal gravitation,
$F_G={{Gmu} \over r^2}$
Let $d$ be the distance between the centers of the planet and the satellite, $R$ be the radius of the planet, and $F_T$ be the tidal force exerted on $u$ by the planet,
$F_T={{2GMur} \over d^3}$
When the satellite is at the Roche limit, we have
$F_G=F_T$
Thus,
${Gmu \over r^2}={2GMur \over d^3}$
From this, we can easily obtain
$d=r(2M/m)^{1 \over 3}$
At this point, we can say that the derivation of the formula is complete, but we can further replace mass with density as a variable.

By simply using some knowledge from middle school, we can easily express the mass of a sphere:

• Mass of the planet
  $M={4π\rho_MR^3}/3$
• Mass of the satellite
  $m={4π\rho_mr^3}/3$
  Substituting the above two equations into the formula we just derived, we get
  $d=r\left({2\rho_MR^3} \over {\rho_mr^3}\right)^{1/3}$
  After rearranging and simplifying,
  $d=R\left(2{\rho_M \over \rho_m}\right)^{1/3}$
  The above equation is the final formula.

Calculation Method of the Roche Limit#

Let $d$ be the Roche limit.
According to the formula we just obtained, we know that under ideal conditions, the Roche limit is approximately
$d\approx1.260R\left(\rho_M \over \rho_m\right)^{1/3}$
(Here, ideal conditions refer to the satellite being a completely rigid, spherical body, with its material completely held together by gravity (universal gravitation), and the planet it orbits also being a spherical body, while ignoring other factors such as tidal deformation and rotation)

For a fluid satellite, tidal forces will elongate it (making it noodle-like). Therefore, under the same conditions, its Roche limit is greater than the calculated result above, approximately
$d\approx2.423R\left(\rho_M \over \rho_m\right)^{1/3}$
Due to various factors (such as viscosity, chemical bonds, friction, etc.), most satellites are not completely rigid or fluid, so their Roche radius lies between the two values.

Other Situations#

Observing the formula we just obtained, when $\rho_m$ is twice $\rho_M$, the Roche radius $d$ is equal to $R$, which is on the surface of the planet.

When $\rho_m$ is more than twice $\rho_M$, we can know that the Roche radius is within the planet's interior, so the satellite will never be torn apart by the gravitational force of the planet it orbits.

• This article refers to the following references or sources:

Wikipedia - Roche Limit
CNKI - "Roche Limit: The Key to Whether Celestial Bodies Are Torn Apart"