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A small cohesionless spherical body of radius $s$ and density $\rho$ has surface self-gravitational acceleration $4\pi G\rho s/3$. The radial stretching of the corotating point-mass tidal field is $3GMs/R^3$. Equating them gives the displayed survival-distance estimate. The size $s$ cancels because both accelerations are linear in $s$ at fixed density. This is a local spherical force-balance estimate, not the exact fluid Roche figure: shape deformation, material strength and nonsynchronous spin can change its numerical coefficient.

The extra radial term. A circular Kepler orbit has ${\Omega }_{\mathrm{orb}}^2=GM/R^3$. In the frame rotating with the synchronously spinning body, centrifugal acceleration increases by ${\Omega }_{\mathrm{orb}}^{2}a$ across a radial offset $a$. This adds the outward differential accelerationEquivalently, in the inertial frame the farther corotating point needs an extra inward centripetal acceleration ${\Omega }_{\mathrm{orb}}^{2}a$, which gravity alone does not supply. The total radial stretching relevant to the static corotating body is therefore $3GMa/R^3$. This is the corotating point-mass tidal field. The source's “force” expression is per unit mass.

Self-gravitational survival. Denote the asteroid radius by $s$, distinct from the orbital radius $R$. Its mass is $m_a=4\pi \rho s^3/3$ and its surface self-gravitational acceleration is $G{m}_a/s^2=4\pi G\rho s/3$. Compare this restoring acceleration with radial tidal stretching $3GMs/R^3$:The corotating tidal-disruption estimate for a uniform sphere isThis is the requested spherical local force-balance estimate, not an exact fluid Roche figure. It neglects deformation and material cohesion.

Why the asteroid radius cancels. At fixed density, self-gravitational surface acceleration scales as $m_a/s^2\propto s$. The linear tidal acceleration also scales as $s$. Their ratio is therefore independent of $s$, provided $s\ll R$ so that the tidal expansion is valid. Density, rather than asteroid size, controls this estimate.

The terrestrial value. Convert $\rho =2\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ to $2000\mathrm{kg}{\mathrm{m}}^{-3}$ and use $M_E=6.0\times 10^{24}\mathrm{kg}$:This is distance from Earth's centre, not height above its surface.