Let us consider a deformable body of mass $ M_b $. Under its own gravitational attraction its shape is a sphere of radius $ R $. Now suppose we introduce another point mass $ M_d $ at a distance $ d > R $. Gravity due to $ M_d $ deforms $ M_b $. The new shape of $ M_b $ is bounded by an equipotential surface. We denote by x the maximum deviation from a sphere of radius $ R $. The difference in the tidal potential due to $ M_d $ is $ \frac{G M_d R^2}{d^3} $, while the difference due the displacement by $ x $ is $ \frac{G M_b x}{R^2} $, hence

$ \frac{x}{R} \approx \frac{M_d}{M_b} \left(\frac{R}{d} \right)^3 $

The mass protruding from the unperturbed radius $ R $ is

$ \Delta M = M_d \left(R/d \right)^3 $

This mass manifests as two lobes, one on the point closest to $ M_d $, and another on the farthest, antipodal point. If $ M_d $ is spinning faster than $ M_b $, then there is a misalignment between the line connecting the lobes and the line connecting $ M_d $ and $ M_b $. This misalignment gives rise to torques. For simplicity, we will assume that the angle between the two lines is of order unity. The torque is therefore given by the difference between the tidal forces acting on the two lobes

$ T \approx \frac{G M_d}{d^3} R \cdot R \cdot \Delta M \approx \frac{G M_d^2 R^5}{d^6} $

This process will exert a torque on $ M_b $ until it rotates at the same angular velocity as that of the orbital motion of $ M_d $, which we denote by $ \Omega $. At point we say that the system is synchronised. The time for synchronisation is

$ \tau_s \approx \frac{M_b R^2 \Omega}{T} \approx \frac{M_b}{M_d} \frac{d^3}{R^3} \frac{\Omega d^3}{G M_d} $