Answer to Question #271310 in Physics for Kai

Question

An asteroid (mass= one-tenth of the Earth's mass, radius= one-twentieth of the Earth's radius) spins with constant angular velocity if 289 revolutions per day and is on a rotational collision with Earth. Assuming inelastic collision, what is the common final angular velocity (in rev/day) of the Earth-asteroid system? Assume the two objects are solid spheres.

m_Earth= 5.97x10²⁴kg, r_Earth= 6.37x10⁶m

Answer 1

Apply conservation of angular momentum:

$$L_E+L_A=L,\\\space\\ I_E\omega_E+I_A\omega_A=I\omega\\\space\\ \frac{2}{5}m_ER_E^2·\omega_E+\frac{2}{5}m_AR_A^2·\omega_A=\\\space\\=\bigg(\frac{2}{5}m_ER_E^2+\frac25m_AR_A^2\bigg)·\omega,\\\space\\ \omega=\frac{m_ER_E^2\omega_E+m_AR_A^2\omega_A}{m_ER_E^2+m_AR_A^2}=1.07\text{ rev/d}.$$