Question

(This is a challenging problem, so don't spend too much time on this one. It's mostly for bragging rights. The physics is not difficult, but the algebra is a bit tricky.) Suppose a 7.2 1010 kg meteoroid struck the earth at the equator with a speed v = 1.7 104 m/s, and remained stuck. What would be the magnitude of the change in the rotational frequency of the Earth (fE = 1 rev/day)? To avoid round-off errors, plug in numbers only after the necessary algebraic manipulation. (For moments of inertia of some common geometries)

Answer 1

if we assume the earth is a uniform sphere (its not... but we'll approximate it as such) then the angular momentum of the earth is

       L for Earth = I ω = (2/5) M R² * ω

= 0.4* 6 x 10²⁴ * (6.38 x 10⁶)² * (2π radians / 86400 sec ) =    71.043 x 10³²

The angular momentum of the meteoriteis

    L = m v r sinθ = 7.2 x10¹⁰ * 1.7 x 10⁴ * 6.38 x 10⁶*sin45

= 55.218x 10²⁰

This angular momentum is imparted to the Earth, and the ratioof the added angular momentum to the Earth's original is

55.218 x10²⁰ / 71.043 x 10³²   =     0.777 x10^-12

Since the angular speed is directly proportional to theangular momentum of the earth, this is also the ratio of Δω / ω    roughly one part in abillion.