If earth had no atmosphere, how would the asteroid's kinetic energy shortly before it impacted the surface compare to its kinetic energy when it was far away from earth? (hint: think about the law of conservation of energy.)

The kinetic energy is the same as in space, which in general is very large

Explanation:

If the Earth had no atmosphere we can use the conservation of kinetic energy in two points

Initial. In space far from the planet

      Em₀ = k = ½ m v₀²

Final. Just before touching the surface of the Earth

      = K = ½ m v²

As there is no rubbing

      Em₀ =

      = ½ m v₀²

The kinetic energy is the same as in space, which in general is very large

When the Earth has an atmosphere we must use the energy work theorem

      W = ΔK

The work is done by the friction forces when the meteor enters the atmosphere, increases in density as it approaches the surface, so the work also increases.

       W = - K₀

        = K₀ - W

        = ½ m v₀² - W

We see that the kinetic energy decreases as the work increases, this makes the impact is higher and part of the meteor also evaporates by friction at the entrance

true tan speed = ang vel x radius

does anyone know or what?