Answer
The ratio of the asteroid's kinetic energy to the Earth's kinetic energy is $~~1.0\times 10^{-4}$
Work Step by Step
Let $M_s$ be the sun's mass
Let $M_E$ be the Earth's mass
Let $d$ be the Earth-Sun distance
We can write an expression for the Earth's kinetic energy:
$K_E = \frac{GM_sM_E}{2d}$
We can find the ratio of the asteroid's kinetic energy to the Earth's kinetic energy:
$K_a = \frac{GM_sM_a}{(2)(2d)}$
$K_a = \frac{GM_s(2.0\times 10^{-4}~M_E)}{(2)(2d)}$
$K_a = (1.0\times 10^{-4})(\frac{GM_sM_E}{2d})$
$K_a = (1.0\times 10^{-4})~K_E$
$\frac{K_a}{K_E} = 1.0\times 10^{-4}$
The ratio of the asteroid's kinetic energy to the Earth's kinetic energy is $~~1.0\times 10^{-4}$
