#### Answer

(a) The recoil speed of the earth is $6.7\times 10^{-8}~m/s$.
(b) The recoil speed is $2.23\times 10^{-10}~\%$ the earth's speed around the sun.

#### Work Step by Step

(a) We can use conservation of momentum to find the earth's recoil speed $v_f$. Let $m_e$ be the mass of the earth. Let $m_a$ be the mass of the asteroid.
$p_f = p_0$
$(m_e+m_a)~v_f = m_a~v_0$
$v_f = \frac{m_a~v_0}{m_e+m_a}$
$v_f = \frac{(1.0\times 10^{13}~kg)(4.0\times 10^4~m/s)}{5.97\times 10^{24}~kg}$
$v_f = 6.7\times 10^{-8}~m/s$
The recoil speed of the earth is $6.7\times 10^{-8}~m/s$.
(b) We can find the earth's speed as it revolves around the sun.
$v = \frac{distance}{time}$
$v = \frac{2\pi ~r}{1~year}$
$v = \frac{(2\pi)(1.5\times 10^{11}~m)}{(1~year)(365~days/year)(24~hr/day)(3600~s/day)}$
$v = 3.0\times 10^4~m/s$
We can find the recoil speed as a percentage of the earth's speed as it revolves around the sun.
$\frac{6.7\times 10^{-8}~m/s}{3.0\times 10^4~m/s} \times 100\%$
$= 2.23\times 10^{-10}~\%$
The recoil speed is $2.23\times 10^{-10}~\%$ of the earth's speed around the sun.