Gravity assists such as this are a form of elastic collision. There's a bit of number crunching here (hopefully no mistakes!), so you'll want to be familiar with the basics of momentum, kinetic energy, and the conservation thereof.
Question: If Ceres (the largest known asteroid and nearly 500 km in diameter) used Earth to perform a gravity assist to increase its own velocity, by how much would this slow the Earth down, and how much larger would Earth's orbit become?
The orbital speed of Earth around the sun is $U = 29.8~\mathrm{km~s}^{-1}$. So at a mass of $$M = 5.97\times 10^{24}~\mathrm{kg},$$
it has a kinetic energy of
$$K = 2.65\times 10^{33}~\mathrm{J}$$ and momentum $$P = 1.78\times 10^{29}~\mathrm{kg~m~s^{-1}}.$$
So let's say Ceres is performing a gravitational slingshot as in the simple diagram below. Ceres has a mass $m = 9.47 \times 10^{20}~\mathrm{kg}$. It approaches Earth at velocity $v$, and after the slingshot its final velocity is (up to, for a low-mass object) a velocity of $2\times U+v$.
The total momentum of the system must be conserved. Ceres has changed direction and thus gained a significant amount of momentum in the leftwards direction: the same momentum that Earth must then lose. Kinetic energy is also conserved. So, we have a system of equations, where the subscripts i and f are initial and final momenta and velocities. M and U are the mass and velocity of Earth, m and v are that of Ceres.
$$MU_i^2 + mv_i^2 = MU_f^2 + mv_f^2$$
which says that the sum of the initial kinetic energies of the two objects must equal the sum of the final kinetic energy. We also have conservation of momentum:
$$MU_i + m\vec{v}_i = MU_f + m\vec{v}_f $$
Solving these equations, the solution is
$$v_f = \frac{(1-m/M)v_i + 2U_i}{1-m/M} $$
If Ceres approached Earth at $v_i = 30~\mathrm{km~s}^{-1}$, I get a solution of $v_f = 89.6~\mathrm{km~s}^{-1}$ - even for such a massive object, the $v_f \approx 2U+v$ approximation is extremely good. This means that Ceres' velocity has nearly been tripled by the gravity assist.
So, the final momentum of Earth is
$$MU_f = MU_i - mv_i - mv_f = 1.78 \times 10^{29}~ \mathrm{kg~m~s^{-1}} $$
In fact, Earth's linear momentum will only decrease by $mv_i + mv_f =
1.13 \times 10^{23} ~\mathrm{kg~m~s^{-1}}$. From this change in momentum and Earth's mass, we find its orbital velocity decreases by
$0.019~\mathrm{m~s}^{-1}$.
Approximating a circular orbit (using $r=GM_{sun} / v^2$), Earth's
orbit widens by 190 km. Sounds like a lot, but bear in mind that's 190
km out of 150 million!
Ceres is many orders of magnitude larger than any satellite that we could launch. So we could never practically use spacecraft to change our orbit significantly, and even an enormous near-miss asteroid would be of little consequence. But, it hasn't stopped some from trying!