The Kepler orbit of the Earth around the Sun is determined by two constants: the
specific orbital energy $E$ and the specific relative angular momentum $h$:
$$
\begin{align}
E &= \frac{1}{2}v_{r,\oplus}^2 + \frac{1}{2}v_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\
h^2 &= r^2\,v^2_{T,\oplus} = \mu a(1-e^2),
\end{align}
$$
where $\mu = G(M_\odot + M_\oplus)$, $r$ is the distance Earth-Sun (at the moment of impact), $a$ is the semi-major axis, $e$ is the orbital eccentricity, $v_{r,\oplus}$ is the radial orbital velocity of the Earth, and $v_{T,\oplus}$ the tangential velocity. Now, suppose that a large asteroid collides with the Earth, with orbital velocity $(v_{T,A},v_{r,A})$ and mass $M_A$. Its relative velocity is then
$$
\begin{align}
v_{T,A}' &= v_{T,A} - v_{T,\oplus},\\
v_{r,A}' &= v_{r,A} - v_{r,\oplus}.
\end{align}
$$
We can express these relative velocities in terms of the total impact velocity $v_\text{i}$ and the impact angle $\theta$:
$$
\begin{align}
v_{T,A}' &= v_\text{i}\cos\theta,\\
v_{r,A}' &= -v_\text{i}\sin\theta,
\end{align}
$$
where I defined $\theta$ as in Fig. 1 of this article. So we obtain
$$
\begin{align}
v_{T,A} &= v_{T,\oplus} + v_\text{i}\cos\theta,\\
v_{r,A} &= v_{r,\oplus} - v_\text{i}\sin\theta.
\end{align}
$$
If we assume that the collision is central, that heat loss is negligible and that the debris remains gravitationally bound to the Earth, then conservation of momentum implies
$$
\begin{align}
M_\oplus\,v_{T,\oplus} + M_A\,v_{T,A} &= (M_\oplus+M_A)u_{T,\oplus}\\
M_\oplus\,v_{r,\oplus} + M_A\,v_{r,A} &= (M_\oplus+M_A)u_{r,\oplus},
\end{align}
$$
with $(u_{T,\oplus},u_{r,\oplus})$ the new orbital velocity of the Earth (and the gravitationally bound debris) after the impact. We get
$$
\begin{align}
u_{T,\oplus} &= v_{T,\oplus} + \frac{M_A}{M_\oplus+M_A}v_\text{i}\cos\theta,\\
u_{r,\oplus} &= v_{r,\oplus} - \frac{M_A}{M_\oplus+M_A}v_\text{i}\sin\theta.
\end{align}
$$
So the orbital energy and angular momentum will have changed into
$$
\begin{align}
E' &= \frac{1}{2}u_{r,\oplus}^2 + \frac{1}{2}u_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a'},\\
h'^2 &= r^2\,u^2_{T,\oplus} = \mu a'(1-e'^2).
\end{align}
$$
(the change in $\mu$ is negligible). Right, let's plug in some numbers. Suppose we start with a circular orbit, with a radius equal to the present-day semi-major axis:
$$
\begin{align}
\mu &= 1.32712838\times 10^{11}\;\text{km}^3\,\text{s}^{-2},\\
r &= a = 1.49598261\times 10^{8}\;\text{km},\\
e &= 0.
\end{align}
$$
For a circular orbit, it follows that
$$
\begin{align}
v_{T,\oplus} &= \sqrt{\frac{\mu}{r}}= 29.785\;\text{km}\,\text{s}^{-1},\\
v_{r,\oplus} &=0\;\text{km}\,\text{s}^{-1}.
\end{align}
$$
The impact velocity of an asteroid will always be at least equal to the Earth's escape velocity $11.2\,\text{km/s}$, which is the speed it takes up as it falls into the Earth's gravitational potential well. The article that I already linked to states that typical asteroid impact velocities are in the range of $12-20\,\text{km/s}$. In theory, the impact velocity can be as large as $72\,\text{km/s}$ in the case of a head-on collision, when the Earth and the asteroid have opposite orbital velocities, thus a relative velocity of ~$60\,\text{km/s}$, augmented with the escape velocity as the asteroid falls into our gravitational potential well. This is very unlikely for asteroids, but it is possible for comets.
So, let us assume a typical impact velocity $v_\text{i}=16\,\text{km/s}$,
a mass $M_A = 0.1M_\oplus$ and an impact angle $\theta=45^\circ$. We find
$$
\begin{align}
u_{T,\oplus} &= 30.813\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= -1.0285\;\text{km}\,\text{s}^{-1},\\
E' &= -411.87\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 2.1248\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.61109\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0788,\\
r_\text{p} = a'(1-e') &= 1.48411\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.73807\times 10^8\;\text{km},
\end{align}
$$
with $r_\text{p}$ and $r_\text{a}$ perihelion and aphelion. Evidently, the influence on the Earth's orbit is substantial.
In the case of a direct-from-behind collision, we get $\theta=0^\circ$, $v_\text{i}=11.2\,\text{km/s}$, so that
$$
\begin{align}
u_{T,\oplus} &= 30.803\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\
E' &= -412.72\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 2.1234\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.60778\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0695,\\
r_\text{p} = a'(1-e') &= 1.49598\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.71958\times 10^8\;\text{km}.
\end{align}
$$
As expected, the radius at impact has become the perihelion, and the change in eccentricity is lowest.
And just for fun, let's try the worst-case scenario: $\theta=180^\circ$, $v_\text{i}=72\,\text{km/s}$:
$$
\begin{align}
u_{T,\oplus} &= 23.239\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\
E' &= -617.10\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 1.2086\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.07530\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.391,\\
r_\text{p} = a'(1-e') &= 0.65462\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.49598\times 10^8\;\text{km},
\end{align}
$$
so that the radius at impact has become the aphelion, and the change in eccentricity is highest. Although I wonder how much would be left of the Earth after such an apocalyptic event...
If the collision isn't central, then part of the energy will be transferred to the axial rotation of the Earth, which should reduce the effect on the orbit. But that will be more difficult to quantify.
Best Answer
From a purely energy point of view, calculate the kinetic and potential (relative to the sun) energies of earth at Mars' orbit minus that in its current orbit.
The kinetic energy is very straight forward, 1/2 m v2. Velocity (approximating to circular orbit) is only a function of the mass of the sun and the orbital radius. Or, you can simply look up the values for Earth and Mars.
Potential energy is more complicated than the mgh approximation we use in everyday lives here on the surface of Earth, because g is no longer a constant since the distance to the sun changes substantially. You have to solve the integral or look up the formula.
However, how to start with power here on Earth and actually cause the sustained force it would take is totally another matter. What I describe above is just that absolute minimum energy it would require without violating conservation of energy. There is a long way between that and a working system.