If an asteroid were threatening the Earth, could I deviate it just by jumping on it? 
An impact by a 10 kilometres asteroid on the Earth has
  historically caused an extinction-level event due to catastrophic
  damage to the biosphere. There is also the threat from comets coming
  into the inner Solar System. The impact speed of a long-period comet
  would likely be several times greater than that of a near-Earth
  asteroid, making its impact much more destructive...

Of course this is a paradoxical situation and question, but: suppose a perfectly spheric asteroid with radius of a few (hundred?) kilometers is on a collision course with the Earth. 
If I were standing on its surface, could I deviate its direction just jumping on it in a direction perfectly perpendicular to the vector of velocity? 
If it is possible, what is the angle of deflection, and the maximum size of an asteroid  I could deviate?
Update


*

*The asteroid has velocity and momentum in the direction of motion. On the normal direction momentum is zero and we know that it takes negligible/non zero energy to change the direcion of motion. Can we consider the action on the perpendicular as if the asteroid were at rest and treat it like a collision, with conservation of momentum etc?

*If the abovesaid is correct, how do we measure the angle of deflection? In what way the push/jump can be calibrate to obtain a desired oucome?
 A: The key here lies in two observations:


*

*If you can jump hard enough to escape the gravitational pull of the asteroid, this will permanently change the momentum of the asteroid; if you just jump and fall back down, nothing of lasting significance happens (conservation of momentum in the system)

*You want to change the angular momentum of the asteroid's orbit relative to the earth; so if you make a small change in the tangential velocity from a sufficiently long distance to earth, you can make a significant change in the angular momentum (which is velocity times perpendicular distance).


Let's do some math:
10 km diameter asteroid, density $5 kg/m^3$ has mass of $2.6\cdot 10^{15}kg$. If you have a mass of $100 kg$, and you make a big jump (say with a force of 2000N for one second - remember there is not much gravity and I'm going to assume Earth sent a strong guy to save the world...) the change in velocity of the asteroid is
$$\Delta v = \frac{F \Delta t}{m} = \frac{2000}{2.6 \cdot 10^{15}} = 7.6\cdot 10^{-13}m/s$$
If you wanted this to result in a deflection at Earth of at least 10,000 km (assuming that is enough of a "nudge", if done in the right direction, to get the asteroid to miss the earth; that is a big if - it is assuming the asteroid was not aiming straight at the earth, but the correct calculation needs to take account of the orbital dynamics a bit more carefully) then you need to give this nudge at a time
$$t = \frac{10^7}{7.6\cdot 10^{-13}} \approx 400 \text{ billion years}$$
Clearly this isn't going to work. We need a new plan to save the earth. "Dispatch war rocket Ajax".
A: The size of the asteroid does not matter: when you jump, the asteroid will move in the opposite direction. 
Even if you jump up on earth, the earth moves backwards - but only infinitesimally so. Remember that $$F=ma$$
The mass of the earth is $5.97219 × 10^{24} kg$. If your mass is $100 kg$, then the earth's acceleration will be $6 × 10^{22}$ (almost 100,000 billion billion times) smaller than yours - in other words, not measurable.
You have a second problem though: when you fall back down you give a reverse push that cancels the first, and the net effect is zero. Of course this assumes the asteroid is small enough so you didn't jump completely away from it.
