As <PERSON> points out the possibility of a man-made force on earth itself is next to impossible. My alternative may be outside your rules of "Can't be an asteroid or object from outer space" but it depends on how willing you are to stretch this.
If you consider <PERSON>'s point about changing orbits then we see that these depend on mass and radius. If the disaster were to involve a planet further out than earth and, depending on what rules you want your universe to have/stretch, less massive than earth then it may be more possible.
If any orbiting object loses it's transverse momentum (along the path of the orbit) it will fall in towards the host star (and it is further out so it can fall past earth on the way). So lets look at the possibilities here.
Pluto at it's lowest momentum : $$M_{pluto} = 1.3 x10^{22} kg , V_{pluto,min} = 3710 ms^{-1} $$ Gives: $$P_{min} = M_{pluto}V_{pluto,min} = 4.8x10^{25}kgms^{-1}$$ So you would need to hit it head on with something very massive to stop it entirely ($\frac{P_{min}}{c} = 1.6x10^{17}kg $ where $c$ is the speed of light). 921
However you needn't use pluto and you needn't stop the object entirely. I don't know much about the future you're creating, perhaps you have some death star like ship which can travel at the speed of light to get from system to system. That would do it I believe (see below).
If you had a large ship (perhaps a mining ship to make the excuse of why the ship was so heavy) crash into pluto or some other fairly slow moving object then this could cause the effect you need. Perhaps a ship is rushed off badly repaired, the trajectory is badly calculated or some such and leads to it coming into the solar system head on with something.
However it happens this method means we can have the huge impact needed but have it occur on another celestial object and so not wipe out those on earth. It does still require a huge impact though.
* Based on the death star being 160km across, a volume of $V = \frac{2\pi 160^{3}}{3} = 8.6x10^{6} km ^{3}$, lets say that is mostly empty space and 10% some material like steel with a density of $\rho = 7850 kg m^{-3}$ so our death star has a mass of $\rho V = 6.7x10^{19}kg$ obviously this is all just speculative b 921