Work is defined as $$W=\int\vec{F}\cdot\mathrm{d}\vec{s}.\hspace{1in}(1)$$ The relationship for torque and work which states $$W=\int\vec{\tau}\cdot\mathrm{d}\vec{\theta}\hspace{1in}(2)$$
is only true if the particle experiencing the torque is at a constant distance from the axial point, i.e., rigid body rotation or circular orbits. Derivations which develop equation (2) are assuming circular motion.
That is not the case here. The meteor is getting closer to the origin (the point chosen for torque calculations). To account for the change in distance one must add a term to equation (2), $$W_r=\int\vec{F}\cdot\hat{r}\mathrm{d}r.\hspace{1in}(3)$$
Edit
For this problem the $x$ cooridinate is fixed at $X$, but $y$ is changing. So we can write $$r=\frac{X}{\cos\theta}.\hspace{1in}(4)$$
$\theta$ is the angle with respect to the horizontal ($x$-axis).
We also note that $\vec{F}$ makes an angle $\pi/2 - \theta$ with the $\hat{r}$ so $$\vec{F}\cdot\hat{r}=F\sin\theta.\hspace{1in}(5)$$
From (4) we can determine d$r$: $$\mathrm{d}r=\frac{X\sin\theta}{\cos^2\theta}\mathrm{d}\theta.$$
Now we can put all the pieces together for (3) and integrate from $\theta_1$ to $\theta_2$ and see that (2)+(3) gives the same result as (1). I'll leave the details of the integrals to the curious person, but it does work out properly.
End Edit
If one considers elliptical planetary orbits, this makes sense: the kinetic energy of the satellite or planet or moon is increasing and decreasing which means work is being done, but the torque is zero which means equation (2) gives zero. The work being done is done by the radial component of the attractive force as the object moves toward or away from the attractive center.
