Fall into the sun

We will use Kepler’s 3rd law: $T_1^2/T_2^2=a_1^3/a_2^3$ . Originally the semi-major axis of the orbit is $a_1=$ 1AU; the orbit is circular. The period is $T_1=$ 365.25 days.

The new orbit is an extreme ellipse: one of the focal points is the Sun, the other one is the point where the collision happened. The semi-major axis is $a_2=$ 1/2 AU. Solving the equation yields $T_2=365.25(2)^{-3/2}=129.2$ days.

Falling to the Sun takes half of the orbital period. The answer is 64.6 days.

The probe moves only along the radial direction. In particular, the spacecraft has zero orbital angular momentum. The energy conservation is in this case $E = T + U = \frac{1}{2} m \dot r^2 - G \frac{M m}{r} = - G \frac{M m}{r_0} = \text{const}$ with the initial distance $r(t = 0) = r_0 = 1\,\text{AU}$ . We solve now for the velocity: $\begin{aligned} & & \dot r = \frac{dr}{dt} &= -\sqrt{2 G M} \sqrt{\frac{1}{r} - \frac{1}{r_0}} = -\sqrt{2 G M} \sqrt{\frac{r_0 - r}{r_0 r }} \\ \Rightarrow & & dt & = - \frac{1}{\sqrt{2 G M}} \sqrt{\frac{r_0 r}{r_0 - r }} dr \\ \Rightarrow & & T = \int_{0}^T dt &= - \frac{1}{\sqrt{2 G M}} \int_{r_0}^{0} \sqrt{\frac{r_0 r}{r_0 - r }} dr \end{aligned}$ The differential equation was solved by separation of variables and expressed by an integral. The integral can be analytically determined by substitution and partial integration: $\begin{aligned} -\int_{r_0}^{0} \sqrt{\frac{r_0 r}{r_0 - r }} dr &= r_0^{3/2} \int_{0}^{1} \sqrt{\frac{x}{1 - x }} d x & &\left| x = \frac{r}{r_0}\right.\\ &= r_0^{3/2} \int_{0}^{\infty} \frac{2 z^2}{(1+z^2)^2} dz & & \left| z = \sqrt{\frac{x}{1 - x }} \right.\\ &= r_0^{3/2} \left[ \left. - z \frac{1}{1 + z^2} \right|_{0}^\infty + \int_{0}^\infty \frac{dz}{1 + z^2} \right] & &|\text{integration by parts}\\ &= \left. r_0^{3/2} \arctan(z) \right|_0^\infty = \frac{\pi}{2} r_0^{3/2} \end{aligned}$ Therefore, the result for the travel time is $T = \frac{\pi}{2\sqrt{2}} \frac{r_0^3}{\sqrt{G M}} = \frac{T_0}{4\sqrt{2}} \approx 64.6 \,\text{days}$ with the period $T_0 = \dfrac{2 \pi r_0^{3/2}}{\sqrt{G M}} = 1 \,\text{year} = 365.25 \,\text{days}$ of the earth.

I solved this problem very differently from the intended "right" way. I used this code to iteratively track the force, momentum, and position of the asteroid with small time intervals. Running the code estimates the time it takes for the asteroid to reach the sun.