A planet's orbit from the law of gravitation

A planet starts $1.0\ \text{AU}$ from the Sun, moving at right angles to the line to the Sun at $5.0\ \text{AU}$ per year. That is slower than a circular orbit needs, so it falls inward on an ellipse. You will make the program step it forward with the law of gravitation, check Kepler's law of equal areas, and find its year.

The idea. In units of AU and years, $GM = 4\pi^2$ for the Sun, and the pull gives $\vec{a} = -\dfrac{GM\vec{r}}{r^3}$. In each small step $\Delta t$, the line to the Sun sweeps the triangle of area $\tfrac12|x_1y_2 - y_1x_2|$. Kepler's third law gives the year: $T = a^{3/2}$, with $a$ half of the nearest plus farthest distance.

Your tasks.

1. Inside the loop, work out the acceleration a from the law of gravitation.
2. Inside the loop, work out the area swept in the step.
3. Find the nearest and farthest distances from the Sun, and the year from Kepler's third law.