This is a supplemental answer to help show how to calculate the speeds and see that the high speed of spacecraft passing close to the sun is only available close to the sun - and not to suggest that the 2011 Centauri Dreams article reprinted in Gizmodo is just a little confusing if not misleading about this by citing the speed of a spacecraft near the sun and comparing it to a spacecraft that has left the solar system. (There are of course slingshot maneuvers, but then that's beyond the spirit of the original article)
Total energy:
$$E_{tot}=\frac{1}{2}mv^2 - \frac{GMm}{r}$$
Divide through by the mass of the satellite, it becomes specific orbital energy;
$$\mathscr{E}_{tot} = \frac{1}{2}v^2 - \frac{GM}{r}$$
Make hand-waving gestures, or read here about basic orbital mechanics:
$$\mathscr{E}_{tot} = - \frac{GM}{2a}$$
where $a$ is the semi-major axis. The total energy is constant, and can be calculated from the semi-major axis alone (don't need eccentricity). Put them together;
$$- \frac{GM}{2a} = \frac{1}{2}v^2 - \frac{GM}{r}$$
...and you get the very handy vis-viva equation.
$$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$
From Wikipedia, Helios-B perihelion and aphelion are 0.29 and 0.98 AU, or about 43 and 147 million km, and the semi-major axis $a$ is half of the sum (half of the major axis), or about 95 million km.
The Standard Gravitational Parameter of the sun is about 1.33E+11 km^3/s^2 (1.33E+20 km^3/s^2 in standard mks units), so you can calculate the velocities are about 69 km/s at periapsis, but drop to about 20km at apoapsis.
