 Solution

Direct fall towards a point Sun. If the Sun is treated as a point target, the projectile must lose its heliocentric angular momentum. The cheapest change is retrograde, cancelling Earth's tangential orbital velocity. It then starts its radial fall with zero heliocentric speed, so $w=v_E$ andThis is the usual radial-fall estimate. Merely launching towards the Sun while retaining Earth's tangential velocity does not remove the angular momentum barrier.

Minimum for impact on a finite Sun. A literal collision with the solar surface does not require zero angular momentum. The limiting solar impact by a retrograde launch instead has apocentre $a$ at Earth's orbit and pericentre $R_{\odot }$ at the solar surface. The vis-viva equation with semi-major axis $(a+R_{\odot })/2$ gives the heliocentric departure speedIt remains prograde about the Sun, but the Earth-relative departure is retrograde with $w=v_E-v_a$. Therefore the surface launch speed for grazing the finite solar surface isHere $a\simeq 149.598\times 10^6\mathrm{km}$ and $R_{\odot }\simeq 695700\mathrm{km}$.^[ref][ref] Slightly reducing the departure speed below $v_a$ makes the trajectory enter the solar surface rather than just graze it. The $31.81$ and $29.15\mathrm{km}{\mathrm{s}}^{-1}$ values correspond to the explicitly different point-target and finite-radius idealizations, not to two ways of adding escape speeds. These are direct-flight estimates; planetary gravity assists and Earth's rotation are outside the model.