Power levels for a 20 ton SE car can be roughly estimated.

P = mv.dv/dt

m = 20,000 kg, v = 50 m/s, dv/dt = 1 g = 9.8 m/s^2

P ~ 9.8 MW, then there's efficiency of the mechanical drive, the power transfer/converter, the attenuation of any beam, and the laser efficiency.

Say that's 0.9, 0.8, 0.9, 0.8 respectively and you need 18.52 MW laser-power. Or for 35% efficiency solar you need ~ 23,000 sq. metres of collectors. Kind of hard to pack into 20 tons of mobile climber and liable to blow away in the atmosphere.

Power required, fortunately, drops off with an inverse square law, just as a laser does, so perhaps solar can take over for the later stages.

On the Moon the power levels are more tractable ~ 1.64 MW for a similar cruising speed. And there's no wind and much longer daylight hours.

Adam