What is the final velocity at, 1 G acceleration for half the the distance between Earth and Mars?

Question

If a vehicle were to travel from Earth to Mars, at 1 G acceleration for one half the journey and 1 G deceleration for the second half of the journey, what would its velocity be at the mid-point?

Answer 1

Due to the nature of orbital mechanics, the distance traveled between planets varies a great deal depending on the position of the plants at the time of launch.  Also be aware that no spacecraft has traveled from earth to mars under constant acceleration/deceleration that I am aware of.

Neglecting these caveats and assuming that such a trip would be made when the plants were physically close to each other (actually, the lowest energy trip is made when they are in opposition, the Hohmann transfer orbit), then the distance to be traveled would be ~0.5 AU, or 46.5E6 miles or 246E9 ft.

After half this trip at constant acceleration of 1G=32.2 ft/sec, the time would be 87,320 sec and velocity would be 2.81E6 ft/sec=1.92E6 mph = 0.29% the speed of light.

Note that the amount of energy and propellant needed to perform this amount of acceleration at 1G is well beyond our current technology.

Answer 2

The distance between the two varies greatly.

Minimum: on same side of the sun

Maximum: on opposite sides of sun

(and remember, the orbits are not circular and they're both moving)

However, if we arbitrarily pick the minimum over the past several years, (August 27, 2003:  3.46 x 10^7 mi or 5.57x10^10 m)

then the problem becomes:

v = (a)(t)

d = (1/2)(a)(t)^2

solving for v in terms of d and a, we have:

v^2=2(a)(d)

v = sqrt(2ad)

substituting

d=5.57x10^10m / 2

a=9.81 m/s^2

we have:

v=5.2x10^5 m/s

(for reference, the speed of sound is around 343 m/s, so this over Mach 1500)