We received a comment on LinkedIn about how fast the “Mars run” could be achieved with a sustained 1 G acceleration. The reader suggested this could be done in 40 hours. What engine parameters would be required to make that happen?

Using a simple constant-acceleration, straight-line analysis, you can indeed compute that the trip should take only a couple of days. Assuming a Mars conjunction, the straight distance is about 0.5 AU. At this speed you can ignore the gravitational effects of the sun and so the distance is a simple integral of the acceleration: d = 1/2 at^{2}. The ship accelerates for half the time then decelerates, and the change in velocity is ΔV = at. Combining the two halves of the trip, at an acceleration of 9.8 m/s^{2}, the trip takes about 2.1 days.

```
% straight line: distance s = 0.5*at^2
acc = 9.8; % accel, m/s^2
aU = Constant('au'); % km
dF = 0.5*aU*1000; % distance, m
t = sqrt(4*dF/acc); % time for dF, s
dV = t*acc/1000; % km/s
fprintf('\nAccel: %g m/s^2\n',acc)
fprintf('Time: %g days\n',t/86400)
fprintf('Delta-V: %g km/s\n',dV)
Accel: 9.8 m/s^2
Time: 2.02232 days
Delta-V: 1712.34 km/s
```

Now, your ship mass includes your payload, your engine, your fuel tanks and your fuel. Assume we want to move a payload of 50,000 kg, somewhat larger than the NASA Deep Space Habitat. The engine mass is computed using a parameter called the specific power, in units of W/kg. The fuel tank mass is scaled from the fuel mass, typically adding another 10%. When we run the numbers, we find that the engine needs to have a specific power of about 1×10^{8} W/kg, and an exhaust velocity of about 5000 km/s results in the maximum payload fraction. We can compute the fuel mass and trajectory using our `MassFuelElectricConstantUE` and `StraightLineConstantAccel` toolbox functions:

```
d = StraightLineDataStructure;
d.dF = 0.5*aU; % 0.5 AU
d.tF = 2*86400; % 2.1 days
d.uE = 5000; % km/s
d.eta = 1; % jet power sigma
d.sigma = 1e8; % W/kg
d.mP = 50000; % kg
dOut = MassFuelElectricConstantUE( d )
StraightLineConstantAccel(dOut)
fprintf('Acceleration: %g m/s^2\n',dOut.a*1e3)
fprintf('Exhaust velocity: %g km/s\n',d.uE)
fprintf('Initial power: %g GW\n',dOut.p(1)*1e-9)
fprintf('Engine mass: %g kg\n',dOut.mE)
fprintf('Payload fraction: %g\n',dOut.mP/dOut.m0)
Acceleration: 10.02 m/s^2
Exhaust velocity: 5000 km/s
Specific Power: 1e+08 W/kg
Initial power: 2832.61 GW
Engine mass: 28326.1 kg
Payload fraction: 0.442172
```

This produces the following plots:

The power needed is… over 2.8 terawatts! That’s about equal to the total power output of the entire Earth, which had an installed power capacity of 2.8 terawatts in 2020. **And** the engine would need to weigh less than 30 tons, about the size of a loaded tractor-trailer truck. For comparison, we estimate a Direct Fusion Drive would produce about 1 MW per ton, which is a specific power of 1×10^{3} W/kg. So, this is why you see us trying to design an engine that can do the Mars transfer in 90 days and not 3 days!

Now, there is another consideration here. Namely, constant acceleration at 1 G is not the optimal solution by any means. The optimal solution for a fast, light transfer is actually a linear acceleration profile. This knowledge goes way back: 1961! Here’s a reference:

Leitmann, George. "Minimum Transfer Time for a Power-Limited Rocket." Journal of Applied Mechanics 28, no. 2 (June 1, 1961): 171-78. https://doi.org/10.1115/1.3641648.

This would mean that the engine changes its exhaust velocity during trip, passing through infinity at the switch point. We compute this in our “straight-line, power-limited” or SLPL function series. While this can’t be done physically, even an approximation of this with a variable impulse thruster will one day be more efficient than constant acceleration or thrust. How much better? The power needed is nearly 1/2 the constant acceleration solution, 1.5 TW, and the specific power needed is reduced by half, to 5.6×10^{7} W/kg. However, those are still insane numbers!

```
mD = 80000; % dry mass: engine, tanks, payload
m0 = 1.5*mD; % wet mass: with fuel
tF = 3*86400;
vF = 0;
[Pj,A,tau] = SLPLFindPower( aU, tF, vF, mD, m0 );
mTank = 0.05*(m0-mD); % tanks, scale with fuel
mLeft = mD-mTank;
mEngine = mLeft - mPayload;
disp('Straight-line Power-limited (linear accel)')
fprintf('Engine power is %g GW\n',Pj*1e-9);
fprintf('Engine mass is %g kg\n',mEngine);
fprintf('Payload mass is %g kg\n',mPayload);
fprintf('sigma is %g W/kg\n',Pj/mEngine);
SLPLTrajectory( A, tau, Pj, m0, tF )
Straight-line Power-limited (linear accel)
Engine power is 1573.26 GW
Engine mass is 28000 kg
Payload fraction is 0.416667
sigma is 5.6188e+07 W/kg
```

The trajectory and engine output are plotted below. The linear acceleration results in a curved velocity plot, while in the constant acceleration case, we saw a linear velocity plot. You can see the spike in exhaust velocity at the switch point, which occurs exactly at the halfway point.

After all, who needs 1G gravity when the trip only takes 2 days?

For even more fun though, we computed a planar trajectory to Mars using the parameters we found – just to confirm the straight-line analysis is in fact a good approximation. This figure shows the paths the optimization takes:

It is in fact approximately a straight line!

In reality though, these power system numbers are not even remotely plausible with any technology we are aware of today. That’s why we are designing engines to reduce the Mars trip time to 90 days from 8 or 9 months – still a big improvement!

Liked the above calculations. What happens when your fuel is limited to 1 gallon and you re use the expelled exhaust by changing the molecular structure back to original and feed back to tank. Thus you will have 1gallon fuel mass giving same. 1G acceleration all the time every second. And not to mention the energy requirement less than 100 watts! Please do reply Stephanie Thomas.

If you capture the expelled fuel you will not gain any net momentum. The rocket only works if the fuel carries momentum away from the spacecraft.

Actually the momentum is there and the expended fuel is captured in such a way that there is no momentum loss. Fibonacci system.

If you recapture the expended fuel, you are reversing the momentum transfer process which gave the spacecraft momentum in the first place, thus bringing the spacecraft back to square one: no momentum transfer.