How much kinetic fuel is needed?

Physical formula for elastic and inelastic bounce of two bodies 1 and 2 is:

But kinetic fuel does not impact on aerobrake all at once. It hits aerobrake as a continuous stream of fuel. While the ship becomes faster, speed difference v between fuel and ship becomes smaller. Hence the ship will never become faster as the kinetic fuel. How much velocity do we get from a certain amount of fuel, if it is applied to the ship "drop by drop"? Here is what I got:

Figure 1: Amount of kinetic fuel mf needed to accelerate the space ship to speed vs (red curves).

mf is given in multiples of the ship mass and is vs is given in multiples of the speed of the kinetic fuel. Mass needed depends on how elastic or inelastic the fuel bounces off the ship. Red curves are for estimated range of realistic bounce factors (see below), green dashed curves are for bounce factors 0 and 1. Dashed blue curves indicate performance if transfer would be possible in a single elastic or non-elastic bounce.

Example: if speed of fuel is 12 km/s then going to orbital speed of 8 km/s (which is 0.66 times 12 km/s) needs about 0.6 to 0.75 times the mass of the ship. For a 20 t ship (mass after reaching orbital height) that is 12 to 15 t of fuel. Compared to traditional rockets, this is very good.

A kinetic fuel velocity of 12 km/s can be achieved if the fuel comes directly from the moon. This will allow for about one launch per week (per robotic swarm). The robotic swarm will swing around earth, doing inverted aerobraking on its way, and returns to the moon, where it rendevouz with new fuel, and starts its cycle again.

Another option would be to keep the robotic swarm in low earth orbit and bring extraterrestial fuel to earth orbit (e.g. by using normal aerobraking). Maintenance of the robotic swarm could be easier in LEO with protection from earths magnetic field and close to where most of the equipment is manufactured. Up to one launch every 90 minutes would be possible. But speed of the kinetic fuel would be only 8 km/s. Using an amount of fuel of the ships mass, the ship can be accelerated to 0.8·8 km/s = 6.4 km/s. The remaining v of 1.6 km/s could be achieved with a traditional rocket engine. But inverted aerobraking could be used for this phase of launch too, if the kinetic fuel would be shot towards the space ship with something similar to a conventional rocket engine. In this case the fuel from space would have to contain chemical energy, i.e. be burnable. (Or energy from other sources like solar or nuclear energy would be necessary.) Chemical rocket engines have an exhaust speed of up to 5 km/s in vacuum. This would give 8 km/s + 5 km/s - 6.4 km/s = 6.6 km/s to hit the ship. 1.6 km/s / 6.6 km/s = 0.24, which needs 0.2 times the ships mass (see figure 1). But to prevent the robotic swarm from deceleration and falling into earths atmosphere, the force from accelerating the kinetic fuel towards the ship must be compensated. So about twice the amount of fuel is needed. So this gives a total amount of extraterrestrial fuel of 1.4 times ships mass. Which is still very good.

Elastic bounce here does not mean we are using rubber balls. When the kinetic fuel hits the ship, its kinetic energy is converted to heat. This heat makes the gas bounce off the aerobrake and expand into space. This gives thrust from the same principle that rocket engines use to generate thrust. The more of this gas will be bounced backwards (and not to the sides, e.g. by desiging the aerobrake in the shape of a rocket nozzle), the more thrust will result, and the more elastic the bounce will be (see OnBoardFuel). Elastic bounce is the best possible case, because all of the fuels kinetic energy (in relation to the ship) is used to generate impulse.

The graphs were computed numerically, using the a Matlab script given below, which simulates the impact of kinetic fuel drop by drop.

A closed solution to the differential equation for a bounce factor of 1 was given by someone as:

Using the logarithm of speeds is quite unusual. But we can rewrite it using only a ratio of speeds. If we furthermore use the fact that the bounce factor linearly safes fuel we get:

The Matlab script:

clf; hold on; % clear and prepare figure

% if all the impulse could be transfered in a single bounce
plot(mf, vs,'b--') % inelastic bounce (blue dashed curve)
plot(mf,2*vs,'b--') % elastic bounce (blue dashed curve)

% if impulse is transfered in many bounces
m=3; % maximum relative fuel mass considered here (relative to ships mass)
n=10000; % number of bounces (enough for 3-digit convergence)
vs=zeros(1,n*m); % velocity ship
ms=1; % mass ship
vf=1; % velocity fuel
mf=1/n; % mass fuel per bounce

% theoretically maximal range of bounce factors
for b=0:1:1, % bounce factor: 0=inelastic, 1=elastic
    for i=1:(n*m-1),
        vs(i+1) = vs(i) + (1+b)*(mf/(ms+mf))*(vf-vs(i));

    plot((1:(n*m))/n,vs,'g--'); % (green dashed curve)

% estimated realistic range of bounce factors
for b=(0.5:0.1:1.0)*0.92, % bounce factor: 0=inelastic, 1=elastic
    for i=1:(n*m-1),
        vs(i+1) = vs(i) + (1+b)*(mf/(ms+mf))*(vf-vs(i));

    plot((1:(n*m))/n,vs,'r'); % (red curve)

% format figure
axis([0 m 0 1]);
xlabel('m_f / m_s');
ylabel('v_s / v_f');

print -dtiff -r90 % save figure


Note: depending on the way to place the fuel in space and depending on how sensitive the aerobrake is to spatial and temporary density changes, part of the kinetic fuel will not impact on the aerobrake and will be lost. This fraction is not included in the above estimates and examples.

What is the range of a realistic bounce factor?

From the above we know that the bounce factor ß is 0 for completely non-elastic bounce and 1 for perfectly elastic bounce.

From estimating the usefulness of added on-board fuel we also got an upper bound of 0.92. Considering that chemical rocket engines claim to have an efficiency of 0.95, this would combine into an upper bound of 0.92·0.95 = 0.87. The shape of the aerobrake is probably the dominant factor for this. But also the thickness of the compressed gas/plasma layer on the aerobrake will influence this. If this layer is very thin, most of the gas is deflected vertically to the surface of the aerobrake. This could result in a bounce factor of up to 0.92. If the layer is thick, much more gas will expand to the sides.

To estimate a lower bound let's assume the aerobrake is flat and gas expands to the sides as well as backwards (which is upwards in figure 2). Consider a small cloud of gas, which hits the aerobrake. It is stopped to the velocity of the space ship, heats up due to this and will expand symetrically as a half-sphere. How much impulse do we get from this?

Figure 2: model to estimate a lower bound for bounce factor

Impulse is velocity multiplied by mass. The relevant velocity is the velocity is the vertical component of the velocity of the gas particles. For now lets just assume all particles have the same speed. In reality they have the same speed distribution, so we can apply a corrective factor later. is the angle of the direction in which a single particle expands.

Horizontal velocity components cancel out because of the symmetry of the situation. To express the relative amount of mass that belongs to the same angle, we relate the circumference L of this angle to the total surface area S of the half-sphere and integrate over all angles. Angles are specified in radian, but using degree would give the same result, of course.

Including the corrective factor that means ß = 0.92·0.5 = 0.46 for a flat aerobrake.

The bounce factor for inverted aerobraking is to be expected in the range of 0.46 to 0.92.



We can use the same approach to compute the bounce factor for an aerobrake that does not allow any gas particles to expand at angles of less than a minimal angle. I still assume a uniform distribution of the deflection angles. This is probably somewhat unrealistic, but it gives us an idea about how much restricting expansion directions helps.

Instead of the area of a half-sphere we have to use the area of a cap on the sphere (for h see figure 2).

The graph looks like this: