# Powered landing, analytical solution

There is a long standing dispute whether winged landing or vertical rocket powered landing is more efficient and overall better for reusable rockets on earth. 🙂
Powered landing can be looked at in a broad overview. The post documents what I doodled in my notebook some day in the summer.

## Summary

Deceleration from terminal velocity to hover usually requires about 5-10% of total landed mass as propellants. The amount is directly proportional to the vehicle’s terminal velocity. Also, the higher acceleration, the better. The powered landing penalty fraction is $N_{penalty}=\frac{2g}{3a}$, so landing at 2 gee acceleration (3 gee felt) gives a penalty of 0.3. The required delta vee is $\Delta v = v_{ter} (1+N_{penalty} )$ where v_ter is terminal velocity. And the required impulse $I=m \Delta v$ and propellant mass $m_{prop}=I/v_{ex}$.

## Derivation

Initially, when the craft is descending through the lower atmosphere at terminal velocity, it is at a steady state where gravity and drag balance. A subsonic approximation with a drag number b is used to get:

$F_{drag} = m g = b_{drag} v_{ter}^2$ (Eq. 1)

When the rocket engine is ignited, the vehicle starts decelerating:

$m a(t) = F_{rocket}(t) + b_{drag} v(t)^2 - m g$ (Eq. 2)

Note that a is real acceleration. The felt acceleration would be one g more since this happens at no significant horizontal velocity or distance from earth.

Note also that since powered landing uses so little propellants, we can consider the mass of the vehicle, m, to be constant.

We can get the drag number b from the terminal velocity and thus get rid of it:

$m a (t) = F_{rocket}(t) + \frac{m g}{v_{ter}^2} v(t)^2 - m g$ (Eq. 3)

Now, if we assume the thrust of the rocket to be constant, and use a(t) = v(t)’, we get a not nice equation (arc tans and all) because of the second order of v(t) in the drag term.
But! If we assume a constant deceleration, ie, a(t)=a, by throttling the rocket accordingly, the problem becomes a bit different:
$F_{rocket}(t) = m ( a + g - g \frac{v(t)^2}{v_{ter}^2} )$ (Eq. 4)

Now, since we have a constant acceleration:

$v(t) = v_{ter}-at$ (Eq. 5)

We combine 4 and 5 to

$F_{rocket}(t) = m (a + g - g ( \frac{v_{ter}-at}{v_{ter}} )^2 )$ (Eq. 6)

$F_{rocket}(t) = m (a + \frac{2ag}{v_{ter}}t - \frac{a^2g}{v_{ter}^2}t^2)$ (Eq. 7)

If we integrate this with regards to t from 0 to t=v_ter/a, we get the total impulse I to slow down the vehicle to a standstill (or hover, really).

$I=\int_{0}^{v_{ter}/a}F_{rocket}(t)dt$ (Eq. 8 )

This yields

$I = m v_{ter} (1 + \frac{2g}{3a})$ (Eq. 9)
also known as the formula of Powered Landing Impulse from now on.

Simple, isn’t it?!

The first term, 1, tells about the terminal velocity momentum that always has to be nulled, and the second term 2g/3a is gravity loss with aerodynamic gain taken into account.

What we can see from this is that the landing impulse is directly proportional to the terminal velocity, which was not so obvious beforehand.

We can still calculate further parameters of the landing penalties:

The delta vee can be calculated (again we use the constant mass approximation):

$\Delta v=I/m =v_{ter} (1 + \frac{2g}{3a})$ (Eq. 10)
And the propellant mass as well:

$I=m_{prop} v_{ex}$ so $m_{prop}=I / v_{ex}$ (Eq. 11)

One last simplification is the Powered Landing Penalty Fraction which is the fraction of delta vee going over the ideal delta vee with infinite acceleration. This is simply:

$N_{penalty}=\Delta v - v_{ter} = \frac{2g}{3a})$ (Eq. 12)

## Example

For a 1000 kg vehicle with a terminal velocity of 100 m/s, and real landing acceleration of 2 gees (about 20 m/s^2), the penalty is $\frac{2g}{3a} = \frac{2g}{3 \cdot 2g} = 1/3 \approx 0.3$ powered langing impulse is thus

$I = m v_{ter} (1 + N_{penalty}) = 1000 kg \cdot 100 m/s (1+1/3) \approx 1.3E5 Ns$

and $\Delta v = 100 m/s (1+N_{penalty}) \approx 130 m/s$

With an ISP of 250 s, the exhaust velocity would be roughly 2500 m/s. Thus the amount of propellant needed would be:

$m_{prop}=I / v_{ex} = \frac{1.3E5 Ns}{2.5E3 m/s} = 50 kg$

The amount of propellant is surprisingly low, 5% of the total mass. Thus we can get away with not using the rocket equation or changing mass. The vehicle feels 3 gees here (because of the one additional earth gravity gee besides the two acceleration gees), so it might be a little hard on a crew.

If the deceleration would have been done at only one gee, then the penalty would have been 0.7, delta vee would have been 170 m/s and the impulse would have been 1.7E5 Ns. The propellant mass would have been 70 kg, or about 7% of the total landed mass.

## Graphs

Included are a few general graphs where you can overview the situation with a quick glance.

## Further Notes

It can be noted that these formulas are probably not very accurate (the drag portion, the total mass portion just to mention a few), but nevertheless take into account most of the important variables in the powered landing problem, and an elaborate finite step analysis can be avoided when doing preliminary design. (I have done such in Simulink). The low terminal velocity and reasonable acceleration are of paramount importance to minimize mass penalties from a powered landing. Also, the craft will require some propellants for hover, as the braking has to be done to a certain height from the surface and a final low velocity landing must be done from there. Height measurements are not sufficiently accurate and the time responses of engine ignition and throttle are too slow so a direct landing to a pad is probably not realistic. This all will be handled in the second part of the powered landing analysis. 🙂

A similar simple result can probably not be reached for launch, because it is most often thrust limited (full throttle), the mass changes significantly (higher fuel fraction) and drag fights against the vehicle as well (drag helps in landing) and the acceleration is to ever higher speeds. Thus it seems unlikely that there is a simple analytical equation taking into account drag and gravity losses for launch, even though there is one for landing.

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