# Estimating thrust and kg fuel for a generation ship's journey to Alpha Centauri

Premise: A generation spaceship leaves Earth around the year 2060 on a journey to colonize Alpha Centauri A (ACA). In this fiction, fusion power is achieved in 2040, improved over 20 years, and used within the solar system. The trip to ACA will take 110 years. The ship will accelerate halfway, flip, and decelerate for the second half.

I understand basic physics equations involving $$F (force) = m (mass) * a (acceleration)$$ and simplified space travel using constant acceleration giving $$d=(1/2)at^2$$, with distance (d) in meters, acceleration (a) in meters per second squared, and time (t) in seconds.

However, this distance traveled does not account for mass loss of Xenon fuel used for propulsion. How do I set up an equation to get (at least a rough estimate of) the Newtons of thrust and kg of Xenon needed for the journey to take 110 years?

Given:

• The ship leaves in 2060: about 40 years more advanced than our current 2021 tech levels.
• The journey takes 110 years (as relatively perceived by those on board the ship).
• Ship launch mass of 1,900,000 kg.
• Each ion drive provides 30 N thrust, averaging 15 kW used per N, fuel use 75 kg of Xenon per 4,000 seconds of burn. (based on advanced versions of current drives)
• Light years to ACA: 4.37.

Edit: thanks to answers and comments : Originally, I thought they would flip the ship to decel halfway, but the ship will want to continue to burn at the same max safe thrust, and so burn near constant fuel during the entire trip. So, the latter half of the trip will see increasingly larger accel, due to decreasing mass but constant thrust Newtons. This changing mass makes the calculation more complex, because they will not simply flip at halfway point... as the decel part will be shorter due to lower mass. I am currently researching rocket equations which account for fuel mass losses but dont have it figured out yet...

Journey with simplified acceleration if time is 110 years: $$a = d/0.5t^2 = (2.06717e16) / (0.5 * (3.469e9)^2) = 0.00343556041 m/s^2 = a$$.

If the ship is 1,900,000 kg at launch from Earth, and $$F=ma$$, $$1900000*a = 6527$$ N (Newtons of thrust). However this is simplified. N thrust will change as fuel mass is lost... My thinking is that the ship will want to continue to burn at the same max safe thrust, and so burn near constant fuel during the entire trip. So the latter half of the trip will see increasingly larger accel, due to decreasing mass but constant thrust.

6527N can be provided by 218 individual 30N drives (around this number may be good even as mass lessens, for redundancy safety). Based on above givens, this requires 861,110 kg Xe fuel. Ship mass would continually decrease as Xe used, until the ship is empty of fuel and about 1,040,000 kg mass remains, requiring less force to move.

I'm not sure how to estimate how much N of thrust and mass of Xe fuel will be needed for this journey. I am imagining two functions, with the force function relying on the lost Xe mass (which is a constant loss over time), but I am unsure how to set that up so that everything results in a 110 year journey. Should I integrate to get areas underneath both functions, then adjust until I get roughly 110 years? Ideally I'd like equations where I can easily adjust the ship mass, thrust Newtons, and so on to calculate with different variables if needed.

Regarding initial velocity: Ideally for the story, the ship would leave from Mars orbit: Linear distance can be expressed as (if acceleration is constant): $$s = v_0 * t + 0.5a t^2$$. With $$v_0 =$$ initial linear velocity (m/s) = Mars mean orbital velocity in (m/s) = $$24070$$

Regarding relative movement of both the Solar System and Alpha Centauri, I found:

Using spectroscopy the mean radial velocity has been determined to be around 22.4 km/s towards the Solar System. This gives a speed with respect to the sun of 32.4 km/s, very close to the peak in the distribution of speeds of nearby stars.

But without knowing ship's max v, because the ship-flipping point is unknown to me, I'm not sure how much 22.4 kps will affect the journey.

Info and chart below from https://en.wikipedia.org/wiki/Ion_thruster#Comparisons

Ion thrusters in operational use typically consume 1–7 kW of power, have exhaust velocities around 20–50 km/s (Isp 2000–5000 s), and possess thrusts of 25–250 mN and a propulsive efficiency 65–80%.[3][4] though experimental versions have achieved 100 kW (130 hp), 5 N (1.1 lbf).[5]

Thruster Propellant Input power (kW) Specific impulse (s) Thrust (N) Thruster mass (kg)
X3 Xenon max 102 kW 1800–2650 5.2 230
AEPS Xenon 13.3 2900 .6 100
BHT8000 Xenon 8 2210 .449 25
NEXT Xenon 6.9 4190 .236 max.
NSTAR Xenon 2.3 3300–1700 .092 max.
PPS-1350 Hall effect Xenon 1.5 1660 .090 5.3

https://solarsystem.nasa.gov/missions/dawn/technology/spacecraft/ Dawn Ion Propulsion System Number of thrusters: 3 Thruster dimensions (each): 13 inches (33 centimeters) long, 16 inches (41 centimeters) in diameter Weight: 20 pounds (8.9 kilograms) each Spacecraft acceleration via ion propulsion at full thrust: 0 – 60 mph in 4 days Thrust: 0.07 to 0.33 ounce (19 to 91 millinewtons)

Fuel https://en.wikipedia.org/wiki/Ion_thruster#Propellants Many current designs use xenon gas, as it is easy to ionize, has a reasonably high atomic number, is inert and causes low erosion. However, xenon is globally in short supply and expensive. VASIMR design (and other plasma-based engines) are theoretically able to use practically any material for propellant. However, in current tests the most practical propellant is argon, which is relatively abundant and inexpensive.

https://en.wikipedia.org/wiki/Variable_Specific_Impulse_Magnetoplasma_Rocket [Higher energy use ok because of fusion power.] Other propellants, such as bismuth and iodine, show promise, particularly for gridless designs such as Hall effect thrusters. Krypton is used to fuel the Hall effect thrusters aboard Starlink internet satellites, in part due to its lower cost than conventional xenon propellant. FUEL USE: The Deep Space 1 spacecraft, powered by an ion thruster, changed velocity by 4.3 km/s (2.7 mi/s) while consuming less than 74 kg (163 lb) of xenon. [=4300 m/s for 75kg Xe?] The Dawn spacecraft broke the record, with a velocity change of 11.5 km/s (41,000 km/h), though it was only half as efficient, requiring 425 kg (937 lb) of xenon.

• A lot of engines for redundancy is actually a trap. NASA found that what you want is as few engines as possible, as with each additional engines probability of engine failure is increasing (as you are rolling more dice). link Also bear in mind each of these engines has mass, so by adding engines you're adding more dry mass to the ship. Commented Jun 26, 2021 at 18:13
• Additional questions: 1. Are you expecting the same delta-v for acceleration and deceleration burns? 2. Tied to 1, what about the relative motion of Sol and ACA? Are we trying to counteract it during the burn or is the path calculated to make use of this relative velocity? Velocity of Mars gives us little, besides calculating Sol escape delta-v, as the frame of reference is with regards to Sol and we're intending to go interstellar, which means it's the Sun's velocity in relation to ACA that interests us. Commented Jun 26, 2021 at 18:24
• @JANXOL you make a great point! now i need to research the relative movements of the 2 systems to see if they make a big enough impact on the journey... If i remember correctly, the 2 are moving towards each other, but im not sure how much in 110 yr timescale Commented Jun 26, 2021 at 18:43
• @JANXOL I just made an edit about how thrust should be fairly constant, but accel will change due to decreasing mass. so decel portion will be shorter; not half of the trip. Commented Jun 26, 2021 at 18:44
• @KoonW please disregard the other comments !and answers! here so far, they either focus on the wrong aspects or completely lie to you. What you need is to dive into the en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation , which is what is used to calculate the mass cost and effectiveness of rocket propulsion. For a 110 year trip to Alpha Centauri, your speed will remain low enough that you could completely disregard relativistic affects, thus avoiding the headache of factoring Relativity into your math. I advise you use omnicalculator.com/physics/ideal-rocket-equation wisely Commented Jun 26, 2021 at 18:56

Since you mention integrals, I know you are familiar with calculus, so I can give you the short and sweet answer. It is not always true that $$F=ma$$. The more complete version of Newton's equations yields $$F=\frac{dp}{dt}$$, where $$p$$ is momentum. Force is the change in momentum over time. Now, since $$p=mv$$, we can quickly see that if mass is constant, we get $$F=m\frac{dv}{dt}$$ which is $$F=ma$$. If mass is not constant, then you have to use the chain rule to get $$F=\frac{dm}{dt}v+m\frac{dv}{dt}$$, which is what is used in rocketry. Integrate that, and you get the answer you need.

You aren't the first to want to do this. the Tsiolkovsky rocket equation is the go-to equation for doing these calculations

$$\Delta V= I_{sp}g_0\ln\frac{m_0}{m_f}$$

Why go here first, rather than integrating? Well we don't have our ship spec'd out yet. We need to understand our mass fraction before daring to integrate to get distance. But what we know is that we have to do two burns. The first burn takes us from an initial velocity (call it 0) to $$v_{\frac{1}{2}}$$, the velocity at the flip point (which, as you note, isn't quite at the half way point in distance, but I'm using the subscript $$\frac{1}{2}$$ anyways). Then, the second burn takes us down to the velocity of ACA with respect the earth..

Once you have this, you just have to use the above complete version of Newton's law to do the integration.

I'm going to give us control of "number of engines" as a variable. Now, I don't recommend actually just stacking more and more little engines. It's not always the most efficient approach. But a multiplier on the existing ion engine you described seems like a pretty good way to go! We'll call this scale factor $$k$$. If your ship has a scale factor of $$k$$, it means it produces $$30k$$ Newtons of thrust, and consumes $$\frac{75}{4000}k\frac{kg}{s}$$ worth of Xenon while active.

We're also going to need the ISP. Now it looks like you mixed the numbers from several ion thrusters, and got one which is actually quite weak. Others can check my math, but I pegged it at an ISP of about 160 seconds, which is extremely low (its lower than a chemical rocket). Typically the ISP is in the thousands for an ion thruster. So let's just leave it as a variable, $$I_{sp}$$, but I'll peg it to the really nice ISP of NEXT, at 4190s. Feel free to adjust from there, but that's really the dominating variable in these thrusters. You can adjust size and flow rate as much as you like, but changing ISP is incredibly difficult.

You should also pick a $$m_f$$. Your question listed a $$m_0$$, but $$m_f$$ is typically easier to work with because its bounded by the need to do something with a payload. For example, it might be all of the life support needed to support 10,000 people, or something like that. It will just be a scale factor on everything, so I won't include it... but you'll need it to turn into the question of "how hard is it to actually make this rocket." For now, I'll just assume a $$m_f$$ of 1,000,000kg.

We can do everything in velocities in the initial frame, so $$v_0=0$$ and $$v_f$$ is the velocity of ACA in our frame, which is roughlyy 21.4km/s towards us, so we'll say $$v_f=-21.4km/s$$ to make all of the signs line up

Now, we know that our total burn is the sum of the speeding up burn plus the slowing down burn. $$\Delta V=v_\frac1 2 + (v_\frac 1 2 - v_f) = 2 v\frac 1 2 - v_f$$. By the rocket equation, we can now see that we can relate this to the propellant mass that we use.

$$\Delta V=v_e\ln\frac{m_f}{m_0}=v_e\ln\frac{m_f}{m_f+m_p}$$

$$2 v_\frac 1 2 - v_f = v_e\ln\frac{m_f}{m_f + m_p}$$

Here I've broken out the initial mass into a final mass plus the mass of the propellant, $$m_p$$. This is convenient because we can calculate the propellant mass from the data you've given. If $$k=1$$, then we know that we consume $$\frac{75}{4000}\frac{kg}{s}\cdot T$$ fuel, where $$T$$ is the duration of the flight, 110 years. A quick unit conversion and a multiplication by k to $$591300kT\frac{kg}{year}$$ points out that this is going to be quite the high mass fraction. At 110 years, you will consume just over $$65,000,000k$$ kilograms of fuel. Thus for

• $$k=1$$, $$m_p=65,000,000kg$$, ($$\zeta=0.984$$)
• $$k=5$$, $$m_p=325,000,000kg$$ ($$\zeta=0.9969$$)
• $$k=10$$, $$m_p=650,000,000kg$$ ($$\zeta=0.9984$$)

I note the mass fraction, $$\zeta$$ because it is a common way to measure rockets. Typical mass fractions are in the 0.8 to 0.9 range, with 0.9 being typical for the single-stage-to-orbit (SSTO). Note that one of the great challenges of SSTO is that its hard to achieve a mass fraction that high. So, when you talk about using current technology, recognize that this is quite far outside of what we're typically working with. You will be bringing a lot of fuel!

Regardless, we can combine these equations to get one overarching solution:

$$2 v_\frac 1 2 - v_f = v_e\ln\frac{m_f}{m_f + \dot m_1kT}$$

Where $$\dot m_1$$ is the above mass flow rate of a $$k=1$$ engine. Or, rearranged slightly,

$$v_\frac 1 2 = \frac{v_f + v_e\ln\frac{m_f}{m_f + \dot m_1kT}}{2}$$

Now this is really neat. It says that if you want to visit ACA, not just fly past it at painfully fast speeds, there's only so many ways you can do it. It says that, for any fuel flow rate ($$k$$), there is exactly one $$v_\frac 1 2$$ that leaves you at exactly the correct velocity you need, $$v_f$$. Any other 110 year long burn will leave you at the wrong velocity.

This means we're really close to having an answer. We can build a plot with $$k$$ as our independent term, and the distance traveled, $$d$$ as our dependent term. All we have to do is calcualte the result of a constant-force 2 stage burn, where we burn out the first stage at the point where $$v=v_\frac 1 2$$, and then we burn in the opposite direction.

At this point, we could solve a bunch of integrals, but I'll leave this as an exercise for the reader. In the spirit of astrophysics, I invoke "shut up and calculate" and throw everything into a really cheesy python simulation. I just do Riemann integration at 1/10th year intervals, and trust that's fine grained enough to cover for the laughably inexact way I handle updating all of the anti-derivatives.

from math import log

ln = log # Python's log(x) is actually the natural log.  Aliasing it

mf = 1000000        # kg - my own assumption
m1 = 75/4000        # kg/s
vf = -21400         # m/s
isp = 4160          # s
T  = 110 * 31556952 # s - trip length
f = 30              # N - force of the reference engine
g0 = 9.8            # m/s^2 - gravitiy on earth

def calcDist(k):
"""Returns distance traveled in light years"""
# step 1: for given k, calculate the ship's stats
mdot = m1 * k
mp = mdot * T

# step 2: compute v 1/2
v05 = 0.5 * (vf - isp * g0 * ln(mf / (mf + mdot * T)))

# step 3: Integrate!
dt = 0.1 * 31556952 # arbitrary decision, dt is 1/10th of a year
m = mf + mp  # kg
v = 0        # m/s
d = 0        # meters
t = 0

# step 3a: Burn 1 (accel)
# stop at v1/2
while v < v05:
a = (f * k - mdot * v) / m # rearrange force equation
d += v * dt
v += a * dt
m -= mdot * dt
t += dt

# step 3b: Burn 2 (decel)
# stop when out of fuel
while m > mf:
a = (f * k - mdot * v) / m
d += v * dt
v -= a * dt # note minus sign: slowing down
m -= mdot * dt
t += dt

return d / 9460730472580800 # meters to light years

k = np.linspace(100, 1000, 100)
d = [calcDist(x) for x in k]

plt.plot(k, d, '-k')
plt.axhline(4.37, linestyle="dashed") # distance to ACA
for kk, dd in zip(k, d):
if dd > 4.37:
plt.axvline(kk, linestyle="dotted")
plt.text(kk + 50, dd - 0.1, "k=%d" % kk)
break
plt.xlabel("Multiple of reference engine")
plt.ylabel("Light years")
plt.show()


So you will need the equivalent of 381 of those 30N ion thrusters to do the job, and 24800 kg of Xe fuel for every 1 kg of payload. (for a mass fraction of $$\zeta=0.9999596$$)

This is consistent with the back of the envelope calculations you did. You calculated 218 to get there without slowing down. Slowing down requires 4x more thrust, so would require just under 900 engines if we didn't account for the decreasing mass. The actual answer is somewhere between the two.

Note, you will have to be mighty creative to achieve that mass fraction. Your fuel tanks are going to have to be very thin, and very large, and yet still survive the 110 year journey!

• Perhaps worth noting: the worldwide Xenon production in 1998 was around 36,000kg. So, at that rate, each kg of payload you send in this generation ship requires somewhere around one year's worth of Xenon harvesting! Commented Jun 28, 2021 at 1:53
• thank you for all of this hard work - even a python program! It is gonna take me a bit to crunch through this. Im glad for the flexibility of the equations, as I just threw some numbers on here that seemed similar but more advanced than current drive tech. Commented Jun 29, 2021 at 2:20
• @KoonW There seems to be two limits. The first is raw power. The more power density you achieve, the better the ISP. The other issue is electrode erosion, which I have a hunch gets nastier as you increase the power, but I haven't confirmed that. But I wouldn't expect magic. If Ion thrusters got much higher ISP with more power, we'd see more satellites pulsing their ion thrusters. However, that might be in the form of a question that Space Exploration.SE can handle. They're the experts at ISP, and you're talking about modern technology. Commented Jun 29, 2021 at 4:40
• The real limit comes from the fact that you're trying to maximize $v_e$. Energy goes up by the square of velocity, so increasing $v_e$ 10 fold requires 100 times the power. At least until relativistic effects come into play, I'm horrible at making good guesses about how things will go when relativity is involved./ Commented Jun 29, 2021 at 4:49
• You can edit the code to: from math import log as ln for brevity Commented Jun 29, 2021 at 5:50

Halfway point mass is average of start and finish.

You state that you lose xenon mass steadily thru the journey.

/relying on the lost Xe mass (which is a constant loss over time)/

Your mass at the midway flip point is the average of full and empty: half full. If you want a solution for the journey taken as a while, work your equations based on the midpoint weight. The increased weight at journey start will be balanced by decreased weight at journey end and the math will work for the journey as a whole.

"But wait!" you object. "I was wrong! It is not steady use! I actually use xenon less fast for the second half of the trip, because the ship is less massive and so requires less force to accelerate than it did on the first half!" True, true. This then becomes a calculus problem to model both the smoothly decreasing rate of use of fuel and smoothly decreasing rate of loss of mass. Which I would like to see worked out but which is beyond my ability.

• Thank you for pointing out the part about Xe depletion being constant - My thinking is that the ship will want to continue to burn at the same max safe thrust, and so burn near constant fuel during the entire trip. So the latter half of the trip will see increasingly larger accel, due to decreasing mass but constant thrust. Commented Jun 26, 2021 at 18:32
• @KoonW - I like the idea of your characters discovering the error because unexpectedly, they have some Xe left when they arrive at their destination. A lucky thing - they have use for it! Commented Jun 26, 2021 at 18:33
• This changing mass makes the calculation much more complex, bc they will not simply flip at halfway point... as the decel part will be shorter due to lower mass. Commented Jun 26, 2021 at 18:36
• its a pretty elementary error - if the ship planners didnt catch it, im sure the mission is doomed haha! though much of the game is about the simulation of the slim chances of success, with many 'alternate endings' of various failures being as interesting as successes Commented Jun 26, 2021 at 18:48
• If the rate of mass burn was constant, you would not turn around at the halfway distance.. More like 75-80% f the way there, as the acceleration of a constant mass usage by the same engine results in exponentially increasing acceleration as your total mass decreases. Commented Jun 26, 2021 at 18:51