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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.

https://www.space.com/38444-mars-thruster-design-breaks-records.html https://www.popularmechanics.com/space/moon-mars/news/a28754/new-ion-thruster-breaks-records-power-thrust/ https://www.space.com/28732-nasa-dawn-spacecraft-ion-propulsion.html https://www.nasa.gov/centers/glenn/technology/Ion_Propulsion1.html https://www.nasa.gov/multimedia/imagegallery/image_feature_2416.html How fast will 1g get you there? http://www.projectrho.com/public_html/rocket/slowerlight2.php http://www.xenology.info/Xeno/17.3.htm Conventional Interstellar Propulsion Systems https://forum.nasaspaceflight.com/index.php?topic=34036.1060 https://www.omnicalculator.com/physics "The Martian" Hermes ship design https://the-martian.fandom.com/wiki/Hermes_Spacecraft https://www.nasa.gov/directorates/spacetech/niac/index.html

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What you are looking for is the rocket equation

$$\Delta v = v_e \cdot \ln\left(\frac{m_{start}}{m_{end}}\right)$$

To get to alpha centauri in 110 years, your average velocity must be $$\frac{ 4.35\cdot c}{110}$$

Which means you need to accelerate up to at least that speed, and down again, so:

$$\Delta v > 2 \cdot \frac{ 4.35\cdot c}{110}$$

Which combined with the rocket equation is:

$$v_e \cdot \ln\left(\frac{m_{start}}{m_{end}}\right) > 2.37 \cdot 10^7 m/s$$

With some considerably more generous mass ratio than the one you give in your question, say 99.9% propellant, we have:

$$v_e \cdot \ln\left(\frac{1000}{1}\right) > 2.37 \cdot 10^7 m/s$$

Or: $$v_e > 3.43 \cdot 10^6 m/s$$

Which is far, far more than any of the engine technologies you are looking at.

Conclusion: your setup doesn't work.


(And here's an attack strategy for an analytic solution in case you want to go ahead anyway)

Assumptions:

  • Halfway flip*
  • Relativistic effects ignored (Lorentz factor of ~1.0013, so that should be okay).
  • Constant thrust.

First, we have to have to clear up that asterisk. In your simplified constant acceleration model, the halfway point is self evident. Half the distance, half the time. But when we want to take into account that propellant is consumed, this is no longer nice and symmetric. In the beginning, the ship is heavy from all the propellant and accelerates slowly. But when almost out of fuel at the end, the acceleration is high.

So the "halfway" point would be our top speed. This point will be closer to the end of the journey.

From that, we can derive that the top speed (half the delta-v, the other half needed to slow down), happens when the mass is:

$$m_{flip} = m_{end} \sqrt{\frac{m_{start}}{m_{end}}}$$

We also know that all propellant should be burned off after a certain time, $t$. In your case 110 years.

$$m_{start} - m_{end} = \frac{F\cdot t}{v_e}$$

With some calculus, we can express the distance travelled after accelerating with constant thrust $F$, with some start mass $m_0$, until some end mass $m_1$, with some exhaust velocity $v_e$:

$$d_{leg}(m_0,m_1) = \left(\left(1 - \frac{m_{0} - m_{1}}{m_{0}}\right) \cdot \ln\left(\frac{m_{1}}{m_{0}}\right) + \frac{m_{0} - m_{1}}{m_{0}}\right) \cdot \frac{v_e(m_0 - m_1)}{F}$$

In your case, however, you have two distances. One up to the flip, and one after. For the first leg, $m_0 = m_{start}$ and $m_1 = m_{flip}$. For the second leg, $m_0 = m_{flip}$ and $m_1 = m_{end}$. You have:

$$d_{leg}(m_{start},m_{flip}) + d_{leg}(m_{flip},m_{end}) = 4.37ly$$

$$t = 110 years$$

This system has only two degrees of freedom, so if you fix two of $F$, $m_{start}$, $m_{end}$ or $v_e$, the system of equations admit an unique solution.

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  • $\begingroup$ Thank you for all of this! I am curious about the 99% propellant mass. I have heard of this for chemical rockets launching from Earth surface, but what about craft only moving between planetary upper orbits? And also what about ion drive crafts? Dawn had Launch mass of 1,217.7 kg and Dry mass 747.1 kg. which is only ~61% fuel? $\endgroup$
    – Koon W
    Jun 27 at 18:49

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