15
$\begingroup$

Many authors describe worlds with higher gravity than our own, from whence super-muscular aliens originate. However, as we increased the size, mass and surface gravity of a world; there would be points at which no chemical-fueled rocket, either practical or theoretical, could achieve escape velocity or a practical orbit, thus necessitating externally-powered launchers or atomic power.

Taking things further, at what point of increased planetary size, mass, and gravity would a practical atomic rocket no longer be able to achieve escape velocity or orbit?

Since this is hard-science, please limit answers to those involving rocketry that is either currently implemented or scientifically feasible.

$\endgroup$

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • 1
    $\begingroup$ My guess is that it depends heavily on your assumption on how the atmosphere scales with planet size. $\endgroup$ – 2012rcampion Apr 18 '16 at 2:51
16
$\begingroup$

Introduction

If you are seriously interested in this question and are willing to devote some time to reading about it, I recommend reading the Atomic Rockets: Engine List page.

It will also discuss the problems you will face as a rocketeer. A partial list is this:

  • For planetary launches, your engine must provide a higher acceleration than the local gravity (e.g. for Earth, it must exceed 1 g). This means you need a high thrust engine.
  • To reduce propellant usage, your engine must provide a high specific impulse ($I_{sp}$).
  • For most engines these two things are mutually exclusive.

Tyranny of the Rocket Equation

The rocket equation shows that a rocket's total propulsive capability is driven by surprisingly few factors.

$$\Delta v = v_e \cdot \ln{\frac{m_0}{m_f}}$$

  • $\Delta v$ - propulsive capability of rocket
  • $v_e$ - exhaust velocity of propellant
  • $m_0$ - starting mass of rocket
  • $m_f$ - final mass of rocket (starting mass minus propellant used)

Stellar Blackhole

Obviously the theoretical limit for anything is the formation of an event horizon (aka Blackhole). This is because the $\Delta v$ requirement exceeds the speed of light and no propellant can exceed that.

You can achieve this formation through many mechanisms. Take a small mass and compress it or keep adding mass to a single object.

A stellar Blackhole forms when several solar masses of matter are brought together under normal conditions (we don't yet know how much mass is required). No amount of fancy rocketry will get you out of a blackhole

3 Stage Chemical Rocket - works up to $M_{Saturn}$

Unlike many types of rocket engines, chemical rockets "burn" their fuel and expel the reaction products as its propellant. This limits your rocket engine to exothermic (energy releasing) reactions.

To keep your $I_{sp}$ as high as possible you must use chemicals that are as low mass as possible. $I_{sp}$ is driven by exhaust velocity, not momentum (increasing propellant velocity decreases propellant usage). So an engine that delivers the same thrust will use less fuel if you expel a low mass at high velocity rather than a high mass at low velocity.

LOX + LH2

The commonly used high performance chemical rocket fuel is liquid oxygen (aka LOX) + liquid hydrogen (LH2). This provides an $I_{sp}$ of about 450 (exhaust velocity of $4,400 \frac{m}{s})$.

LF2 + LH2

An even higher performance fuel would be liquid hydrogen + liquid fluorine. This combination can provide an $I_{sp}$ of about 480 (exhaust velocity of $4,700 \frac{m}{s}$). However, it introduces a host of huge problems:

  • Handling the liquid fluorine before launch is tough.
  • Keeping hot gaseous hydrofluoric acid from destroying your launch complex and poisoning the people on the ground is much tougher.

Calculations

For argument's sake if we limit the equation to $\frac{m_0}{m_f} = 10$ (shuttle has a fraction of $\approx 5$ - meaning it is 80% fuel and 20% everything else)

Plugging the provided numbers into the equation would yield the following:

$$\Delta v = 4,700 \frac{m}{s} \cdot \ln{10} = 10,822 \frac{m}{s}$$

Subtract a reasonable atmospheric + gravity drag value ($\approx 20$ % $= 2164 \frac{m}{s}$). This leaves $8,658 \frac{m}{s}$ available for getting to orbit.

Orbital velocity is calculated using this approximation:

$$v_o = \sqrt{\frac{G \cdot M_{planet}}{r}}$$

Now solve for r (and $M_{planet}$):

$$8,658 = \sqrt{\frac{6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^3}{r}} \rightarrow 7.5 \cdot 10^7 = 6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^2$$

$$r^2 = \frac{7.5 \cdot 10^7}{6.7 \cdot 10^{-11} \cdot 23,039} \rightarrow r = \sqrt{\frac{7.5 \cdot 10^7}{1.5 \cdot 10^{-6}}}$$

  • $r = 6,971 km$
  • $M_{planet} = 7.8 \cdot 10^{24} kg$

Different density planets give different results.

Essentially, the Earth is the limit for single stage chemical rockets.

Staging

But wait a second! Clearly we launch vehicles into space that aren't single stage, so what gives?!

So far we've only discussed doing this as a single stage to orbit. It turns out that by staging a vehicle we actually get better performance and are able to more easily achieve orbit.

How much we actually gain depends upon the number and type of stages. But let's assume we use a 3 stage rocket with each stage has the performance given above. The staging equation is given by:

$$\Delta v = N_{stages} \cdot v_e \cdot \ln{10} \rightarrow \Delta v = 3 \cdot 4,700 \frac{m}{s} \cdot 2.3 = 32,466 \frac{m}{s}$$

All the rest of the numbers stay the same, so solve for r (and $M_{planet}$) again:

$$32,466 = \sqrt{\frac{6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^3}{r}} \rightarrow 1.05 \cdot 10^9 = 6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^2$$

$$r^2 = \frac{1.05 \cdot 10^9}{6.7 \cdot 10^{-11} \cdot 23,039} \rightarrow r = \sqrt{\frac{1.05 \cdot 10^9}{1.5 \cdot 10^{-6}}}$$

  • $r = 26,971 km$
  • $M_{planet} = 4.1 \cdot 10^{26} kg$

This is almost the mass of Saturn ($5.7 \cdot 10^{26} kg$).

Different density planets give different results.

3 Stage Nuclear Pulse Propulsion - works for all planet masses (up to $170 \cdot M_{Jupiter}$, which is actually a star)

The rocket equation doesn't distinguish between engine type. So you can use exactly the same equations.

According to Atomic Rockets: Engine List, you can expect the optimal performance of a nuclear pulse propulsion engine to be the $100,000 \frac{m}{s}$ design on that page.

If you use that configuration, a nuclear pulse propulsion single stage rocket might be able to launch from a planet 6 times Jupiter's mass (Jupiter's mass $= 1.9 \cdot 10^{27}$, this planet's mass would be $1.2 \cdot 10^{28}$).

A three stage version of this ship would be able to generate about 3 times this $\Delta v$. That would correspond to a planet with the mass of $3.24 \cdot 10^{29} kg$ - about 170 times the mass of Jupiter. However, since a body with a mass above 84 Jupiter masses is a star, we can safely say that a technological civilization could develop nuclear pulse propulsion rocketry to launch into space from any planet.

All planets used in this answer assume an Earth density planet.

$\endgroup$
  • $\begingroup$ What about escape velocity? Is it a different thing to reach an orbit than to escape the planet's gravity entirely? $\endgroup$ – Monty Wild Apr 18 '16 at 5:34
  • $\begingroup$ Escape velocity is 2x the orbital velocity. So $v_e = \sqrt{\frac{2 \cdot G \cdot M}{r}}$. What this means practically, is take my planet masses and divide by $\sqrt{2} \approx 1.44$. This means the final planet mass would be about $1.15 \cdot 10^{29}$. $\endgroup$ – Jim2B Apr 18 '16 at 5:49
  • $\begingroup$ You had the meaning of $m_f$ in the rocket equation wrong; I fixed that. Though I find your calculations just above "Essentially, the Earth is the limit for chemical rockets." confusing, particularly the one starting with 8,658. Can you double-check those? $\endgroup$ – a CVn Apr 18 '16 at 7:46
  • $\begingroup$ And given that in practice you would need a rocky planet to launch a rocket from a 3 stage rocket would be able to launch from any planet. See worldbuilding.stackexchange.com/questions/9948/… for maximum rocky planet size. $\endgroup$ – Selenog Apr 18 '16 at 10:11
  • 1
    $\begingroup$ But where do you get the claim that "staging increases any rocket's total attainable $\Delta v$"? It isn't stated in the wiki article you linked to, and my analysis above seems to show that according to the equation in that article (which as I said treats the mass of empty stages as negligible), combining the fuel in multiple stages into a single larger stage results in exactly the same $\Delta v$. If you disagree that this is true in the example I gave with two stages, each 10 times larger than what it's pushing, can you point to where you think I made an error in the equations? $\endgroup$ – Hypnosifl Apr 18 '16 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.