How much methane would have to be expended (think Raptor engine) to propel Titan into Saturn? It's clearly a humongous amount.

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    $\begingroup$ Why do you hate Titan? ;) $\endgroup$ Commented May 17, 2017 at 20:23
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    $\begingroup$ I think this is off topic here, isn't it? It feels like a WorldBuilding question... $\endgroup$
    – Rory Alsop
    Commented May 18, 2017 at 6:29
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    $\begingroup$ @RoryAlsop I looked--hypothetical questions are allowed here, unlike Astronomy. $\endgroup$
    – called2voyage
    Commented May 18, 2017 at 13:05
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    $\begingroup$ Absolutely - it just feels like it isn't about space exploration :-) $\endgroup$
    – Rory Alsop
    Commented May 18, 2017 at 13:06
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    $\begingroup$ How much O2 would you need to burn that methane? $\endgroup$
    – ShadoCat
    Commented May 18, 2017 at 19:42

2 Answers 2


Let's turn the question on its head and see what exhaust velocity we need to if Titan's entire (mostly nitrogen) atmosphere were used as a propellant.

$\Delta v = v_e log(m_i / m_f)$

Wikipedia tells us that the atmosphere of Titan is about 1.19 times as massive as that of Earth so we get about 6.13e18 kg of atmosphere (propellant) in a total mass of about 1.35e23 kg.

Assume that we apply all of our delta-v at apoapsis (1 257 060 km)—either instantly or over multiple orbits—to lower periapsis from 1 186 680 km to the equatorial radius of Saturn (60 268 km). The semi-major axis changes from 1 221 870 km to 658 664 km.

Specific energy $\epsilon = v^2/2 - \mu/r = - {\mu \over 2a} = -{3.793e16 m^3/s^2 \over 2a}$

Thus $ \epsilon_{initial} = -1.55e7 J/kg $ and $ \epsilon_{final} = -2.88e7 J/kg $ which we plug back into the specific energy equations to get an initial speed of 5414 m/s and a final speed of 1662 m/s, $ \Delta v = 3752 m/s$.

$ v_e = {\Delta v \over log(m_i/m_f) } = {3752 m/s \over 4.56e-5 } = 8.24e7 m/s = 0.27c$

(This is a small enough fraction of light speed that we aren't too concerned about relativity.)

The Raptor or indeed any chemical rocket can not achieve this task by any stretch, as Hobbes has already pointed out. Ion engines don't cut it either.

Now, methane makes up about 1.4% of Titan's atmosphere. If all the methane were antimatter and annihilated with some of the nitrogen in an antimatter rocket, we can calculate our exhaust velocity with (p. 54 of The Starflight Handbook: A Pioneer's Guide to Interstellar Travel by Mallove & Matloff, 1989)

$ v_e = \sqrt{\epsilon (2-\epsilon )} c = \sqrt{0.028 \times 1.972} c = 0.23c$

Even if it were antimatter, the methane is not enough.

(Numbers are from the Wikipedia pages on the respective bodies and the gravitational parameter.)


Let's do a Fermi estimate:

Rockets bring about 2-5% of their start mass to orbital velocity. To cancel out Titan's orbital velocity, you're looking at two orders of magnitude more fuel and oxidizer than Titan's mass.

Earth's atmosphere weighs $10^{18}$ kg, or 1/200,000 of Earth's total mass. Titan's is 1.5 times as dense, so if Titan's atmosphere were 100% methane it still wouldn't make a significant difference.

  • $\begingroup$ Upvoted, but if you're utilizing Titan's own methane as propellant, wouldn't Titan's reduction of mass factor in somehow? $\endgroup$
    – called2voyage
    Commented May 17, 2017 at 18:36
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    $\begingroup$ Do we have to cancel out the orbital velocity completely or we just need to dip the perikrone below the surface of Saturn? $\endgroup$
    – Mys_721tx
    Commented May 17, 2017 at 18:55
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    $\begingroup$ Or even just lower the perigee enough to let atmospheric drag do the rest of the job? $\endgroup$ Commented May 17, 2017 at 20:24
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    $\begingroup$ @Mys_721tx Titan is in a fairly high orbit so there wouldn't be a huge difference. A bigger difference is Titan's orbital velocity is only 5.6km/s, compared with 7.8km/s for LEO - that 2.2km/s makes a big difference in the rocket equation, the payload fraction would be about 15%. Doesn't change the fermi estimate though. You'd still have to use 90% of Titan's mass as propellant to get the last 10% into Saturn. $\endgroup$ Commented May 18, 2017 at 7:43
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    $\begingroup$ @BlakeWalsh: High orbit means lower delta-V to deorbit; you can drop a Kuiper belt object into the Sun with ~1km/s of delta-V while deorbiting something from Mercury altitude would be absolutely prohibitive. Getting Titan to litobrake against Saturn ring asteroids would bring the requirements down evern farther, Still, both atmosphere and methane oceans account for too small a fraction of mass of Titan for stuff like Raptor engine. If you were to accelerate that methane to near speed of light, increasing the exhaust mass through relativistic means, that could work... $\endgroup$
    – SF.
    Commented May 19, 2017 at 10:54

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