I encountered sci-fi stories about planets of the twice the mass of Earth, and the characters were able to leave them in spaceships.
I wonder, whether it is possible at all?
$\begingroup$
$\endgroup$
8
-
6$\begingroup$ Your title asks something different than your body. Which of the two you are asking? Hint: since you have got an answer already, you have not many choices. $\endgroup$– L.Dutch ♦Jun 7, 2020 at 20:19
-
1$\begingroup$ The radius and mass of the planet will give the escape velocity, and the tsiolkovsky rocket equation combined with the effective exhaust velocity v_e of a chemical rocket will tell you the ratio of (initial total mass including fuel)/(final payload mass after fuel is spent) needed to achieve that escape velocity. But note that to say how much fuel this actually requires, you need to know the payload mass (the mass of the module that provides life support for the characters along with their own mass). $\endgroup$– HypnosiflJun 7, 2020 at 23:23
-
1$\begingroup$ @Anixx Perhaps you should look up a paper or online version of Habitable Planets for Man by Stephen H. Dole (1964,2007) to find the largest planets that Dole thought that humans would want to spend long periods of time on. Possibly the characters in your story use antigravity to be comfortable on a high g world and their ship has an antigravity drive. I note there are no solid planets with 2 earth masses in our solar system, so if the spaceship in the story was a rocket it had to be at the very least an atomic rocket and not a chemical rocket. $\endgroup$– M. A. GoldingJun 8, 2020 at 17:48
-
2$\begingroup$ See also this discussion. There's a big difference between the often-quoted rocket equation and a (still ludicrous but not out-and-out impossible) approach using a slower build-up of horizontal velocity. $\endgroup$– Adam ChalcraftJun 8, 2020 at 21:34
-
2$\begingroup$ Uh, guys, are we assuming rockets must go straight up? How about launching at equator using basically a horizontal accelerator that arcs up into the sky later down the track? How much would the planetary spin contribute? $\endgroup$– Kasey ChangJun 9, 2020 at 2:39
|
Show 3 more comments
5 Answers
$\begingroup$
$\endgroup$
17
On Space Exploration Stack Exchange, Russell Borogove addressed a very similar question and made some reasonable estimates about the required mass of a Saturn V-like rocket (arguing it's impractical to consider all possible rocket designs, given the variation in performance) to escape a planet of various surface gravities (see also Hippke 2018). Above $10g$, the rocket is effectively the same mass as the planet. At this point, we would need literally quadrillions of Saturn Vs. Even escaping a planet with a surface gravity of $6g$ would require millions of them. On the other hand, that assembly would weigh 38 billion tons, which is tiny compared to Earth's mass. $10g$ is a somewhat generous limit, but beyond that, we do know it's fairly absurd. This corresponds to an escape velocity of $35.4\text{ km s}^{-1}$.
From an astronomical point of view, is it even possible to have a terrestrial planet with that sort of escape velocity? Looking at the mass-radius curves by Seager et al. 2007, it seems like a super-Earth composed predominantly of silicates and iron could achieve that surface gravity with a mass of $M\sim30M_{\oplus}$ and $R\sim2.5R_{\oplus}$. From a compositional perspective, this is certainly reasonable, according to their numerical work.
What if we consider non-silicate compositions? Perhaps we could make the planet less dense, ensuring that the same mass yields a higher radius and therefore raises the mass limit. Say we have a planet made mostly of water. To achieve that same velocity, it appears that we may be able to reach $\sim40M_{\oplus}$.
On the other hand, super-Earths larger than $2R_{\oplus}$ aren't expected to exist. At masses above this (corresponding to $20M_{\oplus}$ for a mainly silicate planet, and larger for an iron planet), the body would accrete a thick atmosphere, beginning to bridge the gap between terrestrial and gas planets (see Lopez & Fortney 2013 for an even more conservative limit!). (Coincidentally, this ends up yielding, actually, a similar escape velocity for the silicate case.)
In the particular scenario you describe, a $2M_{\oplus}$ planet would have a radius of perhaps $1.4R_{\oplus}$, according to Seager et al.'s models, and thus a surface gravity (and escape velocity) barely greater than Earth's. This is obviously quite easy to escape with a chemical rocket.
answered Jun 7, 2020 at 20:22
-
4$\begingroup$ Great answer and nice papers linked to. $\endgroup$ Jun 7, 2020 at 20:48
-
3$\begingroup$ @Anixx The mass of the rocket definitely depends on surface gravity. A higher surface gravity requires more fuel, which in turn requires a much more massive rocket. These calculations takes the fuel into account. $\endgroup$– HDE 226868 ♦Jun 7, 2020 at 22:34
-
9$\begingroup$ @Anixx We can express the escape velocity as $v\propto \sqrt{2gR}$. Therefore, yes, $v$ depends on $R$ as well as $g$. But note that we cannot simply choose any radius we wish for our planet - for a given mass we only have one possible radius according to hydrostatic models. Therefore, we see that surface gravity and escape velocity are indeed intertwined. $\endgroup$– HDE 226868 ♦Jun 7, 2020 at 22:41
-
6$\begingroup$ @HDE226868 "This is obviously quite easy to escape with a chemical rocket" I woudn't say easy since as it's noted in your own links, chemical rockets barely work on this Earth. You need a massive rocket to lift a tiny fraction of its mass to LEO. $\endgroup$– RekesoftJun 8, 2020 at 7:46
-
3$\begingroup$ Such a large planet would likely have a thick atmosphere. If that atmosphere contains appropriate oxidiser (like oxygen) or fuel (say, methane) in sufficient quantities, you may be able to use some form of jet engine to get up to high speed in atmosphere before leaving to orbit, which would reduce the amount of fuel needed. Even in an inert atmosphere, using a propeller design would let you use the planet's own atmosphere as reaction mass until you manage to leave the atmosphere. So I imagine with appropriate designs it'd still be quite possible to leave such a large planet with an atmosphere. $\endgroup$– HearthJun 8, 2020 at 12:26
$\begingroup$
$\endgroup$
11
Rockoon!
https://en.wikipedia.org/wiki/Rockoon
Gravity drags hard on rockets. But a balloon cares nothing for gravity. The lift conferred by a hydrogen balloon will depend on the composition of your atmosphere, but will be the same regardless of the gravity of your planet.
Using balloons to get rockets above the atmosphere and (partly) out of the gravity well is a real thing.
https://www.universetoday.com/tag/rockoon/
Is there a better way to get to space? Current traditional methods using expendable rockets launching from the surface of the Earth are terribly inefficient. About 90% of the bulk and mass of what you see on the launch pad is expended in the first few minutes of the mission, just getting the tiny payload above the murk of Earth’s atmosphere and out of the planet’s gravity well.
Recently, on May 20th, 2016, Zero2infinity lofted Aistech’s first satellite into the upper atmosphere, aboard its Sub-Orbital Platform in Near Space balloon system. Zero2infinity uses these Near Space balloons to carry client payloads up above 99% of the Earth’s atmosphere. This is a cheap and effective way to get payloads into a very space-like environment.
I here assert that one could use this method with a high gravity planet and avoid the need for a prohibitively large and powerful rocket. But you cannot leave the planet in a balloon - you still need the rocket to get away, and so this meets the OP's request for "a chemical rocket to leave it".
-
13$\begingroup$ I'm afraid a balloon doesn't help much to deal with gravity. The altitude record for a balloon on Earth is 53km. Most high altitude balloons are more around 35km. Gravity is about 10% weaker at 250km, so at balloon heights we're taking a difference of only a couple percent. Furthermore, most of the energy required to achieve orbit is needed to attain the requisite velocity, not altitude, so a balloon isn't very helpful there either. $\endgroup$– GeneJun 8, 2020 at 8:15
-
7$\begingroup$ Space isn't up, space is going sideways very far. A balloon lifting up a payload is of no great use if you want to go into orbit. You need ~7.8km/s to stay in LEO, a balloon offers 0. Since you need about 9.4km/s with a traditional rocket launch, you can imagine that lifting a rocket able to produce a delta-v of 7.8km/s with a balloon is infeasible. $\endgroup$ Jun 8, 2020 at 9:33
-
2$\begingroup$ yah yah orbit. But OP did not mention orbit. OP wants to leave planet! $\endgroup$– WillkJun 8, 2020 at 10:55
-
4$\begingroup$ The trouble is there is a "no-mans land" where the atmosphere is too thin for aviation but too thick for orbit. This effectively precludes techniques like strapping an ion engine to a balloon and using it to slowly transition from floating to orbit. $\endgroup$ Jun 8, 2020 at 13:57
-
5$\begingroup$ One thing that rockoons are good for: avoiding drag losses. As has been noted, achieving orbit is more about going faster than higher, and going fast in thick atmosphere means lots of drag, which wastes delta-V. A more massive planet would have a thicker atmosphere, and so this problem would be more severe than on Earth. A rockoon would allow the thick atmosphere to work in your favour by providing bouyancy to your balloon, and let you lift the rocket to altitudes where the air is thinner and more conducive to reaching high speeds with lower drag. $\endgroup$ Jun 9, 2020 at 14:12
$\begingroup$
$\endgroup$
5
The escape-speed of a planet is given as
$v_2=\sqrt{{2GM}\over{r}}$
r = radius of the planet
G = gravitational constant
M = mass of the planet
As long as the spaceship is able to reach this speed, it can leave the planet. Twice the mass of earth and the same radius would result in $v_2$ = 15.7 km/s, that's absolutely possible.
answered Jun 7, 2020 at 20:15
-
1$\begingroup$ please make yourself familiar with MathJax for typing formulas. $\endgroup$– L.Dutch ♦Jun 7, 2020 at 20:18
-
1
-
2$\begingroup$ 22.4 km/s is not possible with a chemical rocket $\endgroup$– SlartyJun 7, 2020 at 22:31
-
1$\begingroup$ They actually can, it's just very far away from beeing efficient. Here it is described in detail: quora.com/… $\endgroup$ Jun 8, 2020 at 3:56
-
2$\begingroup$ An interesting article, but it suggests 17.7 km/s which is less than 22.4 km/s. And even then assumes the use of vacuum level thrust at take off (5295m/s should read 4498 at sea level). It also assumes that 33g acceleration on a minimalist tank thickness and a thrust to weight ratio which is impractical, the rocket would be uncontrollable and fall over at launch. And and... $\endgroup$– SlartyJun 8, 2020 at 8:55
$\begingroup$
$\endgroup$
3
I suggest that a deltaV of 13.5km/s is about the most that might be expected for a chemical rocket.
This is calculated from the rocket equation and assumes a high (hydrogen-oxygen based) exhaust velocity and a high mass ratio (20:1) giving a deltaV of 4500*In(20) = 13.5km/sec but is probably optimistic. Although using hydrogen provides a good exhaust velocity, it is very low density and so needs a very big tank which in turn makes a mass ration of 20:1 very difficult.
But taking this as a best case, it is possible to put this number into the escape velocity equation holding either the radius constant or the mass constant to see what we get.
Keeping the radius constant whilst increasing the mass: A mass of 1.4 x Earth mass would produce an escape velocity of 13.2km/s
Keeping the mass constant whilst decreasing the radius: A radius 0.7 x earths would have an escape velocity of 13.4km/s
Alternatively the mass can be increased whilst the radius is also increased. Case in point as suggested by HDE where mass = 2 x Earths mass and radius = 1.4 x Earths mass: In this case the escape velocity would be 13.4km/s
In each of these cases it may be just barely possible to escape from the planet using chemical rockets alone. But I would stress only with the greatest of difficulty.
It should also be noted that it might be possible to escape from a planet using an orbital rather than an escape velocity. This might be achieved by using gravitational assists from moons and planets.
Alternatively this might be achieved by retanking the rocket in orbit. Refilling the propellant tanks in orbit provides a practical method for effectively increasing the mass ratio.
Minimum DeltaV for low Earth orbit is 7.8km/s (9.4km/s allowing for air resistance and gravitational losses), whereas Earth escape velocity is 11.2 km/s so there is a bit of wriggle room here.
-
7$\begingroup$ 13.5 km/s might be the limit for a single-stage chemical rocket, but it's far from the limit for chemical rockets in general. You need a minimum of 15.7 km/s for a Jupiter gravity assist, or 17.6 km/s for a lunar landing and return, both of which we've done, so it's clearly possible to pack more delta-V into a rocket. $\endgroup$– MarkJun 8, 2020 at 22:13
-
$\begingroup$ What about increasing the mass while keeping the density constant? $\endgroup$– AnixxJun 9, 2020 at 8:56
-
$\begingroup$ Increasing the mass with constant density equates to increasing the mass and radius . $\endgroup$– SlartyJun 9, 2020 at 9:24
$\begingroup$
$\endgroup$
5
There is a limit to the size of a planet, above which it is a star. Jupiter is nudging that. Jupiter doesn't exactly have a surface, but at the cloudtops it has a little over 2.5g, which is I think escapable with a suitable chemical rocket.
The problems of assembling a suitable rocket and suitable chemistry are left as exercises ;)
Hal Clement wrote about a solid and dense planet, Mesklin, with a high surface gravity, but with a very high rate of rotation. Launching from near Earth's equator reduces the delta-V for escape by a 25th - for orbit by an 18th. Spin your planet fast enough, and you should be able to jump off the equator. It may be a bit windy.
-
$\begingroup$ According to this article a planet would have to be at least 11 times the mass of Jupiter before it would start undergoing fusion and become a brown dwarf. discovermagazine.com/the-sciences/… $\endgroup$– naschJun 9, 2020 at 23:40
-
1$\begingroup$ @nasch Yes, but such proto-stars would not have a diameter much larger than Jupiter. The gravity compresses the planet to such an extent that a heavier planet stays around the same size as Jupiter. Only when fusion starts does the fusion energy blow a star up again to star-like sizes. $\endgroup$– JanKanisJun 10, 2020 at 10:45
-
$\begingroup$ At 10 times the mass and about the same radius, it would have a far higher "surface" gravity, yes? It sounded like you were saying the escape velocity of a gas giant couldn't be much higher than Jupiter's. $\endgroup$– naschJun 10, 2020 at 14:19
-
-
$\begingroup$ Not so much that the escape velocity of the brown dwarf wouldn't be high, as that apart from wondering why you were on the surface of the proto/near/failed star (was it your fault, are you an exponential growth of monoliths?) you would have worse troubles than gravity, and be unlikely to be escaping. $\endgroup$ Jun 25, 2020 at 17:37