# When a planet has a high gravity, is it impossible to build and launch a successful chemical rocket to space?

Just recently a a large rocky planet has been discovered. "Astronomers have discovered a new type of rocky planet beyond the solar system that weighs more than 17 times as much as Earth while being just over twice the size"

http://www.reuters.com/article/2014/06/02/us-astronomy-exoplanet-idUSKBN0ED29V20140602

Then I was thinking, if aliens lived on this planet, could they get off the planet with a normal chemical rocket, or would the rocket have to be so massive that it could not be built? or so heavy that the amount of fuel required would push the weight too far so that it could never reach space?

I wonder is there are sad planets where aliens never made it into space due to the large amount of gravity.

It never becomes impossible per se, but at some point there could be so much gravity that construction of a working rocket would be beyond our current ability to engineer something that could work. That is, it might take impracticably huge quantities of fuel, or require materials stronger than we can construct.

There are just a couple of amazingly simple equations that govern this (under some reasonably idealized conditions). First, understand that to break free of the gravity of any planet, a craft much reach escape velocity ($v_e$). For a spherical body, this is given by

$$v_e = \sqrt{2GM\over r}$$

Where:

• $G$ is the gravitational constant
• $M$ is the mass of the body
• $r$ is the distance from the center of gravity (assuming we start from the planet's surface, this is just the radius of the planet).

Now if we want to leave this planet, we need to go from not moving at all to moving at least this fast. Otherwise, we either fall back to the planet and crash, or (if we are lucky) get stuck in an orbit around the planet. This change in velocity is called "delta v" or $\Delta v$.

A craft's available $\Delta v$ is given by the Tsiolkovsky rocket equation:

$$\Delta v = v_x \ln \frac{m_0}{m_1}$$

Where:

• $v_x$ is the effective exhaust velocity (essentially, the fuel efficiency of the rocket)
• $m_0$ is the initial total mass of the craft with fuel
• $m_1$ is the final mass of the craft with no fuel

So, we must engineer our craft to have at least enough $\Delta v$ to reach escape velocity (plus enough to do something interesting afterwards, like land on the destination planet, or overcome any atmospheric drag...), but at a minimum we require:

\begin{align} \Delta V &> v_e \\ v_x \ln \frac{m_0}{m_1} &> \sqrt{2GM\over r} \end{align}

This means:

• launching a smaller payload reduces fuel required
• more efficient engines help a lot

In any case, there's no way to increase the mass $M$ of the planet such that this equation can not be solved by carrying more fuel, making more efficient engines, making a lighter payload, etc. You just might end up with an absurd solution.

• You mention "orbit" but you've used the escape velocity. I wouldn't tell you to do it one way or the other, but I feel like your answer needs some editing for consistency. – Alan Rominger Jul 11 '14 at 17:36
• @AlanSE err..yeah. I'll edit... – Phil Frost Jul 11 '14 at 17:39
• I haven't really tried to work it out, but does this take into account that the mass must come from within the planet itself? – user541686 Aug 6 '14 at 5:06
• @Mehrdad I guess not....but if you get to a point where the rocket contains a significant fraction of the mass of the planet, you will run into the question of who's orbiting who, and how you can "leave" a planet if you have destroyed it. In any case, if you required half the planet's mass as fuel, this falls into the category of "absurd" solutions that are not feasible, but which could in theory be overcome through better technology, making more efficient use of fuel, thus requiring less of it. – Phil Frost Aug 6 '14 at 11:01
• True, good point! – user541686 Aug 6 '14 at 11:07

It would be possible, though much more complicated than on earth.

Chemical rockets on earth deliver ~3% of it's launch weight to low earth orbit. On this superheavy planet some 0.1-0.01% of launch weight might be delivered to low orbit.

While human exploration would be extremely hard in such conditions, it would be possible to explore the universe with tiny automated probes.

Such a dramatic change would indeed make it harder to reach space, in a precisely defined sense. The easiest measure of how 'hard' it is to get so space is the escape velocity, which is determined by the planet's mass $M$ and radius $R$ as $$v_\text{esc}=\sqrt{\frac{2GM}R}.$$ For a planet like the one you mention, the escape velocity goes up by a factor of $\sqrt{17/2}\approx3$ with respect to the Earth's escape velocity $v_\text{E}$.

The reason this matters is that the total mass $M$ of a rocket that will send a payload of mass $m$ into space depends exponentially on the escape velocity. (More accurately, on the change in velocity required, $\Delta v$, which is of the order of $v_\text{esc}$.) This relation is known as the Tsiolkovsky rocket equation and it is one of the fundamental principles of rocket science; it states that $$M=m\exp\left(\frac{v_\text{esc}}{v_\text{exh}}\right),$$ where $v_\text{exh}$ is the exhaust velocity.

Because of this exponential dependence, if $v_\text{esc}$ goes up from $v_\text{E}$ by a factor of three, the rocket mass will increase by a factor of $$\left(e^{v_\text{E}/v_\text{exh}}\right)^{3},$$ which can be a lot more than three. To put some numbers in, $v_\text{E}\approx11.2 \,\text{km}/\text s$ whereas liquid propellants can get up to about $v_\text{exh}\approx 5 \,\text{km}/\text s$. When exponentials are involved the details do matter, but if you put this in you get an overall factor of the order of $$(e^2)^3\approx 400.$$

This would mean, for example, that a behemoth like the 3,000-ton Saturn V would be able to transport about 100kg of payload - about the size of a 'minisatellite', instead of the 45 tons of a fully-fledged Apollo mission.

Now, there are a number of ways to get around this restriction, of which many are technological but some are physical. The most obvious one, to me, is that the size of the atmosphere will also change. This obviously depends on the amount and composition of the exoplanet's atmosphere, but with all other things equal, a more massive planet will compress its atmosphere into a thinner layer. This change on length scale is linear: as a first approximation, it is inversely proportional to the surface gravitational acceleration, $$g_0=\frac{GM}{R^2}$$ which is about 4 times that of Earth for the exoplanet in question.

Thus, if all other things were equal - if the atmospheric composition and surface pressure were the same as Earth's - then you'd only have to go up, say, 40 km for low exoplanet orbit, instead of the ~160 km of low Earth orbit. This is important, because it radically reduces the $\Delta v$ required to get into orbit, and this goes again into the exponential dependence of the Tsiolkovsky equation. The $\Delta v$ to get to a height $h$ is roughly $$\Delta v=\sqrt{\frac{2GM}R-\frac{2GM}{R+h}}=\sqrt{\frac{2GMh}{R+h}},$$ and this has now gone down, slightly, for Kepler-10c. (You still need to accelerate to stay in orbit, but that depends on the planet's rate of rotation which is yet another completely unknown variable.)

To summarize, then, it will indeed be harder to go to space from such a planet, but under certain circumstances it may be easier to get to orbit. The problem with all this, though, is that details - about the specifics of the planet and its atmosphere, and also about what you want to do - do matter, because of the exponential dependence, which is hard to understand until you run into a good number of walls like this one. As Phil Frost mentions, xkcd what-if is a good place to read about this, but in general, it pays to sit up and pay attention when a variable of interest is on the exponent.

• I think worth adding that the thinner atmospheric limit doesn't "get around" the escape velocity, but just reduces the additional energy one must expend to overcome drag. That is, Earth's escape velocity at the surface is about 11.2 km/s atmosphere or no -- and then the atmosphere means the drag your craft will experience on the way up requires additional fuel to overcome. – Phil Frost Jul 11 '14 at 17:52
• Yes, that is another mechanism which helps. What I had in mind, though, is that if you only want to orbit (instead of escaping), then a thinner atmosphere also opens up lower altitudes which are easier to get to and which are now safe from orbital drag. – Emilio Pisanty Jul 11 '14 at 18:07
• Ah, I see. I was misled by "Now, there are a number of ways to get around this restriction", where I assumed "this" to mean "escape velocity". – Phil Frost Jul 11 '14 at 18:08
• One wonders how much rocket stages can help cc/ @PhilFrost (+1 anyway, so it's not like you'll get more points from me if you do more calculations ;) – user10851 Jul 11 '14 at 18:12
• @ChrisWhite We use rocket stages on Earth anyway, so it's a question of when rather than whether you start needing them. The key point remains - you start needing them "exponentially sooner in $\sqrt{GM/R}$", which is the same as the considerations in my and Phil's answer. – Emilio Pisanty Jul 11 '14 at 18:29