On a Super-Earth 1.5x the volume and mass of Earth, would our rocket technology allow us to reach orbit? [duplicate]

Question

To try and make parameters clear, can we say we are talking about 50% 'more Earth'? As in, Earth, but 1.5 times as big and heavy? And let us include the atmosphere. If there is 50% more of it by volume, then that would also make it thicker at sea level, right? And then it is also affected by higher gravity?

I considered using a known Super-Earth as a reference, but the reputation of many is in question, and none seem at all promising as possible homes of intelligent aliens. Yet, the perennial question is where is everybody? Perhaps we have an unusually easy time getting to orbit from our nice liquid-water rocky planet. Maybe that is a partial answer to that question. Just a thought.

Answer 1

I'm going to go with "1.5x the size" as referring to volume, so combined with 1.5 times the mass results in the same density as Earth.

The alternative of 1.5 times the radius or circumference combined with 1.5 times the mass would result in a Super-Earth with a sub-Earth density of 2.4 g/cc. That would be like an icy moon, not a rocky planet, and would result in a lower surface gravity than Earth (2/3 g), and the same orbital velocity as Earth, making it overall a little easier to get into orbit. That does not sound like the intention of the questioner.

So with 1.5 times the volume and mass of Earth, we would have about a 14.5% higher surface gravity, $1.5\over {1.5}^{2\over 3}$, and a 14.5% higher orbital velocity, $\sqrt{1.5\over {1.5}^{1\over 3}}$. There should be no problem at all with our current rocket technology to get into orbit from the planet, so long as its atmosphere isn't any thicker than ours. In fact, there should be no problem to stick with two stages to orbit with that modest increase in orbital velocity. The increase in surface gravity could be accommodated with a modest 14.5% increase in thrust to maintain the same thrust to weight ratio. Perhaps just somewhat larger engines or one or two more engines added to the first stage cluster.

However this Super-Earth isn't very super as Super-Earths go. Most of the discoveries are two to ten times Earth mass, as opposed to the 1.5 times Earth mass posed here.

As for an explanation for where everybody is, no, this isn't it. First off, our discoveries of predominantly Super-Earths is observational bias. It is easier to find larger planets than small ones. I'm sure that there are plenty of Earths and sub-Earths out there. Second, you can just add stages to your chemical rocket to get off of a larger planet. It will simply cost more, so you will need to be more determined. Third, you don't need to use chemical rockets. You could use electromagnetic rail launchers, nuclear fission high Isp engines, or thermonuclear bomb propulsion (see Orion). All within our current technological capability.

Answer 2

There's a very large gap between what's theoretically possible in rocketry and what's practical and what an entity has the resources and political will to accomplish.

In the staged version of the rocket equation, the possible achievable delta-V is proportional to the number of stages when mass ratio and engine efficiency is held constant. In short, you can get any delta-V you want by adding additional stages.

If you needed twice the delta-V to reach orbit (because of a larger planet and denser atmosphere, say), then instead of a 1.5 stage launcher for a one-man orbital mission (like the R-7/Vostok or Atlas-Mercury missions), a 3-stage launcher like a Saturn 1B (counting the Apollo service module as a third stage, here) might be required. (I'm handwaving a bit here; R-7 and Atlas aren't literally 1.5 stages for purposes of the rocket equation.)

Typically, each stage is 4-5 times as large as the one above it in the stack, so linear increases in delta-V requirements lead to exponential increases in overall rocket size. Complexity and cost probably increase faster than linearly with stage size, considering the assembly facilities, transport, etc. required to support them. Eventually, as you turn up the gravity, you might hit a point where a nation wouldn't be willing to take on a Saturn-Apollo sized development program to achieve the goal of a single person in orbit.

Answer 3

There is a great blog post on the NASA homepage about "the tyranny of the rocket equation". It is going into quite some detail, and it gives in one of the last paragraphs:

If our planet was 50% larger in diameter, we would not be able to venture into space, at least using rockets for transport.

However, you don't want to venture in space, but reach orbit, and your Earth has 1.5 times the mass and not the diameter, so there's still hope. Just in case that the link goes dead, I am listing some of the interesting details of the blog post, as well as some additional information. The rocket equation gives:

$$M_f = 1-\frac {m_1} {m_0}=1-e^{-\Delta v\ / v_\text{e}}$$

where $M_f$ is the fraction of propellant by total mass of the spacecraft, $\Delta V$ is the necessary escape velocity and $v_e$ is the exhaust velocity. This can be rewritten to

$$\Delta v=- \ln(1-M_f) v_e$$ The orbital velocity of a body (without adjusting for aerodynamic drag) is given by

$$\Delta v=\sqrt{\frac{G M}{R}} $$

Assuming that you're interested in an Earth with $M'=1.5M$, this gives $R'=(1.5)^{\frac{1}{3}} R$, so $\Delta v'= \sqrt{\frac{1.5}{1.5^{1/3}}} \Delta v=1.145 \Delta v$, just like Mark Adler wrote in his answer.

Now we need some maximal possible values for exhaust speed and the ratio propellant/mass, which are handily given in the NASA post:

Hydrogen-oxygen is the most energetic chemical reaction known for use in a human rated rocket. Chemistry is unable to give us any more. In the 1970’s, an experimental nuclear thermal rocket engine gave an energy equivalent of 8.3 km/s. This engine used a nuclear reactor as the source of energy and hydrogen as the propellant.

And:

The common soda can, a marvel of mass production, is 94% soda and 6% can by mass. Compare that to the external tank for the Space Shuttle at 96% propellant and thus, 4% structure.

So, if we use $M_f=0.96$ and $v_e=8\;\mathrm{km/s}$, we get a maximum possible $\Delta v = 25.8\;\mathrm{km/s}$. The orbital velocity of Earth is about $7.9\;\mathrm{km/s}$, so the orbital velocity of your bigger Earth would be about $9.0\;\mathrm{km/s}$. So it would still be possible to reach orbit, also with a less efficient propellant.