If the Earth had the mass of Jupiter, would we be able to launch a space missile ?

Is there a formula to calculate how the launch of 1-Kg payload increases with gravity?

  • $\begingroup$ Since the question has been reopened, I've removed the comments discussing whether the closure was appropriate. $\endgroup$
    – David Z
    Commented Nov 23, 2015 at 5:55

4 Answers 4


Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then:

  • radius of Earth = $6.4 \times 10^6~\text{m}$
  • mass of Jupiter = $1.9 \times 10^{27}~\text{kg}$
  • Escape velocity, $v_\text{escape} = \sqrt{\frac{2GM}{r}}$

This gives a value for $v_\text{escape}$ of $200~\text{ km}/\text{s}$. For comparison, the actual value (for the real Earth) is $11~\text{ km}/\text{s}$. Incidentally, the surface gravity on this new Earth is about $300~\text{g}$.

To work out how this could be done, we need the rocket equation, which is $\Delta v = v_\text{exhaust} \ln \frac {m_0} {m_1}$.

We need a delta-V of $200 ~\text{km}/\text{s}\; ,$ with a chemical rocket (exhaust velocity about $4400 ~\text{m}/\text{s}$). Solving for $m_1 = 1~\text{kg}\;,$ we get mass of fuel ($m_0$) required of about $5 \times 10^{16}$ tonnes.

That's about $5\%$ of the mass of all the oceans on Earth. If you used hydrogen and oxygen as the fuel, you'd need to convert a volume equivalent to the Mediterranean Sea.

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    $\begingroup$ The rocket equation pertains to a single stage rocket. Multiple stages changes the game a bit, but not nearly enough to escape from Jupiter. $\endgroup$ Commented Nov 19, 2015 at 9:14
  • $\begingroup$ Thanks, Owen, could you say up to what gravity we can launch 1-Kg net payload, and the total weight of the missile $\endgroup$
    – user96370
    Commented Nov 19, 2015 at 9:29
  • $\begingroup$ It's interesting to imagine the kind of life forms that might live on the surface of a planet with this surface gravity so that there would be intelligent beings there able to take the challenge in the first place! Not quite the Cheela but certainly would be radically different from Earth life. $\endgroup$ Commented Nov 19, 2015 at 9:40
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    $\begingroup$ @DavidHammen The rocket equation pertains to a single stage rocket with only payload and propellant mass, i.e zero empty mass for tanks and engines. In practical terms that's equivalent to an infinite stage rocket. The reason staging helps in real rockets is it means we don't have to carry empty tanks and oversized engines all the way. But any real rocket (regardless of number of stages) will have a worse mass/payload ratio than the rocket equation (unless, of course, you consider your empty tank and engine to be payload). $\endgroup$ Commented Nov 19, 2015 at 10:25
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    $\begingroup$ Note that a planet with the radius of Earth and the mass of Jupiter would have a mean density of $1.7 \times 10^6\,{\rm kg/m^3}$. That's over 75 times the density of solid osmium, making such a planet pretty much physically impossible. (White dwarfs and neutron stars achieve far higher densities, around $10^9\,{\rm kg/m^3}$ and above, but there doesn't seem to be any way to have a planetary-size body with a density like your calculations imply.) A (slightly) more reasonable scenario might be a Jupiter-mass planet with the same density as Earth. $\endgroup$ Commented Nov 19, 2015 at 18:44


The question keeps getting edited! Make up your mind!

You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not?

What if Earth had the mass of Jupiter? More impossible. It would have a surface gravity of $g \frac{M_J}{M_E}$ or something like 3100 m/s2. I don't think you could even build two-story buildings on that kind of planet

However, here is the math for Mars.

Answer for Mars

Gravity differs, yes, but Mars also has a 0.6 kPa surface pressure, compared to Earth's 100 kPa. This makes comparisons between Earth and Mars practically impossible. Fortunately, the math is easier on Mars.

Tsiolkovsky's rocket equation tells us the answer for general rocket maneuvers.

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

For Mars to LMO (low Mars orbit), the $\Delta v$ is about 4.1 km/s. This is just a function of the gravitational potential that you are escaping. For comparison, Earth to LEO is about 9.3-10 km/s, and Kerbin to LKO is about 4.6 km/s.

The value $v_e$ is the effective exhaust velocity, which could be about 4.4 km/s for a bipropellant rocket.

The values $m_0$ and $m_1$ are the masses of the rocket before and after the maneuver.

We will pretend that Mars has no atmosphere.

Suppose that 75% of your rocket is fuel, then $\frac{m_0}{m_1} = \frac{1}{1 - 0.75} = 4$, and your $\Delta v$ is 6.1 km/s, which is more than enough to get into orbit. But it's not enough to escape Mars! For that, you need to double the $\Delta v$.


Actually you can go to the orbit of Jupiter with a $\approx 2500$ tonnes rocket and a $3$ tonnes payload. From there you can use an ionic engine. A rocket launched from the equator of Jupiter that turns at $12.6~\text{km}/\text{s}$ needs just an increase in speed $v = 29.5~\text{km}/{\text{s}}$.

$$v_{rj}:= 12.6~{\text{km}}/{\text{s}} \;\;\; R_j := 71492~\text{km}\;\;\; g_j:= 24.79~\text{m}/{\text s^2}$$ Given $$\frac{(v + v_{rj})^2}{R_j}= g_j \\ \text{Find}(v)\rightarrow \left[\frac{1}{5}\cdot \frac{\left[-63~\text{km} + \sqrt{44307167}\cdot (\text{km}\cdot\text{m})^\frac{1}{2}\right]}{s} \; \frac{1}{5}\cdot \frac{\left[-63~\text{km} - \sqrt{44307167}\cdot (\text{km}\cdot\text{m})^\frac{1}{2}\right]}{s}\right] \\= \left(2.95 \times 10^4 \;\; -5.47\times 10^4\right)~\text{m}/\text{s} \\ m_L:= 3000~\text{kg}\;\;\; v_e := 4400~\text{m}/\text{s} \;\;\; v:= 2.95 \times 10^4 ~\text{m}/\text{s}$$ Given $$v=v_e\cdot \ln \left(\frac{m_r}{m_l}\right)\\ \text{Find}(m_r)\rightarrow 2448320.9528130687939~\text{kg}= \; 2.448 \times 10^6~\text{kg}$$

  • $\begingroup$ Why don't use mathjax? $\endgroup$
    – user36790
    Commented Nov 19, 2015 at 12:57
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    $\begingroup$ The answer of Owen Boyle is fundamentally wrong. He compressed Jupiter to the size of the earth obtaining an impossible average density. The many pluses this user received are 80-90% bogus points. I do not believe there are so many people on this site that were fooled by his calculations. $\endgroup$
    – PeterS
    Commented Nov 19, 2015 at 13:17
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    $\begingroup$ @PeterS: I answered the question the OP asked. I'm not trying to fool anyone; I'm just applying the maths to the (highly!) hypothetical case presented. I think most people get that it's just a bit of fun. $\endgroup$ Commented Nov 19, 2015 at 14:11
  • $\begingroup$ Being able to solve the rocket equation for a particular situation and being able to build a rocket with the parameters that pop out are very different things. Your proposal allows 12 parts in 1000 of the mass for the superstructure, tanks, pumps, computers, etc. That's more than an order of magnitude less than achieved designs $\endgroup$ Commented Feb 7, 2018 at 15:47

could you say up to what gravity we can launch 1-Kg net payload, and the total weight of the missile

The escape velocity of Jupiter is just 59.5 km/s. Doing the math using the formula posted by others above you will get:

$m_{rocket} = 746.2$ tonnes

which is well below the mass of the rocket that transported people on the moon between 1969 and 1972.

So, it is possible to launch a 1 kg payload from Jupiter.

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    $\begingroup$ Hi Darius. The 1kg includes the mass of everything that isn't fuel i.e. the rocket engines, the skin and all the pumps and electronics needed to make the rocket work. So the rocket you describe would have to 99.99987% fuel. No such rocket could be built. $\endgroup$ Commented Nov 19, 2015 at 11:05
  • $\begingroup$ The problem from launching from the real Jupiter is deciding where you're starting from. There's no "surface", AFAIK. Also, I rather imagine atmospheric drag on Jupiter might be a bit a.... drag. $\endgroup$ Commented Nov 19, 2015 at 12:25