# What is the value of SpaceX Starship 1200-ton propellant capacity? Would a fully refueled Starship be able to accelerate to Jupiter (for example)? [duplicate]

2200000N (1 Raptor) / 1,300,000*0.4 kg (Total weight adjusted to fuel loss) = 4.2 m/s^2 (acceleration)

Delta-V (to travel to Jupiter) = 9000 m/s

9000 m/s / 4.2 m/s^2 = 2,142 s (1 Raptor engine burning time: 1.8 hours)

SpaceX Raptor engine mass flow = 650 kg/s

2,142 s * 650 kg/s = 1,392,300 kg

Gravity assist could probably reduce required propellant by half, but is it still too heavy?

When traveling to other planets, maybe a classical 3-stage design is better than SpaceX's revolutionary 2-stage design?

• Earth comparison: 14-ton fuel truck - can bring - 28-ton of fuel (1/2 ratio) - compare to 1/20 in Starship. Commented Feb 17, 2023 at 22:44
• close-voting as duplicate. You've tried to use this method before on the site and been told about the rocket equation. Commented Feb 17, 2023 at 23:32
• This question is about the design of spacecraft limitations, when increasing fuel capacity would not increase your payload capacity. Commented Feb 17, 2023 at 23:49
• @TheMatrixEquation-balance that is sorta what the rocket equation tells you, for any chemical rocket getting to Jupiter involves getting deep into the exponential curve of increasing fuel mass. Commented Feb 17, 2023 at 23:55

You're not accounting for the weight of the Starship decreasing as fuel burns off.

You need to use the Tsiolkovsky rocket equation:

$$\Delta v = v_\text{e} \ln \frac{m_0}{m_f} = I_\text{sp} g_0 \ln \frac{m_0}{m_f}$$

$$m_0$$ is the initial mass of Starship, fully fueled (1300 tons). $$m_f$$ is the final mass after propellant consumption, i.e. the dry mass of Starship (100 tons). $$v_\text{e}$$ is the rocket exhaust velocity, about 3600 m/s for the Raptor engine, and $$\ln$$ is the natural log function.

So for those figures (gleaned from a very quick glance at Wikipedia and probably wrong), the potential delta-v works out to about 9233 m/s. Payload mass would add to both the initial and final mass figures, reducing the mass ratio and thus the $$\Delta v$$.

• Now I can see why it is not recommended to reference Wikipedia as a source. They have changed Starship's propellant specifications recently. Commented Feb 18, 2023 at 1:14
• Wikipedia has had the 1200 t number for years now. Commented Feb 18, 2023 at 2:30
• @ChristopherJamesHuff - They did have "Starship Interplanetary" with 1900-ton propellent capacity. Now I don't see it on Wikipedia. Commented Feb 18, 2023 at 3:40