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The SpaceX's Starship is in orbit, empty payload. Magically, 100,000kg of payload appears inside the ship, setting aside all the logistical questions of how it got there. Now it needs to deorbit and land back on Earth.

Given that it didn't carry any payload to orbit which reduces the fuel usage - how much fuel does it need to land that payload? Can a Starship land 100,000kg on Earth?

I'm just looking for rough numbers here.

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  • $\begingroup$ The dry weight being 85T, you need to land 185T, so you need > 1.8MN of thrust. Starship has 13.9MN of thrust (vacuum), so landing 100T is plausible. $\endgroup$ – Antzi Jan 18 at 3:12
  • $\begingroup$ You say de-orbit and land, how do you plan to do that matters. If you are going to use a thermal protection system and parachutes all you need is the fuel for the retro-burn. On the other hand if you want some sort of powered landing it's a very different story. $\endgroup$ – GdD Jan 18 at 9:03
  • $\begingroup$ Your answer is at Could you take a Cessna from the ISS to Earth? you need about 30 times the weight of fuel as your ships loaded weight. So 5,550 tons of fuel $\endgroup$ – James Jenkins Jan 18 at 15:12
  • $\begingroup$ @JamesJenkins this does not account for the propulsive landing $\endgroup$ – Antzi Jan 18 at 15:17
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    $\begingroup$ @h22, well I meant the SpaceX Starship specifically, not colloquially, "a starship". So no wings, and a lot more specifics (though perhaps still not enough) to hang the question off of. $\endgroup$ – Chris B. Behrens Jan 21 at 18:25
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Apart from the fact that there's almost no information about the latest version of the vehicle, there's also little information about the differences in capabilities between Earth and Mars landings. We do know it's intended to be able to land on Mars fully loaded. Earth has a few times the gravity and a much deeper gravity well, but a far thicker atmosphere.

So, assuming you need around 1 km/s of delta-v to do the final deceleration and landing burn, roughly equivalent to what's been shown for Mars. You have a dry mass of around 185 t, and engines that have a specific impulse of 330 s at low altitude. The propellant requirements work out to:

(exp((1000 m/s)/(330 s*9.8 m/s^2)) - 1)*185 t = 67 t

That's for landing. The Starship is to use active cooling for reentry, and it's unknown how much methane that will consume.

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    $\begingroup$ Jeez...in retrospect, I should have been able to figure that out myself. Thanks. $\endgroup$ – Chris B. Behrens Jan 21 at 18:28

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