1
$\begingroup$

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.

Improve this question
$\endgroup$
8
  • $\begingroup$ 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, 2019 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, 2019 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, 2019 at 15:12
  • $\begingroup$ @JamesJenkins this does not account for the propulsive landing $\endgroup$
    – Antzi
    Jan 18, 2019 at 15:17
  • 1
    $\begingroup$ @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, 2019 at 18:25

1 Answer 1

Reset to default
2
$\begingroup$

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 $1 \frac{km}{s}$ 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:

$$(exp{\left(\frac{1000\frac{m}{s}}{330\ s * 9.8 \frac{m}{s^2}}\right)} - 1) * 185\ t = 67\ t$$

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

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Jeez...in retrospect, I should have been able to figure that out myself. Thanks. $\endgroup$
    – Chris B. Behrens
    Jan 21, 2019 at 18:28
  • $\begingroup$ Is Starship still using active cooling? I thought they dropped that a while ago. $\endgroup$
    – Dan Hanson
    May 25, 2022 at 19:17
  • $\begingroup$ They've mostly dropped the transpiration cooling, they still may use some form of active cooling for spots that are hard to shield. $\endgroup$
    – Christopher James Huff
    May 25, 2022 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.