2
$\begingroup$

So I was looking into the payload mass a fully fuelled Centaur V could send to Jupiter and used the formula in presented in this answer to create a C3 graph. For a bit of fun I also looked into the payload capacity of Starship and was surprised to find out that it had 0kg capacity to that orbit; it reaches 0 at 63.37 $km^2/s^2$, well short of the ~82 $km^2/s^2$. The graph is based on the C3 graph presented here. Is my working out incorrect or does a fully fuelled Starship seriously not have the ability to send a payload to Jupiter? This is somewhat pertinent given questions like this.

enter image description here

On a side note, I also did delta-v calculations for it and found that it has plenty of payload to Jupiter; 86 tons. Are the delta-v assumptions from this graphic too optimistic or can delta-v not be used to predict payload mass or is this graph correct and the C3 graph incorrect?

enter image description here

edit: So I also looked into the EUS deep space capabilities and this one confuse me. The engineering parameters come from this document. The lunar payload of the EUS here has greater capacity then what is shown in this C3 graph, which makes sense because EUS when riding on SLS B1B is deployed on a suborbital trajectory and would lose some performance. However once we get to Jupiter it ends up being 2.83 tons, well short of the 8 tons of SLS B1B, which doesn't make sense given that it's performance should be boosted. Is this method of calculating C3 performance just inaccurate? (@uhoh help please)

enter image description here

$\endgroup$
8
  • 1
    $\begingroup$ Is Starship really planned to have >160t to Mars? What are these numbers for the empty/full weight? They don't seem to match what I find in Wikipedia. $\endgroup$
    – asdfex
    Jan 11, 2021 at 11:32
  • $\begingroup$ @asdfex I now realise that I was using the wrong number for total mass (1200 tons which is propellant mass vs 1320 tons for total mass) which brings up the max C3 to 69.6 but this is still short for Jupiter (graphic been updated as such). In SS PUG, they have 100+ tons to Mars surface, so >160t to TMI isn't that out there. $\endgroup$ Jan 11, 2021 at 12:05
  • 4
    $\begingroup$ This doesn't seem inherently surprising. Starship has a much higher dry mass, which will make it ineffective for very high delta-V (the same reason you wouldn't take a Space Shuttle to GEO or the Moon). The most obvious way round this is refueling -- launch two or more Starships into a very elliptical Earth orbit and refuel one from the other(s). The fueled one then heads out, while the dry tankers reenter at their next perigee. Even more extreme, so the same with an elliptical solar orbit. $\endgroup$ Jan 11, 2021 at 12:24
  • 1
    $\begingroup$ @SteveLinton: Indeed, the Starship architecture is built around robotic refueling. That's how the "payload to LEO is payload to anywhere" works. $\endgroup$ Jan 11, 2021 at 13:58
  • $\begingroup$ @asdfex That's a Centaur V stage, which is a bigger™. However you did encourage me to look a bit more for numbers (originally based on NSF estimates) and I found a better number for the propellant mass in ULA's upper stage comparison which reduces total mass from ~75 to ~58.5. $\endgroup$ Jan 11, 2021 at 15:01

1 Answer 1

7
$\begingroup$

This is only a partial answer, but there are two major factors that make the performance of the three vehicles not comparable at all.

There are two reasons why Starship is so much worse for missions with a large C3, and both are design features:

First, it's built to land (repeatedly) on planets. This makes it a lot more sturdy than other second stages. The dry-to-wet weight ratio is about 10 for Starship and around 20 for the other two. 90% fuel compared to 95% fuel makes a huge difference: Compared to F9SS it produces 30% less $\Delta v$.

Note that this is not related to the sheer size of the rocket, Starship being almost 20 times heavier than the other two, but only the ratio of fuel to dry weight is relevant.

Second, it uses $\rm {CH_4 / LOX}$ as fuel compared to $\rm{LH_2/LOX}$ for Centaur. Their effective exhaust speed is 3700 m/s and 4400 m/s, respectively. This gives another huge hit in the performance by 16%.

Both factors together give Centaur a performance that is 40% higher than Starship if they both start fully fueled from LEO. Obviously this is not a fair comparison because only Starship can be refueled, but it gives an impression about the differences between the vehicles. On the other hand, we're discussing the small payload performance limit, which gives the first stage of each launcher the chance to take over a large part of the propulsion needed to get to LEO, i.e. the second stages will be rather full when reaching LEO.

Here's a plot comparing the estimated real performance of a half full F9SS (in yellow) and a refueled Starship (in purple). If we add refueling capability to F9SS we end up with the blue curve. Likewise, if we remove the sturdiness for landing from Starship and make it 50% lighter we end up with the green curve. C3 of Starship and F9SS

Please note that this plot is only valid for the case of LEO as a starting point, which is not correct. With a light payload the second stages can get to a much higher orbit before refueling and therefore can reach a much higher C3.

$\endgroup$
2
  • 3
    $\begingroup$ Also, if Starship ends up dominating the LEO market and there's a lot of interest in sending stuff to Jupiter, there's no fundamental reason why one couldn't load a modified Centaur onto a Starship and let the hydrolox workhorse handle the second halfway to anywhere. It nearly happened with the Shuttle. $\endgroup$
    – TooTea
    Jan 11, 2021 at 15:53
  • $\begingroup$ @TooTea That sounds like a vehicle reaching Pluto within a few months? $\endgroup$
    – asdfex
    Jan 11, 2021 at 16:00

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.