# What would be the incremental cost of launching 1 ton of LOX/LH2 into LEO? [duplicate]

I am asking this question in a context of feasability of orbital fuel depots.

I understand that cost/kg estimates exist (2-14 k$/kg): What is the current cost-per-pound to send something into LEO? ]), but the post dates back to 2013 and asked for an update on the values. But what about larger masses ie 1 metric ton and more? What would the launching cost be, considering the structural requirements for storing the LOX/LH2? • This has already been answered in the very link you put in your post. The costs do not change. – GdD Apr 27 '16 at 10:17 • What about considerations regarding the weight of the storage system for the LOX/LH2? – Tatane Apr 27 '16 at 10:22 • Cost to launch per kg is the total cost of the whole payload, including the storage and delivery system. If the storage and delivery system is 2 tons for 1 ton of LOX then the payload is 3 tons. – GdD Apr 27 '16 at 11:22 • The more interesting question is how you'd store LOX and LH2 long-term in LEO. Is a sunshade enough, or would you need active cooling? – Hobbes Apr 27 '16 at 12:12 • @Hobbes: Cryocooler on board, using LOX and LH2 as working liquids? :) A much more practical approach would be delivery of a large block of ice, and infrastructure to split it on demand, using solar energy. – SF. Apr 27 '16 at 13:33 ## 1 Answer Assuming that: • regarding the storage system, we are considering an aluminium-lithium SLWT Super Lightweight external Tank, which was used for the space shuttle launches ; • the mass ratio of LOX to LH2 is approximately 1 to 2.7 ; • that the mass of the storage tank is 0.03% of the total fuel mass (based on the ratio for the space shuttle, which is not entirely accurate) ; Then for 1,000kg of fuel (730kg LOX/270kg LH2), the total payload is 1,030kg. Considering this 2013 cost per kg to launch in LEO, the average cost of launching 1 ton of fuel is \$7,862,000 (\$2,266,000 to \$13,464,000).

I have no idea on how scaling the cryogenics or other systems would work though, but this seems like a decent enough estimate.