Delta-v required to get to Medium Earth Orbit?

Does anyone know what's the delta-v you need to get to Medium Earth Orbit (the delta-v for 7,000 km high orbit for example)?

I know that to get to LEO you need around 9.5 km/s (including gravity and air drag) but I couldn't find the delta-v for MEO anywhere. Feel free to make any assumptions about launch site or orbital inclination.

A follow up question is: what's the deorbiting delta-v (from MEO)? From LEO it's only 100 m/s.

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You may want to try Space Exploration. – HDE 226868 Dec 2 '14 at 1:43
Do you want a theory of how these calculations are done, or practical numbers? If the latter, then this question is more appropriate for space exploration. If the former, then this is the site. Your last sentence implies you want more practical: the deorbitting will be set by detailed models of the Earth's atmosphere. – WetSavannaAnimal aka Rod Vance Dec 2 '14 at 2:01
Yes, I was looking for a more practice answer. You know, just an estimate. Thanks though. I'll ask the question on the "Space" stack. – Pema Dec 2 '14 at 2:24

1 Answer

A good estimation would be an Hohmann transfer from LEO to MEO. It might be slightly more efficient if the burn would be performed in the upper atmosphere, due to the Oberth effect.

When you calculate the total required $\Delta$v for the two burns of the Hohmann transfer, you get about 2.3 km/s, so the total $\Delta$v to get from Earth to MEO would be about 11.8 km/s.

And to answer your followup question, you can use the same equations of the Hohmann transfer, but do one burn which brings your periapsis into the Earth's atmosphere. This will be slightly bigger than the circularisation burn of the Hohmann transfer at about 1.1 km/s, however the reentry would not be that much higher as for LEO. Namely 9.2-9.3 km/s for MEO, compared LKO of 8 km/s (note that the increased velocity is roughly equal to that of the initial burn of the Hohmann transfer).

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Thanks! That helped. – Pema Dec 2 '14 at 2:45
As an relevant aside, earth's escape velocity is about 11.2 km/s. So that estimate would allow you to leave earth as well--that sorta makes sense since you've a long way to go. – Kyle Kanos Dec 2 '14 at 2:52
@KyleKanos I usually find these a bit skewed comparisons, since this neglects drag and the circularisation burn, which is about 1 km/s. – fibonatic Dec 2 '14 at 2:59
@Pema glade that I could help, I also added an answer to your followup question. – fibonatic Dec 2 '14 at 3:27