# What's delta V and propellant required to go from LEO to HEO?

I know others have touched on this, but I'm looking for the LEO to HEO case specifically.

Here are the parameters I have: LEO 454 km circular orbit at 61 degree inclination; HEO 1100 x 9000 km at 64 degree inclination (I know, two problems, orbit "height" and inclination change)

I have a choice of two engines, one produces 16680 kN vacuum thrust with isp of 328 sec while the other produces 3870 N with isp of 281 sec.

• Do you mean 16680 N instead of kN? In any case, you're presenting a choice that has both higher thrust and higher ISP, so that's quite easy. For maneuvers made once in orbit, if time is not essential, ISP is the most important factor by far. Commented Jul 28, 2017 at 14:26
• Have you tried calculating this yourself? Are there specific things you're getting hung up on? Commented Jul 28, 2017 at 15:29
• Even if it was 16680N of thrust, I am having a hard time finding conditions where thrust is going to matter at all. Isp is always going to be a determining factor until we venture into the realm of the extreme (ion thrusters vs chemical engines) Commented Jul 28, 2017 at 16:14
• +1 because any question that earns such a nice answer into which has been put so much work shouldn't have net-negative votes.
– uhoh
Commented Jul 28, 2017 at 16:57

This problem is a good exercise in orbital mechanics. For my calculations, I assume a spherical earth of radius 6371km. A bunch of orbital mechanics terminology will be used and I make assumptions that all burns are done optimally.

Taking the parameters specified, we first calculate the LEO semi-major axis to be 6825km. The HEO semi-major axis is 11421km. For this calculation, I will assume a Hohmann transfer with a transfer orbit semi-major axis of 11098km. Using the vis-viva equation, we can calculate the delta-V for the initial burn.
$$v_{LEO,Pe}=\sqrt{GM_{earth}(\frac{2}{r}-\frac{1}{a})}=7642\mathrm{m/s}$$
Same for HEO: $v_{Transfer,Pe}=8994\mathrm{m/s}$
This gives a delta-V of $1352\mathrm{m/s}$ for the first burn. A similar calculation for the second burn at apogee of the transfer orbit yields a delta-V of $4119\mathrm{m/s}-3993\mathrm{m/s}=126\mathrm{m/s}$.

Now for the inclination change: Inclination changes are not as simple as you might assume. It is possible to have an orbit at 61 degrees inclination be completely, horribly misaligned with an orbit at 64 degrees inclination. But I will assume best case scenario, where either the Ascending or Descending nodes are at Apogee and the relative inclination is the difference between the two orbits' inclinations (i.e 3 degrees). In this case, the inclination change burn is given by
$v_{IC}=2v \cdot sin(\theta/2)$ with $\theta=3^o$. At apogee, before the prograde burn, $v=3993\mathrm{m/s}$ and so $v_{IC}=209\mathrm{m/s}$.
Note as well that the inclination change burn does not change our total orbital velocity.

Now we have a total delta-V of $1352+209+126=1687\mathrm{m/s}$. In designing a spacecraft, we need to have a small amount extra since burns are never optimal. With a 5% buffer, our total delta-V is $1770\mathrm{m/s}$.

Finally, we can calculate the mass fraction of your spacecraft in LEO:
$\frac{m_i}{m_f}=e^{\Delta V/(I_{SP}\cdot g_0)}$
For 328 sec Isp, $\frac{m_i}{m_f}=1.73$
For 281 sec Isp, $\frac{m_i}{m_f}=1.90$

I hope this helps in your calculations.

Edit

Using a spacecraft initial mass of 30,000lbs, we can calculate the total burn time. $30000\mathrm{lb}=13600\mathrm{kg}$, and so this means there is $5379\mathrm{kg}$ of fuel given a mass fraction of 1.73. If there is $16680\mathrm{N}$ of thrust, then: $$\dot m=\frac{T}{I_{SP}*g_0}=5.2\mathrm{kg/sec}$$ This gives a total burn time of $\frac{5379\mathrm{kg}}{5.2\mathrm{kg/sec}}\approx 1000\mathrm{sec}$.

For engine 2, I get a burn time of 4600 sec.

The times for the individual burns will not be exactly in proportion to their delta-V out of total, but it will be pretty close.

• It's a huge help, but unfortunately, I'm not sure I framed the question properly. How long a burn with each engine would be required to achieve the necessary delta V. And thanks for working on this! Commented Jul 28, 2017 at 16:24
• @WeaselPilot There is information missing to complete your question. Without the initial mass of the spacecraft I cannot answer your question. I recommend tidying up your question to include everything you want to know. What I can tell you is that your burn times will be around 4 times less with the stronger engine. Commented Jul 28, 2017 at 17:04
• Unfortunately I don't know for sure. According to available information about 30,000 lbs. Commented Jul 28, 2017 at 17:22
• @WeaselPilot I strongly recommend that if you have more questions as to various aspects of the mission profile, that you find some mission planning software. It might be a steep learning curve but can provide answers on the spot and adjust for any changes you make. I use KSPTOT. Commented Jul 28, 2017 at 18:02
• I agree, it's a good answer...suspect I'm going to have to find and read the FAQ, but (stupid new guy question) how do I accept an answer? Commented Jul 29, 2017 at 20:05