I've seen charts and tables, etc., that list the delta-V to Low Earth Orbit as 9.3-10 km/s. But LEO could be anywhere below 2000 km. So how do I calculate the delta-V to reach a specific altitude in this range, or to any altitude in Earth orbit?


Without a specific rocket launcher design in mind, the exact ∆v can't be determined because it is not dependent only on the altitude, but on the characteristics of the launcher.

The largest variable factor is what's called "gravity losses". In the early part of the launch, the rocket is mainly expending ∆v to accelerate vertically. As it does so, it's fighting Earth's gravity. As an extreme case, consider a rocket that produces only 1% more thrust than its weight; it will accelerate very slowly and most of the ∆v it produces will be wasted countering the force of gravity.

A rocket with a very high thrust-to-weight ratio, on the other hand, spends less time on vertical acceleration and therefore loses less ∆v to gravity.

The Saturn V, launching into a 185-190km circular orbit, expends ~9200 m/s ∆v, of which ~1750 m/s is lost to gravity and ~50 m/s lost to atmospheric drag. It is a moderate-to-high acceleration launcher.

Once in low Earth orbit, with gravity losses no longer a concern, you can calculate the ∆v needed to transfer from a low orbit to a higher one. In most cases, the (launcher variable) ∆v needed to launch directly to a high orbit is approximately the same as that needed to launch into a low orbit and then transfer to a high one.


$\Delta v$ is just a quantity masking the real effort that needs to be put into a change of orbits:
The potential Energy $E_p$. What is usually done is that the difference between Earth's potential energy and the potential energy of your target orbit $\Delta E_p = E_p(r_{\oplus}) - E_p(r_{target})$ is equated with the kinetic energy $\frac{1}{2}m \Delta v^2$. This comes simply from energy conservation. So now we get
\begin{align} \Delta v = \sqrt \frac{2 \Delta E_p}{m} \end{align}

Now to get any further from there, we have to realize that the potenial difference we seek must be coming from the work $W = \int \vec F \cdot d\vec s$ that one needs to invest in order to fight the force $\vec F$ at any given altitude.
With that we get \begin{align} \Delta v = \sqrt {\frac{2}{m} \int\limits_{r_{\oplus}}^{r_{target}} F(r) dr} \end{align} where I already have taken the liberty of simplifying the scalar product in the integral.

Now there are two different regimes for this Force:

At low altitudes both gravity and friction play a role. Gravity is well known to be $F_g = \frac{GM_{\oplus}m}{r^2} $ with the usual notation (do tell if you need explanation for this) and subsonic friction can be approximated as $F_f= - c_w \rho v^2 $. Here $c_w$ is a geometrical factor for the rocket quantifying it's coupling to the airflow, and $\rho$ is the air density.
This formula for the air-friction is not the correct one, as rockets can reach the supersonic friction regime, but it illustrates the point that was only mentioned in textual answers: geometry and density both play a role into the calculation of $\Delta v$.
Thus we'd have $F = F_g + F_f$ at low altitudes and we'd need to solve the work-integral for this given force.

This leads us to the realization that at high altitudes $F_f$ is negligible and thus at high altitudes we have $F = F_g$.
For simplcity I'll assume now friction to be negligible from the Earth's surface into the orbit, so that we can solve the above integral directly into

\begin{align} \Delta v = \sqrt {2GM_{\oplus} \left( \frac{1}{r_{\oplus}}- \frac{1}{r_{target}} \right)} \end{align}

which is our final result for $\Delta v$ with the gravitational constant $G$, Earth's Mass $M_{\oplus}$, and launch and target distance from Earth's center, $r_{\oplus}$ and $r_{target}$.
Plugging in some numbers into that (e.e. for the 300->500km transfer orbit and keeping in mind that we have to add this always to Earth's radius of 6370 km).


Usually the official papers will site an inclination and an orbit with the specified payload capacity. The orbit will typically be quite a bit lower.

The old Falcon 9 user's guide shows plots of payload vs both orbital altitude and inclination. The more current one is available upon request only. You can see some of the factors that lead to LEO. The reported LEO number tends to be in a low orbit at optimal inclination, or in other words, best case scenarios.

  • $\begingroup$ Such sources don't normally use delta-v figures, do they? $\endgroup$ – Russell Borogove May 20 '16 at 15:01
  • $\begingroup$ No, butty the same principal applied. The Delta v depends on the orbit. $\endgroup$ – PearsonArtPhoto May 20 '16 at 16:03

Depending on T/W and other factors, delta V to LEO can vary. But once you've reached LEO, gravity loss and atmospheric drag are no longer factors. So I'll give you some delta Vs going from a 300 km altitude circular orbit to higher altitude orbits.

300 km to 500 km - .11 km/s

300 km to 1000 km - .38 km/s

300 km to 2000 km - .83 km/s

300 km to 4000 km - 1.51 km/s

300 km to 8000 km - 2.37 km/s

300 km to 16000 km - 3.22 km/s

300 km to 32000 km - 3.83 km/s

300 km to 64000 km - 4.1 km/s

300 km to 128000 km - 4.13 km/s

300 km to 256000 km - 4.02 km/2

300 km to 512000 km - 3.87 km/s

300 km to 1024000 km 3.72 km/s

I got these numbers from my Hohmann spreadsheet which can be downloaded from Cosmic Train Schedule

Screen capture for going from 300 to 1000 km. I add the apoapsis and periapsis circularize burns.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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