I'm trying to estimate the delta-v (and, by applying the Rocket Equation, the propellant) required to lift a payload from LEO to a higher orbit. As I understand it, the standard way to do this is with a Hohmann Transfer: first, boost from a circular orbit at r1 (which in this case is LEO) to an elliptical orbit with apogee at r2 (the altitude of my target orbit); then, boost the elliptical orbit into a circular orbit at r2.

Using the equations described here:



r1 = 6451000m (i.e., an orbital altitude of 80km + the radius of the earth)

r2 = 95896000m (orbital altitude of 89,500km + earth radius)

I come up with a delta-v of about 4,215 m/s.

However, if I am instead transferring from my 80km LEO altitude to 179,000km, I get a delta-v of about 4,150 m/s. I.e., increasing the altitude of my target orbit by 80,000km decreases the total delta-v by about 75m/s.

Mathematically, I can see why it's happening. The value of Δv2 is decreasing faster than Δv1 is increasing. But it doesn't seem right. Am I misunderstanding the purpose of the delta-v value in the Hohmann Transfer equation? Or is that actually the way it is?


3 Answers 3


While the lower destination orbits don't require a big insertion burn, they do need a big circularization burn at apogee of the transfer orbit.

The higher the apogee, the longer and skinnier the ellipse and the more parabola-like the transfer orbit becomes. So, as the ellipse's apogee goes to infinity, perigee velocity approaches escape velocity. That would be sqrt(2) * circular orbit velocity at departure orbit. So as apogee goes up, insertion burn approaches (sqrt(2) - 1) * circular orbit velocity (assuming you're already in the circular departure orbit).

Circularization burn at apogee approaches zero as apogee goes to infinity.

Maximum Hohmann delta-v occurs when the destination orbit radius is 15.58 times the departure orbit radius.

If the radii of the departure and destination orbits differ by a factor of 11.94 or more, the bi-elliptic transfer is more economical (in terms of delta-v, but not in transfer time) than a Hohmann transfer.


There are already good answers here, but I'll discuss this without being too mathematical.

Let's start from a circular orbit. The velocity in this case is constant, let's call it $v_c$.
When the necessary delta v is imparted to a body transferring to a higher orbit, the transfer orbit always have its periapse at the circular orbit and apoapse at new orbit. From basic orbital mechanics you can tell that the speed of a body is fastest at the periapse.
Let's say the new periapse speed is given by $v_t$.
For a body in elliptical orbit that velocity will be given by $v_t = \sqrt{(\dfrac{2\mu} {r_p}-\dfrac{\mu} {a})} $
Which as you can see will increase as the semi major axis of the orbit increases. Which concludes that the necessary delta v needed at that point will decrease when transferring to a higher orbit.


I may have found the answer to my own question. From the "Example" section of the same page I linked to:

It is interesting to note that this is greater than the Δv required for an escape orbit: 10.93 − 7.73 = 3.20 km/s. Applying a Δv at the LEO of only 0.78 km/s more (3.20−2.42) would give the rocket the escape speed, which is less than the Δv of 1.46 km/s required to circularize the geosynchronous orbit. This illustrates that at large speeds the same Δv provides more specific orbital energy, and energy increase is maximized if one spends the Δv as quickly as possible, rather than spending some, being decelerated by gravity, and then spending some more to overcome the deceleration (of course, the objective of a Hohmann transfer orbit is different).

So, if I'm understanding this correctly, a Hohmann transfer to a higher orbit is more energy-efficient because approximately half of the "lifting" work happens at or near the apogee of the elliptical orbit, where the gravitational force of the earth is a bit weaker. Therefore, if the apogee of the elliptical orbit is higher, less delta-v is required because there is less gravitational force to fight against.

Is that a reasonably accurate explanation?

  • 2
    $\begingroup$ No, that is wrong. All burns in a Hohmann-Transfer are perpendicular to the gravity vector so you don't have to "fight" gravity at all. If anything the burn at the lower orbit is more efficient because you are faster, see Oberth effect. $\endgroup$ Mar 14, 2016 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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