My spaceflight mechanics lecture script has this formula, without derivation or reference.

da=2a^2/sqrt(mu*p)*(1+e*cos nu)*dv

But I think it does not work. For example let us assume we find ourself in a circular 250km orbit around earth and want to raise our apoapsis to the height of a geostationary orbit at 35786 km. I think the formula doesn't work for this example:

We know the radius of earth $R = 6371\cdot 10^3 m $

and it's standard gravitational parameter $\mu = 3.9860044188\cdot 10^{14} m^3/s^2$

The alteration in the semi major axis: $\Delta a = 35786\cdot 10^3 m-250\cdot 10^3 m = 35536\cdot 10^3 m$

The semi major axis $a = 250\cdot 10^3 m + 6371\cdot 10^3 m = 6621\cdot 10^3 m$

I'm assuming p is the semilatus rectum which is the same as the semi major axis, since we are assuming a circular orbit $p = 6621\cdot 10^3 m$

and since the orbit is circular the eccentricity is just $e = 0$

now we can rewrite the formula to solve for the required delta v:

$\Delta v = {\Delta a \sqrt{\mu\cdot p}\over{(2a^2\cdot(1+e\cos(\nu))}}$

$= {35536\cdot 10^3 m \sqrt{3.9860044188\cdot 10^{14} {{m^3}\over{s^2}} \cdot 6621\cdot 10^3 m} \over{ (2\cdot (6621\cdot 10^3 m)^2\cdot (1+0\cos (\nu)))}}$

$= {35536\cdot 10^3 \sqrt{3.9860044188\cdot 10^{14} \cdot 6621\cdot 10^3} \over{ (2(6621\cdot 10^3)^2)}}\cdot{ {m^3}\over{s\cdot m^2}}$

$= 20821 {m\over s}$

But that can't be right because the entire transfer to geostationary orbit should take only about 3900 m/s of delta v, and we're only half way there, because we didn't even raise the periapsis yet.

So is the formula just wrong or am I misinterpreting it? Has anybody ever seen this formula?

  • $\begingroup$ definitely something wrong with the first line of your rewritten formula - it says delta v is proportional to delta a -- as does the given equation $\endgroup$
    – user20636
    Nov 16, 2019 at 0:50
  • $\begingroup$ My best guess is $v_t$ isn't as obvious as one might think $\endgroup$
    – user20636
    Nov 16, 2019 at 1:04
  • $\begingroup$ @JCRM oic what you mean now. $\endgroup$
    – uhoh
    Nov 16, 2019 at 1:13
  • 1
    $\begingroup$ Things like 6371e3m are very hard to read for me. You can try things like $6371 \times 10^3 \ m$ for $6371 \times 10^3 \ m$ or at least 6371E+03 m or 6371e+03 m for "computerese" versions of scientific notation. $\endgroup$
    – uhoh
    Nov 16, 2019 at 1:14
  • 1
    $\begingroup$ but regardless of the worked example, something is wrong with the formula printed in the script - It has $\Delta a$ directly proportional to $\Delta v_t$ - which isn't the case, given the assumptions made about the various values. $\endgroup$
    – user20636
    Nov 16, 2019 at 1:17

2 Answers 2


The formula is only usable for calculating changes in $a$ in response to small tangential impulses. That is, when the shape of the orbit does not significantly change.

It's an important detail to omit, although $\Delta a$ being proportional to $\Delta v_{t}$ should give a strong hint.

The $20821m/s$ number you get is the cost per meter at a circular LEO orbit, multiplied by the distance up to GEO, which isn't particularly useful since the cost per meter will change as you get farther out.

If you wish to apply this formula to greater $\Delta a$ values, you would have to perform an integration, which isn't very practical since all of $a$, $p$, $v$ and $e$ are changing during the burn.

  • 1
    $\begingroup$ Is it a "well known" formula? Is there a write-up of it somewhere which explains it's limits? $\endgroup$
    – user20636
    Nov 16, 2019 at 17:15

Your error is in the calculation of the alteration of the semi-major axis.

If you are going from 250 km circular altitude, and perform a maneuver to achieve an apogee of 35,786 km altitude, your new post-burn semi-major axis will be (250 + 35,786 )/2 + 6,371 km = 24,389 km.

Your change in semi-major axis is then 18,018 km.


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