In this comment I used a "trick" to double check a calculation in the post above it.

Using the vis-viva equation I first determined that if the ISS lost 10 meters of altitude in 86400 seconds it would gain 5.68 mm/sec of orbital velocity over that time period.

But I then treated the ratio of those two numbers as if it were an acceleration and then equated that ratio to $F/m$:

$$\frac{\Delta v_{\text{orb}}}{\Delta t} = \frac{F_{\text{retro}}}{m}$$

where $F_{\text{retro}}$ is any average retrograde force that would have produced that loss in altitude (in this case drag), $\Delta v_{\text{orb}}$ is the change (increase) in orbital velocity and $m$ is the mass of the ISS.

If you have prograde force then you will raise the orbit and $\Delta v_{\text{orb}}$ will be negative.

I learned this "trick" a while ago, probably from some @MarkAdler answer, and for nearly circular orbits and small or slow changes in velocity it tends to work well.

Question: What are the limitations of this "trick"? Can it be extended to elliptical orbits in some way? Can it be used with large impulses? Can it be used with slow spirals seen in solar sail or electric propulsion calculations? Will it work in other universes?

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    $\begingroup$ So you assumed a circular orbit and used Newton's second law to estimate the net force that would have produced the deceleration given by vis-viva? If the ellipse is close to circular, then I can't imagine the error being significant, and even if it were, you could probably still get a useful order-of-magnitude figure, it seems? $\endgroup$
    – user39728
    Mar 14, 2021 at 15:30
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    $\begingroup$ But it seems the only difference in the drag force would come from a difference in velocity and air density. Unless the orbit is highly elliptical, it seems the change in air density (a function of altitude) would be small. And while the instantaneous velocity would vary, on long time scales it's the average velocity that matters. Wouldn't that average velocity be the same as for a circular orbit with radius equal to your semimajor axis? I forget. I can't imagine the error being very large for just slightly elliptical orbits like that of the ISS. $\endgroup$
    – user39728
    Mar 14, 2021 at 15:49
  • $\begingroup$ @user39728 Yes, it works exactly in the very specific and restricted example I've linked to, and I don't know how quickly it breaks down as constraints are relaxed. An answer will hopefully address how fast and in what ways that breakdown unfolds in a mathematical way. $\endgroup$
    – uhoh
    Mar 14, 2021 at 23:41

1 Answer 1


TL;DR It works for a very large amount of orbit changes

First, let's take your equation and rearrange it to get rid of inconvenient values like force, time and mass.

$$\frac{\Delta v_{\text{orb}}}{\Delta t} = \frac{F_{\text{retro}}}{m}$$

$$\Delta v_{\text{orb}} = \frac{F_{\text{retro}} \Delta t}{m} = - \Delta v_{\text{maneuver}}$$

$\frac{F\cdot t}{m}$ is nothing else than a velocity change, this time due to firing some rockets. Thus, you claim that the change in orbital velocity has the same magnitude as the change of velocity required for the orbital transfer, just opposite sign.

Let's assume the transfer is done using a Hohmann transfer - for small changes in the orbit this should be close to optimal and very similar to a low-thrust maneuver. Further, we assume two circular orbits.

The vis-viva equation for circular orbits is a simple $v = \sqrt{\mu \over r}$ and the difference between two orbits is $\Delta v_O = \sqrt{\mu \over r_2} - \sqrt{\mu \over r_1}$

The total $\Delta v$ of a Hohmann transfer is given by

$$\Delta v_H = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2 r_2}{r_1+r_2}} - 1 \right)+ \sqrt{\frac{\mu}{r_2}} \left(1 - \sqrt{\frac{2 r_1}{r_1+r_2}}\right)$$

If we rearrange that:

$$\Delta v_H = -\sqrt{\frac{\mu}{r_1}} + \sqrt{\frac{2 \mu r_2}{r_1^2+r_1r_2}} + \sqrt{\frac{\mu}{r_2}} - \sqrt{\frac{2\mu r_1}{r_2^2+r_1r_2}}$$

The first and third term combined are the same as $\Delta v_O$!

$$ \Delta v_H = \Delta v_O + \sqrt{\frac{2 \mu r_2}{r_1^2+r_1r_2}} - \sqrt{\frac{2\mu r_1}{r_2^2+r_1r_2}} $$

The difference of the two additional terms is very small for small changes in the orbital radius, but how small? Let's assume an initial orbit with $r = 6700~\text {km}$:

enter image description here

Actually, the difference is so small that there is no single purple pixel in this plot. Let's check the ratio:

enter image description here

In fact, we need to transfer between a 300 km and a 3000 km orbit to reach an error in the order of 1%!


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