Loren Petchel wrote:

gravity loss is 9.8m/s * (1 - your fraction of orbital speed) so you want to build horizontal speed as soon as it won't cost you too much drag.

My "feel" in Kerbal Space Program tells me otherwise. A low-TWR upper stage (order of TWR 0.3) is doomed to plunge into the atmosphere if it's missing 1/4 of its orbital velocity, even if burning at an angle "upwards". Actual reduction of gravitational drag "feels" to be non-linear, only last 200-300m/s out of Kerbin's 2200m/s really changes things. This is just a subjective feel in a quite inaccurate space simulator. How is it for real? Is Loren's equation accurate or is my "feel" supported by actual hard maths?

  • I think considering an atmosphere may confuse things here. If we are talking about gravity loss, I think it's best you consider not a launch from Kerbin, but from mun to "calibrate" your gut-feel. How would you say you feel gravity losses in this scenario? – Rikki-Tikki-Tavi Feb 6 '17 at 13:49
  • @Rikki-Tikki-Tavi: On Mun, I always have TWR so high gravitational drag is unnoticeable since moment one. – SF. Feb 6 '17 at 13:56
  • How would you say are gravity losses noticeable to you then? – Rikki-Tikki-Tavi Feb 6 '17 at 14:00
  • 1
    In short, since KSP's planets are tiny and impossibly dense, their gravitational influence drops off much quicker than the earth's would. – Rikki-Tikki-Tavi Feb 6 '17 at 14:04
  • 1
    @SF I hyperedited a ship into a 70km orbit and set the orbital velocity to 75%, then did a radial burn and adjusted the throttle until the ship was hovering: the required acceleration to hover was 3.36m/s^2 while the gravitational force was 7.87m/s^2. So looks like you're right: the ship is still receiving 42% of gravity at 75% of orbital velocity - now that's a big reduction, but definitely not linear. I wonder what the actual relationship is. – Blake Walsh Feb 6 '17 at 18:56
up vote 3 down vote accepted

Since "gravitational losses" is a rather loose term with no firm definition, let's introduce a term (already common in the Kerbal Space Program community) which allows quantifying them well: gravitational drag.

Gravitational drag can be understood as weight of a body in motion over planet's surface, adjusted for centripetal acceleration caused by motion relative to the planet's curvature. For immobile craft it will be its weight; for a craft in circular orbit it's zero.

Gravitational drag is an especially handy tool for calculations regarding spaceplanes - since it's the force directly counteracted by airplane's lift, it makes a very natural appearance in equations regarding spaceplane aerodynamics, in place where airplane equations use weight.

For rockets, it's offset by the vertical component of thrust, and rocket's total delta-V wasted on fighting gravitational drag is directly proportional to value of the gravitational drag and time spent under its influence; that way it's a good measure of the rate at which delta-V is lost.

It seems I was right; the relation between gravitational drag and difference between current and orbital speed is quadratic:

Centripetal force in circular motion is $$ F_c ={{m v^2} \over r} $$

Gravitational force:

$$ F_g = G {{m M} \over r^2} $$

Gravitational drag will be the difference:

$$ F_d = F_g-F_c = G {{m M} \over r^2} - {{m r v^2} \over r^2} = \left( {m \over r^2} \right) (G M - rv^2) $$

So, with constant altitude and masses, gravitational drag is proportionla to $(G M - rv^2)$

In practice, the effect is exacerbated by quite a few more factors:

  • mass of the craft is not constant - it drops with fuel burnt, making the part $ \left( {m \over r^2} \right)$ drop along with $m$ - extra drop of gravitational drag.
  • Mass dropping causes craft's TWR - and acceleration - to increase, so $v$ growth is greater than linear; thus rate of gravitational drag drop over time is growing in more than quadratic proportion.
  • The craft is frequently not moving horizontally, but still climbing, so the $r^2$ in nominator keeps growing $ \left( {m \over r^2} \right)$ drops; as factor of $v^2$, $r$ grows too, increasing the argument. This effect is not as pronounced, as $r$ is counted from center of Earth, and nearing apoapsis the craft loses vertical velocity.
  • the craft escaping atmosphere loses atmospheric drag (increasing acceleration again)
  • the engine operates in vacuum on a higher pressure differential than in atmosphere, resulting in higher thrust (higher acceleration again!)
  • the craft burns nearly horizontally, about all of its thrust contributing towards the horizontal speed, instead of being split between climb component and horizontal acceleration component.

While any of these side effects doesn't contribute as much as sheer horizontal velocity, they all add up, the result being gravitational drag vanishing way more rapidly near the end of the orbital insertion burn than during the earlier phase of flight - its drop in relation to burn time (as opposed to just velocity) becoming a good bit steeper than just quadratic.

Since the term 'gravitational drag' causes some controversy, let me present some arguments in its defense:

  • Atmospheric drag is a valuable tool in case of spaceplanes. At low airspeeds it degrades to weight, and so substituting it for weight in classic aviation equations allows extending them to cases where orbital mechanics begin to matter.

  • With these equations, the four fundamental forces in aviation become thrust, aerodynamic drag (opposing thrust), lift, and gravitational drag (opposing lift).

  • Atmospheric drag is linearly proportional to air density and drag coefficient of the craft. Gravitational drag is proportional to gravitational field strength $GM \over r^2$ (think "density of gravity") and mass of the craft.

  • Atmospheric drag is inversely proportional to square of speed relative to air. Gravitational drag is inversely proportional to square of speed relative to orbital speed.

It's pointless to use above orbital speed, and not needed at low speeds, but it's a very helpful quantity for ascent calculations.

  • 3
    Gravitational "drag" is a really misleading term for his. – Russell Borogove Feb 6 '17 at 21:42
  • @RussellBorogove: What's the correct one then? – SF. Feb 6 '17 at 22:45
  • 2
    @uhoh: In the KSP community it's already caught on. – SF. Feb 7 '17 at 11:38
  • 1
    @uhoh: Since the quantitative change of r is relatively small between launch and low orbit (and especially small when nearing apoapsis), its effect is very minor; approximating it with a constant doesn't make much of a difference. When already in orbit, using gravitational drag would be quite confusing; there's orbital mechanics for that. But it's a great tool for calculations of ascent, especially in case of spaceplanes (where it's directly opposed by aerodynamic lift). In normal airplanes, it's plain 'weight', but using 'weight' when mixing orbital mechanics in causes unending confusion. – SF. Feb 7 '17 at 13:20
  • 1
    (and I believe this is where the name comes from - I didn't invent it. As aerodynamic drag is offset by thrust, gravitational drag is offset by lift.) – SF. Feb 7 '17 at 13:25

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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