How do I find the delta-v loss of a rocket while ascending to its apogee?

Specification of Atlas V 541 rocket:

  • Specification for Atlas V first stage:

    • Dry Mass: 21054 kg
    • Fuel and Oxidizer(RP1/LOX) Mass:284,089kg
    • (Isp)(SL) = 311 seconds
    • (Isp)(vac) = 338 seconds
    • Thrust(SL) = 3,827 KN
    • Thrust(vac) = 4,152 KN
    • First stage wet mass: 305,143 kg
    • Liftoff TWR: 1.179
    • Maximum TWR: 9.06 (Will be throttled down due to G-Force)
    • Operating time: 253 seconds

  • Specification for Atlas V second stage:

    • Dry Mass: 2,243 kg
    • Fuel and Oxidizer (Hydrolox) = 20,830 kg
    • Fairing: 1,500 kg
    • Equipment and imaginary payload: 600 kg
    • (Isp)(vac):449.7 seconds
    • Thrust(vac) = 101.8 KN
    • Second stage wet mass: 25,673 kg
    • Minimum TWR : 0.404
    • Maximum TWR : 3.105
    • Operating time: 926 seconds

  • Rocket Total Mass (Both Stages): 330,816 kg

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    $\begingroup$ Can you please edit that wall of text for readability? $\endgroup$ Commented Mar 27, 2017 at 19:45
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    $\begingroup$ @OrganicMarble I have updated, do you mind check it again? $\endgroup$
    – Raze
    Commented Mar 27, 2017 at 20:53
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    $\begingroup$ Much more readable. I see what you did there :) $\endgroup$ Commented Mar 27, 2017 at 20:54
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    $\begingroup$ You'll get better answers if you try to show what work you have done so far. Just "How do I do X?" may not inspire many answers. One problem would be that it is hard to tell exactly what it is you know already and don't know, and it's no fun taking time to write an answer to then find out "oh I know that already, what I really want to know is Y!" Can you outline how you think you might calculate it? Or at least link to the definition of $\delta v$ loss that you are using? It makes it easier to give an answer that you'll find helpful. $\endgroup$
    – uhoh
    Commented Mar 28, 2017 at 2:26
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    $\begingroup$ @uhoh -- Trajectory matters. $\endgroup$ Commented Mar 28, 2017 at 3:14

1 Answer 1


This is the problem known as launch trajectory optimization. That is a complicated problem mostly requiring a numerical simulation. That does however not mean we can obtain some useful figures and insights from an analytical approach!

To begin, we can with an inaccurate but usable estimate. The flawed assumption is as follows: "Our rocket is fighting against gravity for a time of $T$, so the lost $\Delta v=Tg$"

The Atlas rocket is flying for $253s+926s=1179s$, so the lost $\Delta v=1179s \cdot g=11500\frac{m}{s}$. That is more than the total $\Delta v$ of the rocket...

So what is the problem? Turns out force is a vector quantity and not a scalar. Our rocket is mostly not accelerating in the direct opposite direction of gravity, in fact, we want to gain velocity normal to gravity to reach an orbit.

If we draw some force vectors we can clearly see that the total acceleration is larger than the value we get from simple subtraction, depending on the angle. (Accelerating downwards, we even get a small bonus from gravity).

force vectors

Complication 1: $\Delta v$ loss depends on trajectory

But there are more things to consider than this. For instance, why does the rocket not have to fight gravity anymore? Well, it has gained orbital velocity. But "orbital velocity" is not a binary toggle, everything is an orbit, the slower ones just happen to intersect Earth, not desirable for a satellite.

Instead, we are thrusting against a "vertical acceleration" (The direction of course varies while we are orbiting, so we are viewing this in a rotating frame of reference). We can express this from the circular orbital velocity, $v_c$:

$$a_{vertical} = 1 - \frac{v_{horizontal}^2}{v_c^2}$$

Complication 2: The vertical acceleration depends on velocity

So, an ideal strategy would be to gain horizontal velocity as fast as possible, with just enough vertical acceleration to counteract gravity. This is not very useful though as the drag is going to be extremely large and your rocket will end up crashing into the side of a mountain.

A compromise is to start in an almost vertical position, and always accelerate parallel to the current velocity vector. This is efficient as adding together vectors in parallel gives the longest resulting vector. Thir trajectory also has the nice property that the gravitational acceleration is going to slowly turn the rocket horizontal. With some planning, the rocket ends up with orbital velocity at the target altitude. This is known as a gravity turn.

You now have:

  • Horizontal acceleration, depending on gravity and vertical velocity
  • Gravity, depending on altitude, depending on vertical acceleration over time
  • ...
  • And a ton more.

The resulting system of differential equations is way to complicated to solve properly.

On top of that, you have to consider the atmosphere, as that affects your trajectory (and $I_{SP}$...)

TL;DR You must use a numerical simulation.

So then, how could such a simulation be implemented? Here is a quick sketch: (svg source in answer source for anybody wanting to improve it)

ascent trajectory

  1. Initial ascent (red). Accelerate vertically, and simulate drag losses. Most of those should be here.
  2. Gravity turn (blue). Start to pick up some horizontal velocity, and let gravity turn your rocket over. Repeat this for multiple angles to get your preferred final altitude. Ignore drag for this part.
  3. Final orbit (pink). No more acceleration required.
  • $\begingroup$ Gravity turns don't simply burn along the direction of movement. There's an initial purely vertical ascent, a small adjustment to get a small horizontal velocity component, an ongoing adjustments to keep the trajectory optimized. All of this is very nearly aligned with the direction of movement, but it's a tightly controlled process. $\endgroup$ Commented Mar 28, 2017 at 14:13
  • $\begingroup$ @Schlusstein You are completely right, you can generalize the manoeuvre in that way. It is equivalent of saying that the initial velocity can be something else than zero. $\endgroup$ Commented Mar 28, 2017 at 14:16
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    $\begingroup$ I thought the tl;dr goes at the beginning, or are you being clever rendering it moot by putting it at the end, where one sees it only after reading? Or does it in fact stand for too late; did read? :) $\endgroup$
    – uhoh
    Commented Mar 28, 2017 at 14:58
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    $\begingroup$ @uhoh Let us assume that for a GTA (general text area), somebody has some RC (reading coefficient) between 0 and 1, where 0 is simply looking at the length and 1 is completely reading the text. Let us further assume that there exists a threshold of RC so that for any value below it somebody does not see the bold-text "TL;DR". Your claim is then that the introduction sentence ∉ GTA so everybody can see the "TL;DR" part regardless of their specific RC at the moment, the purpose of such a text is to have a threshold close to 1. Is that an accurate analysis of your comment? TL;DR :P $\endgroup$ Commented Mar 28, 2017 at 15:40
  • $\begingroup$ I'll do some Monte Carlo simulations, and get back to you. ;) Right now I'm not sure anymore. $\endgroup$
    – uhoh
    Commented Mar 28, 2017 at 15:42

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