When you search up in google gravity assist formula, it comes up with the simplest version: enter image description here

Which means that: Final velocity = Initial velocity + 2 (velocity of the planet)

Then if we do some more research we get a formula that takes into account the angle of approach.

enter image description here

But how do we come up to this formulas? What is the reasoning behind it?

And is there any formula that takes into account the mass of both bodies and the distance between those (altitude)?

The source: http://www.mathpages.com/home/kmath114/kmath114.htm

After trying for different values of v and U, this is what we usually get (a V shape). This graph shows the final velocities for different angles of approach.enter image description here

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    $\begingroup$ I added the history tag because the origin of the formula may be much older than spaceflight, it could probably be derived from simple kinematics. Can you add a link to the source where you've found this formula? It might help if you look for other sources as well. I think the distance of closest approach (altitude) can be calculated from the initial conditions; once you choose $u$, $v$ and $\theta$ I think the distance of closest approach is already predetermined. It would be nice to have a diagram that shows what $\theta$ represents in your formula. $\endgroup$
    – uhoh
    Commented Oct 7, 2017 at 11:14
  • $\begingroup$ @uhoh θ represents an angle between 0 and 360, simple as that. $\endgroup$
    – Matthew
    Commented Oct 7, 2017 at 16:31
  • $\begingroup$ It looks like $\theta$ is the angle between the velocity of the planet and the velocity of the spacecraft; $cos(\theta) = \mathbf{v_x}\centerdot \mathbf{v_y} / |v_x||v_y|$ really just varies between -1 and +1, so $0° \le \theta \le 180°$. The equation and the final velocity don't care if $\theta$ is 90° or 270° for example, you'd get the same answer. $\endgroup$
    – uhoh
    Commented Oct 7, 2017 at 16:41
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    $\begingroup$ It's not "different angles of approach". The angle $\theta$ as shown in the link is $\pi/2$ minus half of the total bend angle resulting from the swingby. The extreme case shown in the diagram is $\theta=0$ or a bend angle of $\pi$. $\endgroup$
    – Mark Adler
    Commented Oct 8, 2017 at 4:22

3 Answers 3


You get that formula simply using the Pythagorean theorem, also known as vector addition. The link already provides $v_{2x}=v_1\cos\theta+2u$ and $v_{2y}=v_1\sin\theta$. Then you simply compute the magnitude $v_2=\sqrt{v_{2x}^2+v_{2y}^2}$.

Usually the body shown as a red disk in the diagram is much, much, much more massive than the thing following the black curved trajectory. In that case, the mass of the smaller object has no bearing on the trajectory.

As for the "altitude", the closest approach distance from the center of the body, along with the approach velocity and mass of the large body, determines $\theta$. See this answer for how to calculate it. ($\delta$ there is $\pi-2\theta$ here). It is of course important that the closest approach distance from the center of the body be greater than the radius of the body, lest you have a very violent and extremely short-lived swingby.

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    $\begingroup$ You do get a much more significant delta-v from the lithobraking approach, albeit probably not in the direction you want. $\endgroup$ Commented May 16, 2021 at 2:39

We need to have in mind the term relative velocity. Where Vi is Initial relative velocity and Vf is final relative velocity.

Vi = V1 + U

Vf = -V2 + U

Since Vi = -Vf then V1 + U = V2 - U ; after we do some simple algebra we get that V2 = V1 + 2U Where V2 is the final velocity of the space shuttle, V1 is the initial velocity and U is the orbital velocity of the planet that the space shuttle has approached to do a slingshot.


It is instructive to analyze the encounter from the inertial system at rest relative to the planet2; this inertial system is moving with the planet's velocity $U$ relative to the "resting" observer. This is possible because the physics don't change when we look at them from another inertial system.

From the planet's point of view the encounter is unspectacular: A small probe will approach, perform a U-turn and then leave with the same speed,1 which we call $v_{rel}$ (vor relative velocity, relative to the planet).

Now transform the probe's relative speeds to the ones a "resting" observer sees, simply by compensating for the planet's movement by subtracting its velocity vector. This means

  • subtract the planet's speed from the approaching probe (because its approach seems faster watched from the planet than it "actually" is since the planet is "actually" moving towards it);
  • and add the planet's speed to the leaving probe (its departure appears slower from the planet because the planet is following it).

Voilà, the result: The approaching probe's speed from an observer's point of view is $v_{rel}-U$ and the departing speed is $v_{rel}+U$, a difference of 2U.

This explanation is not a sleight of hand but perfectly valid physics. For a more intuitive approach, let's replace the gravity U-turn with an elastic collision, say with springs between the probe and the planet. (The physics are essentially the same because ideally no friction is involved.) The impact of the planet's motion then is two-fold: Not only do the springs get loaded more; the motion also adds more *oomph* to the back-bounce because it is pushing "in addition" to the springs.

This mechanism is what makes stacked balls bounce high, and as the linked page shows, can be understood with the same change of inertial systems.

1 We look at speeds "far away", ignoring the acceleration when entering and leaving the planet's gravity well which is a zero sum game.

2 The planet is on an elliptic orbit, so strictly spoken not the origin of an inertial system; but the error during the brief encounter is small. After all we usually consider physics classrooms inertial systems, even though they most certainly aren't ...


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