Why does a gravity capture require a loss of velocity?

I've read a gravity capture of an object requires the captured object to lose velocity by some means (impact, third body tidal force, etc).

Is there not a possible encounter that happens at a low enough speed to be below escape velocity? Even for objects on nearly identical orbits?

• Several of Saturn's and Neptune's moons (and probably those of some other planets) are believed to be captured bodies. So you must be correct - it's possible to be captured without needing any form of braking.
– Andy
Apr 11 '16 at 14:01
• I would suggest this - on entering a planetary system which also has moons. It should in theory be possible to enter the system, closely approach a moon and obtain a gravity-assisted reduction in speed, sufficient to remain in the system. Far beyond my maths abilities to calculate though!
– Andy
Apr 11 '16 at 14:10
• Early moons would form from collisions in the gas cloud that was condensing to form a planet; later moon captures would occur via gravity assist from the earlier-formed moons. Apr 11 '16 at 16:12
• @andy Incoming and outgoing Vinfs are the same wrt to a gravitating body, it is the direction that is changed. However a change of direction wrt to one of Saturn's moons can result in reduced velocity wrt Saturn. Thus it's possible for a body with substantial moons to capture asteroids drifting by. However I remain perplexed over Phobos and Deimos. Apr 11 '16 at 17:24
• Long story short: the gravity of the Primary always gives the approaching object exactly enough Velocity to return it to the same distance it started, but going in the opposite direction. So any approaching object (not already captured) is always going too fast to ever be captured. (Unless you add something else to take away that extra velocity). Ultimately, this is because gravity well's are symmetrical: the velocity they take away as you move away from them is exactly the same as it gave you when you approached it. Apr 11 '16 at 20:11

No, actually. Imagine that you launched a rocket straight up, to the point where it wasn't moving at all just where the Earth's gravity was essentially 0. Dropping it, it would gain the Escape Velocity of Earth, and in fact it would always have that escape velocity. It would have to somehow slow down for it to not eventually leave Earth's gravitational field.

Essentially, a rocket that goes straight up and just barely hits escape velocity will slowly hit 0 velocity (Asymptotically) If when it was approaching that 0, something nudged it back, that would be the slowest an object could intersect, which would be just over escape velocity.

• Thanks for the response! I'm not quite sure I understand your given example. Why would the falling object necessarily have the escape velocity? Because the gravity is effectively 0 at its location? Apr 11 '16 at 14:16
• Escape velocity is the speed to escape the pull of gravity. An object entering with basically no speed will be at escape velocity the entire way down, although in the wrong direction. Apr 11 '16 at 14:19
• because 0 m/s relative velocity at gravity 0 is already the escape velocity, so even if it was slightly higher relative speed but moving closer to the planet, any pull of gravity on the object from the planet would further increase its speed which has already met the escape velocity needed... did I follow that correctly? Apr 11 '16 at 14:29
• Yeah, pretty much. Apr 11 '16 at 14:31

The speed of something in a hyperbolic orbit is $\sqrt{V_{esc}^{2} + V_{inf}^{2}}$ where $V_{esc}$ is escape velocity and $V_{inf}$ is the object's speed when it's an infinite distance from the gravitating body.

At infinity incoming and outgoing speeds are both the same on the arms of a hyperbola.

If you drop $V_{inf}$ to zero, you have a parabola. In which case $\sqrt{V_{esc}^{2} + 0^{2}}$ reduces to simply $V_{esc}$. At every point of a parabolic orbit, the object is traveling escape velocity.

The parabola is sort of a boundary conic between hyperbolas and ellipses.

To be in an elliptical orbit (in other words, a capture orbit), the velocity must be less than $V_{esc}$.

Illustrations are from my coloring book Conic Sections and Celestial Mechanics.

• To head off comments on Comic Sans, I like the font. It's informal and legible, just what I want in many cases. I do not care what The Oatmeal thinks. Apr 11 '16 at 18:53
• Now justify the bubble letters ;). Joking, great info-art. Dec 11 '18 at 19:36

The short answer is yes. There is a way for a planet to capture an object without aerobraking or rockets.

The long answer is, It depends on what model you use to describe the system. And it depends on what you mean by capture. Engineers will use various simplifying assumptions to model gravitational systems. They will switching between models as needed to get practical results.

Using a restricted 2 body model: If Mars was in an inertial plane and I was sitting at a distance of Earth's closest approach of 78.3 million km the escape velocity would only be 75mph(120kph). If I were to drop a Satellite from this height it could form a highly elliptical orbit who's apogee would be 78.3 million km.

A more sophisticated model: A more practical model would be to launch a satellite to earth escape velocity and then launch it on a Hohmann transfer orbit whose apogee just barely gets close enough to Mars that Martian gravity can capture it. The resulting orbit would be highly elliptical and this maneuver would require small navigational thrusts to get it just right. This maneuver can be broken down into 3 steps: 1 Earth-centered restricted 2 body model, one solar and one Martian. It is unlikely this would happen naturally with a random asteroid.

In a restricted 3 body model a planet and a moon can create a gravitational bottle that can capture asteroids naturally. These bottles have openings around the 2nd and 3rd Lagrangian points but only for satellites with a certain energy. These Orbits are chaotic however. A captured asteroid in this manner will have a half life before it is ejected through one of the holes or crashes into a planet or moon.

I suspect that most of the moons you see in highly circular orbits are formed from aggregate ejecta.

Here is a video:

I recommend reading this first http://dev.whydomath.org/node/space/index.html

• What is a "homen transfer orbit" ? A Hohmann transfer orbit?
– Uwe
Dec 10 '18 at 15:33
• "...whose apogee just barely gets close enough to Mars that Martian gravity can capture it." At a near Mars aphelion a ship from earth would be moving about 2.7 km/s slower than Mars. Mars capture isn't possible without aerobraking or a burn. Dec 12 '18 at 2:24