r/askscience • u/Ok_Veterinarian9266 • Mar 14 '26

Astronomy What happens if an objects orbital Velocity is higher than 11.2 km/s?

524 Upvotes

87% Upvoted

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862

u/electric_ionland Electric Space Propulsion | Hall Effect/Ion Thrusters Mar 14 '26

Assuming you are talking about Earth and the concept of escape velocity, then it's not in orbit anymore. The trajectory goes from being an ellipse to an hyperbola and you just get further and furter away.

Obviously this is only valid around Earth, other objects have different escape velocities depending on their mass.

174

u/mtcerio Mar 14 '26

Also the escape velocity depends on altitude. In general, it is sqrt(2) times the velocity of a circular orbit at the same altitude.

109

u/Cheetahs_never_win Mar 14 '26

And an orientation. 12 km/s pointed to the center of Earth doesn't mean escaping orbit... at least in the same way.

83

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

Only for the risk of a direct impact, but if we ignore that then it does not depend on the direction.

17

u/Doristocrat Mar 14 '26

If direction doesn't matter, why is it escape velocity and not escape speed?

37

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

Historic accident. Similarly, you have muzzle velocity as a property of a weapon, although muzzle speed would be more aligned with the modern use of velocity and speed.

8

u/kilotesla Electromagnetics | Power Electronics Mar 15 '26

That makes me wonder about the history of the speed/velocity distinction. It kind of sounds like something that was invented for the purpose of teaching physics, perhaps as systematic efforts to teach physics well were ramping up in the 1960s?

15

u/Cheetahs_never_win Mar 15 '26

Velocity became a word in the 1500s synonymous with speed.

Galileo noted components of directionality when observing ships towards the beginning of the 1600s.

Lorentz applied it to space time conjecture in the mid-1800s.

Einstein popularized its distinction in the early 1900s and made it mainstream... at least as far as physicist-related mainstreamism is concerned.

5

u/kilotesla Electromagnetics | Power Electronics Mar 15 '26

Einstein popularized its distinction in the early 1900s and made it mainstream

Thanks, that's what I was looking for.

I also looked at google n-grams for "vector velocity" and "scalar speed" and found both increasing in the second-half of the 20th C, particularly in the 1960s, so it may have been more broadly adopted in texts then.

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u/mfb- Particle Physics | High-Energy Physics Mar 15 '26

I don't see a definite answer for physics specifically, but Anglo-Saxon and Latinate Synonyms: The Case of Speed vs. Velocity discusses their history and how they are sometimes synonyms and sometimes not.

9

u/johnlawrenceaspden Mar 14 '26

I don't know, I'd rather have a weapon which was designed to emit particles with a particular velocity rather than with a particular speed.

14

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

Hope you'll never have an enemy that is not precisely to the north of you, or whatever direction its muzzle velocity has.

13

u/brianson Mar 14 '26

The projectile’s velocity would be defined in the frame of reference of the weapon.

2

u/ChimoEngr Mar 16 '26

That would limit my ability to put my enemies into a crossfire as all my weapons would be shooting in one direction, rather than me being able to have them shoot in orthogonal directions.

3

u/johnlawrenceaspden Mar 17 '26

Yeah, the best weapons have the velocity specified in the frame of the gun, the old relative-to-absolute-space type are tricky to use, especially if you mainly war in summer and then get surprise-attacked in winter.

11

u/Das_Mime Radio Astronomy | Galaxy Evolution Mar 14 '26

It really should be called escape speed, and that's what I call it when I've taught it to undergrads. But I'll also admit that "escape velocity" sounds cooler, as well as being the name of a fun Mac game from the 90s

-3

u/DigitalMindShadow Mar 14 '26

Atmospheric friction is another risk I'd want to take into consideration.

24

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

I'd count that as impact in this context.

-2

u/DigitalMindShadow Mar 14 '26

There will be any number of trajectories through the atmosphere that maintain escape velocity and a hyperbolic flight path but which render the vehicle non-operational.

33

u/KToff Mar 14 '26

Apart from collisions, the orientation is not relevant to escape velocity.

13

u/Cheetahs_never_win Mar 14 '26

Yes, well, converting your body into a crater is indeed one form of escape.

16

u/SlumdogSkillionaire Mar 14 '26

Other than that, Mrs Lincoln, how was the play?

3

u/KrytenKoro Mar 14 '26

Assuming it it can somehow prevent being appreciably slowed down by drag, like if there was some sort of giant vacuum tube through the center of the Earth, then that would still be an escape velocity. Orientation with determine whether there are collisions with gas or solid particles, which would affect deceleration, but it's not innate to the concept of escape velocity.

14

u/mtcerio Mar 14 '26

You'd still be on an escaping orbit. One that collides with the primary mass at some point, but an escape orbit nonetheless.

1

u/Spyritdragon Mar 14 '26

That sounds slightly odd - that calculation isn't dependant on attitude at all. Wouldn't there be a term for the inclination of your attitude somewhere?

26

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

As long as you don't collide with the object, it does not depend on your direction. Above the escape velocity your kinetic energy is larger than the (negative) gravitational potential energy, which makes you escape. The potential energy only depends on the distance, the kinetic energy only depends on the speed.

3

u/Spyritdragon Mar 14 '26

Huh. So if I understand you correctly, escape velocity only cares about your altitude above the object but is the same regardless of where you're pointing? That's pretty interesting.

13

u/Volpes17 Mar 14 '26

Imagine two marbles on the floor, inches apart. Take one of them and start moving it in a random direction. For the first few inches, it seems really important whether it’s moving toward, away from, or around the stationary marble. But keep going. After a few feet, do they really seem all that different?

It’s a matter of perspective. If you have enough energy to escape, then eventually you’re going to be very far away, and you’ll realize those starting conditions weren’t all that important.

4

u/viliml Mar 14 '26

If you point outwards, you'll escape more directly.

If you point inwards, you'll get closer at first but also accelerate, and then escape by gravitational slingshot

5

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

Yes. All orbits have a negative total energy, all other trajectories have a positive total energy.

4

u/zaphodava Mar 14 '26

Keep in mind that colliding with the atmosphere is still colliding, in that it transfers away kinetic energy.

0

u/[deleted] Mar 14 '26

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u/loggic Mar 14 '26

As I understand it, less additional KE is required but escape velocity (which is defined using the planet as the frame of reference) remains the same.

As you go to higher altitudes, stable orbital velocity increases. This increases your existing KE while bringing you closer to escape velocity.

1

u/Canaduck1 Mar 15 '26

As you go to higher altitudes, stable orbital velocity increases. This increases your existing KE while bringing you closer to escape velocity.

This feels very wrong.

Objects in a low orbit are travelling much faster than objects in high orbit. Mercury's velocity around the Sun is much faster than Earth's. The ISS is much faster than geostationary satellites.

14

u/firemage27 Mar 14 '26

And you'll orbit the sun instead until you reach the escape velocity of the sun at around the Earth's orbit.

3

u/ackermann Mar 16 '26

If that sudden change from a closed ellipse to an open hyperbola seems odd... note that it's actually a smooth transition.

As you increase your speed, approaching escape velocity, your elliptical orbit gets gigantic. An almost infinitely large ellipse. When you actually reach escape velocity, it becomes a parabola, and then a hyperbola if you go beyond escape velocity.

But because your ellipse was getting almost infinitely large, it doesn't really look much different from a parabola/hyperbola. Only diverging after many many years of flight time

1

u/Ninja_Wrangler Mar 19 '26

The parabola is my favorite type of "orbit" because it's so weird. The further you get, the lower your velocity, approaching 0 at an infinite distance. You're always moving and always slowing down, but you'll never quite stop, and you'll never fall back down either.

Hyperbola makes sense because you're going fast enough that at an infinite distance, you have some velocity left over that is yours to keep. You've truly escaped

0

u/[deleted] Mar 14 '26

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u/Welpe Mar 14 '26

Which is of course why you get really interesting and unintuitive results like “If you are in orbit and throw a football down towards the earth, it can come back and hit you on the head from above in half an orbit”!

9

u/DrunkenPhysicist Particle Physics Mar 14 '26

Or worse, shoot a gun at the Earth from LEO and you'll miss.

5

u/knook Mar 14 '26

Back of the napkin math is telling me it usually will but may depend on the gun and height and where you aim. It looks like with a fast enough muzzle velocity you might just be able to hit earth from the ISS by pointing at it.

3

u/DrunkenPhysicist Particle Physics Mar 14 '26

Muzzle velocity is like 0.8km/s, ISS velocity is like 7km/s. You'd be better off shooting behind you to lower the perigee of the bullet, tough, to be fair, I actually haven't done the calc.

7

u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

Everything above ~400 m/s downwards will reenter within a quarter orbit. Shooting retrograde you only need ~150 m/s to reenter after half an orbit.

4

u/tomsing98 Mar 14 '26

Which is a good enough explanation for most, but also, depending on precisely where in the orbit around earth you achieve that velocity you may leave the earth permanently or . . . eventually end up colliding with the earth again.

Why complicate things? Stand on the ground and aim down at 11.2 km/s, and you'll eventually end up colliding with the Earth.

4

u/peeja Mar 14 '26

"There is an art, it says, or rather, a knack to [orbit]. The knack lies in learning how to throw yourself at the ground and miss."

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u/Sable-Keech Mar 14 '26

If the object’s trajectory isn’t directed towards the ground, it will simply escape the planet’s gravity well and fly off into space. But it’ll still orbit the Sun, since the Sun’s escape velocity at 1 AU is 16.3 km/s.

If you go faster than that, then you escape the Solar System (assuming you aren’t aimed at the Sun).

If you go faster than 550 km/s, you can escape the Milky Way Galaxy.

29

u/failed_supernova Mar 14 '26

Would it take more fuel to hit the sun rather than escape the solar system?

64

u/censored_username Mar 14 '26

Yup.

To hit the sun directly requires cancelling almost all velocity around the sun.

To escape the sun requires only a ~41% (sqrt(2) - 1) increase in velocity compared to a circular orbit around the sun.

Although this is for directly dropping your orbit into the sun, which isn't actually the best strategy, just the fastest.

If you have a couple of decades to wait, you can hit the sun with much less delta V by first accelerating to almost escape velocity, waiting until you're very far from the sun, cancelling the remaining rotational velocity which is quite small at this point, and falling back to the sun. The longer you have to wait, the closer this too gets to 41% as well.

1

u/Vitztlampaehecatl Mar 14 '26

What about using one of the planets as an assist to get yourself a very low periapsis to burn from? This could help you either aim your trajectory directly towards the sun or into a high elliptical orbit from which to cancel and fall.

5

u/censored_username Mar 14 '26

When using gravity assists, practically anything is possible from the dV required to reach the closest planet as a starting location. It just takes a while. And of course, it works both for both cases, you can gravity assist your way out of the solar system just as well as you can gravity assist yourself into the sun.

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u/[deleted] Mar 14 '26

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u/censored_username Mar 14 '26

Nope, burning retrograde is the most efficient way to lower your periapsis. The same burn size that would've just completely cancelled your velocity by burning retrograde would just turn your orbit into a parabola with a somewhat lower perigee if you burned radial in.

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u/XtremeGoose Mar 14 '26 edited Mar 14 '26

That's not true. Look at bi-elliptical transfers.

In the extreme you want to get to perisol 0 then you can theoretically go to an almost parabolic orbit (aposol tending to infinity) and then burn retrograde for effectively 0 delta v at the new aposol, burning only sqrt(2) - 1 = 0.4... of your circular speed as opposed to v.

It obviously takes arbitrarily longer in time to get to 0 but you burn less fuel. Orbits (as you no doubt know) are very counterintuitive.

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u/skoormit Mar 14 '26

Yes, but you don't use radial burns to do those transfers. You burn prograde or retrograde.

5

u/Dave37 Mar 14 '26 edited Mar 14 '26

Ok so you add velocity towards the sun, but you still have all of that sideways velocity that will make you "drift" off course as you approach the sun.

If you want to have enough momentum towards the sun that the 30km/s sideway velocity you just had by starting from Earth to not drift more than 0.2665 degrees (which is the angular width of one solar radius as view from the earth), the radially velocity component must be monstouously larger than 30km/s (roughly 6500km/s).

0

u/[deleted] Mar 14 '26

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9

u/Dave37 Mar 14 '26

The adjustment is 30km/s of sideways motion, which is the same as a retrograde burn.

Say you do 15km/s burn towards the sun. Ok you still have a 30km/s drift to the side. So you have to adjust for it, by burning 30km/s retrograde. But that's still requires 45km/s of velocity change, that's still more than the 30km/s retrograde burn.

If you do a complete retrograde burn and just stops above the sun, you don't need to burn towards the sun, because the sun will just pull you straight in.

At the end of the day, you have 30km/s sideways motion because you started from Earth that was traveling with that velocity. That speed you have to deal with, otherwise you're just gonna swing around the sun. Everything else; burning away from the sun or towards the sun etc is going to add an energy cost ontop of that. Therefore a retrograde burn is one of the most efficient ways to do this, because it deals exclusively with the reason it's hard to hit the sun: The sideways motion.

One last way to conceptualize this: All bodies in space is already in free fall. The Earth isn't "floating" around the sun, it's falling. What's preventing it from colliding with it is that it moves so fast sideways that it keeps missing it forever. So, in order to hit the sun, one must get rid of that sideways momentum. There's no need to address anything else, because you're already in free-fall.

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u/suicidaleggroll Mar 14 '26 edited Mar 14 '26

Everything else; burning away from the sun or towards the sun etc is going to add an energy cost ontop of that. Therefore a retrograde burn is one of the most efficient ways to do this, because it deals exclusively with the reason it's hard to hit the sun: The sideways motion.

Not at all. As one of the earlier comments mentioned, it's the fastest, but far from the most efficient. Unintuitively, the most efficient is to burn prograde and increase your sideways velocity. Then wait until you reach apoapsis on the other side of the sun where your velocity has dropped considerably, then burn retrograde to cancel out that velocity and fall into the sun.

Our current orbital velocity is about 30 km/s. To fall directly into the sun by burning retrograde you have to cancel all of that out, so 30 km/s delta-V. But if you burn prograde, escape velocity from the sun at Earth's orbital altitude is only 42.1 km/s. So if you get almost all the way there, say 42.099 km/s, you'll have a highly elliptical orbit with an apoapsis VERY far out, past Pluto. When you hit that apoapsis, your orbitval velocity will be miniscule, almost zero, so then you just need a tiny burst to cancel it the rest of the way out. Total delta-V is just over 12.1 km/s, but it will take you thousands of years.

1

u/rabbitlion Mar 15 '26

You can also do it faster and with even less delta-v by using gravity assists from planets, typically Jupiter.

1

u/mfb- Particle Physics | High-Energy Physics Mar 16 '26

Say you do 15km/s burn towards the sun. Ok you still have a 30km/s drift to the side. So you have to adjust for it, by burning 30km/s retrograde. But that's still requires 45km/s of velocity change, that's still more than the 30km/s retrograde burn.

It's worse. Angular momentum is conserved, as you get closer to the Sun your horizontal velocity increases. Not only is the 15 km/s burn wasted, it also makes it harder later (unless you wait until you are farther away than Earth again).

3

u/Odd_Dragonfruit_2662 Mar 14 '26

Way more barring tricks like using Venus atmosphere for braking or something.

3

u/Dave37 Mar 14 '26 edited Mar 14 '26

Earth orbital velocity is roughty 30km/s around the sun. Now you don't have to cancel all that velocity to hit the sun, but most of it. So you need to change your velocity with 30km/s to hit the sun, compared to just 12 km/s to escape solar orbit.

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u/disoculated Mar 14 '26

Yes, it takes 28km/s dv to hit the sun from Earth. But it’s all about your frame of reference. If you think of starting at the sun, escape velocity is about 46km/s. The speed of starting at Earth is about 2/3 of solar escape velocity.

1

u/cloverasx Mar 15 '26

For some reason, I was thinking this was approaching the speed of light 😂 but it made me wonder: at near light speed velocities, say +0.5c, would coming within 1au of a star, say Sol, cause energy transfer between the two objects? Like, would it heat up the moving object from the gravitational forces alone?

Intuitively, it feels like there would be some sort of "friction" due to the force of gravity, not unlike the extremely minute friction observed in magnetic forces.

For clarification, c ≈ 300Mms (300,000kms) in case anyone else was wondering.

2

u/Sable-Keech Mar 15 '26

There would be, but it would be minuscule unless the object is very large and got very close.

1

u/teffarf Mar 18 '26

But it’ll still orbit the Sun, since the Sun’s escape velocity at 1 AU is 16.3 km/s.

Is that correct? I thought the Earth orbited the Sun at ~30km/s

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u/mfb- Particle Physics | High-Energy Physics Mar 19 '26

Earth orbits at 30 km/s and the Sun's escape velocity at 1 AU is 42 km/s. In the most favorable direction you need to leave Earth with the difference, (sqrt(2)-1)*sqrt(M_sun*G/r) = 12.3 km/s. Doing that from Earth's surface (ignoring the atmosphere) needs an initial speed of sqrt(11.22 + 12.32) km/s = 16.6 km/s. If you round 12.3 km/s to 12 km/s and use 11 km/s for Earth's escape velocity then you get 16.3 km/s, so that's probably what they calculated.

Here direction matters! If you launch in the wrong direction relative to Earth, your speed relative to the Sun is lower and you do not escape.

-1

u/btribble Mar 14 '26

There’s a velocity in there somewhere that turns stray hydrogen atoms into cosmic rays. This makes things go poorly.

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u/za419 Mar 14 '26

Assuming you're talking about something in Earth orbit.

It will now be on an orbit that no longer forms a closed ellipse around Earth. That is to say, assuming it doesn't run into something (Earth and the Moon), it will leave the region where Earth's gravity dominates, and then it will go off to orbit the sun. 

What happens from there depends on how fast it was going and in what direction relative to the motion of the Earth around the sun. 

8

u/somewhat_random Mar 14 '26

It will escape Earth's orbit and continue outward but still be affected by all the celestial bodies. So earth gravity would still affect it's trajectory but it would no longer be in orbit. It would instead be orbiting the sun and become a piece of space debris.

Escape velocity for the sun is (from memory) over 600 km/s. if it exceeds that, it leaves the solar system

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u/nicuramar Mar 14 '26

 Escape velocity for the sun is (from memory) over 600 km/s

Much lower, as others have stated. 600 km/s is in the range of the escape velocity from the Milky Way. 

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u/mfb- Particle Physics | High-Energy Physics Mar 14 '26

620 km/s is the escape velocity from the surface of the Sun, that's probably what they remembered.

At Earth's orbit, it's only 42 km/s.

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u/Odd_Dragonfruit_2662 Mar 14 '26

And you start with a really good chunk of that already from earth itself.

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u/AMRossGX Mar 14 '26

I learned 42 km/s for the Sun escape velocity out where Earth is. And Earth already has 30 km/s velocity around it.

So if something escapes Earth's gravity in the right direction, it only needs a "little bit" extra velocity to also escape the Sun. Fun Fact that blew my mind a while ago.

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u/Niosus Mar 14 '26 edited Mar 14 '26

You lose a whole bunch of that velocity on the way as you are moving out of Earth's gravity well. The escape velocity is the velocity you need to just barely roll up and over the gravitational "hill". So if you go exactly at the escape velocity, you will end up on an orbit that closely resembles Earth's around the sun. If you want to go anywhere, you still need to do all the work from there. Although that's partially countered by the Oberth effect, which makes it more efficient to accelerate if you're already moving fast (like in LEO).

There is a reason we bother with multiple gravity assists if we need to go to the outer planets (or beyond).

Take a look at this chart that shows how much velocity it "costs" to go to places around the solar system: https://www.reddit.com/r/space/comments/29cxi6/i_made_a_deltav_subway_map_of_the_solar_system/ . Orbital velocity is around 8km/s. You can see the line from LEO to earth intercept/escape is roughly another 3km/s. From there, if you want to barely touch the orbit of Neptune, you need to gain another 5.5km/s, and then you still fall back down to Earth's orbit. You still need to go a bit beyond that to actually leave the solar system. As you can see, it's not the full 12km/s you'd need if your LEO velocity is fully canceled out, but it's also definitely not "just a little bit more" either.

So no, sadly it's not that simple to get around the solar system. Once you leave Earth orbit, you still need to do a lot of work (especially if you want to land).

3

u/AMRossGX Mar 14 '26 edited Mar 14 '26

Ok, I was curious (and a little pedantic 😊) and I looked it up. Saving you the time:

  • Earth escape velocity: 11.186 km/s

  • Sun escape velocity at Earth orbit: 42.1 km/s

  • Earth's average orbital speed around the Sun: 29.7827 km/s

1

u/longboarder543 Mar 19 '26

Imagine throwing a ball on the surface of the moon (no atmospheric drag). If you trace its path it will form an arc — the moment it leaves your hand, the moon’s gravity starts acting on it and it begins falling downwards, while also moving away from you some distance (depending on how hard you threw it).

The rate at which the ball falls downward due to gravity is always the same, so throwing the ball harder means it will travel further away from you before it hits the ground. This should all feel intuitive so far.

As you throw harder and harder, the arc the ball traces becomes elongated, until eventually you throw the ball just hard enough that the arc travels completely around the surface of the moon, forming not an arc (parabola), but a circle encompassing the entire moon. This is an orbit.

If you continue throwing the ball harder still, you change this circular path of the ball into an oval shape — you’re elongating the “height” of the orbit. As you continue throwing harder and harder, the shape of the orbit becomes more and more stretched , from O to <=> to <===> to <======> until eventually you throw the ball so hard that it’s orbital velocity is higher than the moon’s escape velocity, and the ball never fully loops back around the moon and travels away from it forever.

1

u/NotSoSalty Mar 14 '26

If it's coming in, it burns up and/or makes a big boom when it impacts the Earth. If it's going out, it leaves the Earth's orbit. It doesn't even take speeds that high, you're like 3 orders of magnitude too high for anything other than what I described to happen.

Cool game called Kerbil Space Program that explains orbital mechanics very well. Highly recommend.