Orbital Mechanics on Paper 3 - Escape Velocity

[@NJP695]

You should bring this series back! I mean, studying Physics in college makes me kinda bias, but still calculating interplanetary travel or even calculating gravity assists are ridiculous fun!

[@louismenke8002]

NJP695 I totally agree to that! Lovely stuff. Awesome stuff.

[@Jamesdavey358]

Biased*

[@corajallen9471]

Scott Manley I wanted to thank you for your video "Orbital Mechanics on paper" Your video helped my friend and I (Mostly him) build a calculator that accurately calculates the velocity and deltaV requirements for the altitude you desire. We will be uploading the calculator with instructions for use on the reddit for KSP when it is ready. When used correctly The excel will be able to calculate both velocity and delta V for all of the planets and moons in KSP as of this date. We hope to help others create rudimentary flight plans like we have been! This was super exciting for us because we discovered that the KSP wiki has incorrect escape velocities for at least Kerbin. We will keep you posted when we release the calculator if your interested! Again thank you so so much for your videos and all of your contributions to KSP, KZbin, education and science!

[@firestar509]

I was wanting to learn orbital mechanics, but was having a hard time. But you Scott have helped me understand them quite a lot better then I was, so all I can say is thank you Scott.

[@burningcusader]

It's fun to see that the stuff i'm learing in physic class actually has some real life application. Really nice video Scott!

[@Laylander]

As a mechanical engineer i am used to this kind of language, so i was able to understand, but in my opinion you where going a bit too fast to let it sink in for the 'non engineer'. Loved the video, definitely going to calculate more by myself instead of using tools from now on! :)

[@is_it_zeek]

I'm doing A Level Physics at school now and in a few months time will be looking into orbital mechanics, videos like these I really like because it helps me get an early grasp of what I need to learn. Thank you Scott Manley!!!

[@cengizteouluyurt7053]

Only thing i left to do on this goddamn world is visit you in person and have a chat for few hours. You simply shaped my today form scott. I really respect you. In all seriousness if i ever get a chance, i wasn't joking, i would actually want to visit you in person. Take care scott, and please get this serie back online. It really sharpens my knowledge above what i learn from school

[@breno.r]

I just graduated highschool, have been in college for two weeks, but I found this pretty easy to understand. You're awesome, Scott.

[@Freak80MC]

This one kinda went over my head vs the other 2 but I'll definitely be back to rewatch it, really appreciate these videos

[@themistoclesleonidas]

Awesome video. I've always understood the effects of burning near a body. But it's cool to see exactly why it makes a difference.

[@DeepakGuzzula]

Great videos scott!! Do keep dropping these nuggets of info. Makes for a much more interesting take on the gameplay.

[@Jbenneballe]

orbital mechanics is the subject in my physics class atm. got them all to play KSP as a studdy aid. :) i'll direct them to these videos aswell. Scott is so much better than my teacher.

[@gameboxfreak]

Are you ever going to continue this series Sott? I'd love to learn about calculating rendez vous maneuvers. and how to calculate the most efficient way to get to other planets. Ohh! and calculating the delta-v of a spacecraft! :D

[@Hilgy2911]

I get it now. It's all magic!

[@Justy42069]

I'm taking college prep physics in high school and I love it. I love videos like this! Thank you for sharing!

[@styk0n]

I'm really enjoying these videos, even though I have trouble following (I'm not a maths or science student and neither of those subjects have ever been my strengths). Just playing Kerbal Space Program makes these equations seem so much more relevant to my life than any equation I learned while in high school.

[@TheJFisherman]

Please keep doing these! I'm a senior in high school, and my physics teacher refuses to cover anything relating to orbital mechanics!

[@Krags47]

Hey Scott, it would be great if you did one on calculating phase angles. This series has helped me tremendously learn obital mechinics and has ecen gotten me to use less and less mechjeb and more of my own math. Im just missing the knowledge on how to time burns! Please consider bringing this series back!

[@DrKramshaw]

I enjoy living vicariously through others' abilities with algebra :)

[@youtubeaccount9636]

Scott you should do more math videos, like algebra and geometry and stuff, you would make an amazing teacher.

[@MisterBlaargh]

Learning any type of math...has not ever been fun. Until now. How the hell does Scott make it interesting?

[@damienkurast]

I know the feeling. While i find Scott very interesting and an excellent teacher, i bet the reason you find math interesting is your own shift in focus to a focus on learning new things? Thats whats happened to me, hated school, but with age, i've become more knowledge seeking, u know?

[@MisterBlaargh]

damienkurast It's because he doesnt have the typical boring voice. Which means I can pay attention to it. It's not just white noise. Never had a shift in focus, I've always been an information sponge.

[@ThaFuzzwood]

You have a reference frame which isn't a boring blackboard but a game in which you can test said equations. :)

[@ConeRanger5]

***** I don't think it's that subtle, You can even see it in this comment section, people who say he lost them at 0:00 are getting more thumbs up than any other posts, it's fun to us when people don't know things and some might even make fun of people who understand instead of encouraging the ignorant (or themselfes) to find out why they don't understand certain things YET.

[@gameboxfreak]

damienkurast O man are you saying exactly what I am thinking. If I could show my past self (say 3 - 5 years ago) all the stuff I've been trying to teach myself about orbital mechanics and such. I think my past self wouldn't believe I was being serious. It's funny how age really does change your perspective and interests.

[@HerraTohtori]

Scott, it would be nice to see an energetic approach to same problem - that is, looking at the gravity potential U of a body with mass M and the kinetic energy of a small object nearby, and then just basically looking at how much kinetic energy you need at any given distance r to escape to potential where distance r is infinite - ie. another way of defining escape velocity. I'd be interested in seeing your take on popularizing the common notation of defining potential energy to be zero at infinity and negative infinity when r tends toward zero... it's such an important aspect of physics, and not just orbital mechanics.

[@zolxo]

torille nyt!

[@ThaFuzzwood]

So to summarize the last part: what you are saying here is that it is more efficient to do all required burns at 70 km up from Kerbin instead of escaping SOI first and then adjust orbit to reach whatever orbital body you wish to go to? If so I have to figure out how to sucker my minmus refueling base and kerbin into reaching other planets with crafts using only little amount of fuel. I reckon from a quick drawing in my head that I should go for inner planets when kerbin eclipses minmus so that my periapsis ends up on the prograde side of kerbin and for the outer planets burn when mimus eclipses kerbin so that my periapsis ends up on the retrograde side of kerbin. In both instances Kerbin would slingshot me in the direction I need to go.

[@rodavetrsmisse7428]

It would be nice to see this series continued! Great lesson, thanks! Although some aspects might make more sense if you'd explain this transfer orbit to Mars in a next video.

[@toolkit71]

As always, excellent tutorial....I studied engineering dynamics in U but it dealt with cars and traction on a road...KSP will be the next spring board for our futur aerospace engineers....

[@pppppaaaaaccccchhh]

On my A Level class they had only told us that v^2 = GM/r and was having difficulty as I thought about going towards infinity for the escape velocity. Now that I know where v^2 = GM/r comes from and what it actually means, it makes sense now. Thanks!

[@migueldp9297]

Hey Scott! Thanks for have done this video, it's pretty awsome :D Could you do a video where you explain how to calculate the orbital period of an a circular orbit and a excentric orbit? It would help me a lot.

[@JohnSmith-en9yb]

why not just use the V^2=GM(1/a) and then using pi, calculate the distance you would go in one orbit and after that divide the distance by the velocity and you get your orbital period... well thats just for circular orbits.

[@slpk]

Please explain the Oberth effect. You keep mentioning it every time but I can't get a grasp of how it actually works.

[@zarblitz]

If you look it up on Wikipedia there's a pretty good explanation that explains the basics of why it works the way it does.

[@scottmanley]

Oberth effect says that the energy gained from a rocket thrusting is proportional to the velocity, so if you want to get the most energy from a rocket you perform maneuvers when it's travelling fastest.

[@johncrichton6078]

***** Total energy being potential + kinetic energy. Higher orbits have high gravitational potential energy, lower orbits have high kinetic energy. If you apply thrust when you're at periapsis, you retain more of the speed because you ascend out of the gravity well faster and the planet has less time to slow you down.

[@whsanders78]

You can also think of it explicitly in terms of kinetic energy. When you burn prograde (or retrograde) you are adding (or subtracting) kinetic energy to your orbit. Kinetic energy goes as velocity squared. Say you're on a highly elliptical orbit with a periapse speed of 2000 m/s, and apoapse speed of 200 m/s. And say you want to burn enough fuel to change your velocity by 100 m/s. Where should you burn in order to change your /velocity squared/ by the greatest possible amount?

[@slpk]

Scott Manley zarblitz John Crichton whsanders78 I get how the practical aspect of it works: It's best to burn at the highest speed possible. I just can't wrap my mind on the mechanics. I can't see how burning a rocket engine at a higher speed can increase your kinetic energy further than if you did the burn at a lower speed. I read the Wikipedia article and one of the things that messes with my brain is: "Most of the work done by a rocket early in flight is "invested" in the kinetic energy of the propellant not yet burned, part of which they will release later when they are burned..." I don't see how the kinetic energy, gained by the propellent, can be release later, when burning. Is the Oberth effect present only in orbital mechanics or would it apply the same way if we were talking about a point-object rocket traveling, unimpeded by any forces, through a vacuum? In this scenario, burning at a higher speed would also increase the rocket's velocity at a higher rate? Is the effect something intrinsic to rocket engines or is it simply a manipulation of gravitational, chemical and kinetic energy? That's what boggles me.

[@raspyextreme]

as a 17 years old science high school student, I love to push my limits and try developing myself on something I want to study in the future, thank god there is Scott Manley, because there isn't any information in my country about space physics and stuff like that, amazing!

[@nonnormalhemauers]

Please do more of these!!

[@MrWilliam932]

I would like to see how to calculate the transfer windows. Can you please resume this series?

[@kalebbruwer]

and I'd like to see a vector formula that can give you direction and position at any given moment. I also want to see a formula for the orbital period of an elliptic orbit. I want to see the formula for the things above for a sattelite orbiting multiple bodies. and the thing you mentioned.

[@fiveoneecho]

Wouldn’t you just composite a function that tells you the orbital period of the transfer orbit into a function that represents the difference in theta between the two orbiting bodies? Like have an angular velocity equation with another one composited to find the difference, then composite the period equation so that it provides the time parameter.

[@IwishiknewMinecraft]

Scott, can you do a video explaining all the terms? Delta V, Oberth Effect, Ect? I am trying to follow and pay attention, and I think that a "deffinition" video might help understanding a fair bit.

[@LeLucky90]

Scott this is very interesting, but for most people including myself a little bit hard to follow. Next time you should try to make it more visual.

[@zarblitz]

I think the point of this is to get into the mathematics for those who are interested, not necessarily for everyone.

[@scottmanley]

Right, people already have the visual demonstration from KSP, this is the mathematics.

[@LeLucky90]

Ok I get it.

[@marksman1416]

Never did orbital mechanics in school but did learn about escape velocity.

[@Hale444]

I'd love to see one on how gravity assists work.

[@scubateve007]

Made my day learning from you.

[@marioornot]

i feel like i lacked a lil information of why things are. but apart from that i would like to strongly urge you to keep making these

[@bssam1998]

Thanks for the video. You make it sound so easy

[@Yokopo_]

I really like videos about this kinda stuff. I don't see them often on youtube.

[@johnytest464]

Why does lessons at school are boring and learning physics with Scott isn't? Why lessons at school cant be as interesting as yours? For me it's even funnier because I'm not from english speaking country, I'm just learning english and i've never thought that some day I'll be learning physics in english :D

[@NNOTM]

Here, one of MIT's physics courses in English ;-P Lec 1 | 8.01 Physics I: Classical Mechanics, Fall 1999

[@matsv201]

because teachers hate physics and they try to make you hate it to.... basically

[@damienkurast]

Come on, give ppl who wants to teach others some credit. I bet its a completely different reasons from what you ppl are suggesting. Its not that they don't want to or don't know how, but rather the bureaucracy and politics draining their passion, to the point were they just give up. Don't blame the teachers.

[@Sean_735]

I see you've never met a rather young university physics professor.

[@johnytest464]

Seanathon Balkmenistan I'm fifteen, so that's my excuse :P :D

[@AuroraCypher]

Thanks so much for this.. I build multirotor and robotics but I never really understood or liked physics in high school until now, no joke.

[@MrPotterrable]

Some Principles Of Flight maths would be great too, Scott!

[@JesterSpeight]

Could you do a video on better explaining interplanetary transfers, both from Earth SOI and Sol SOI?

[@spacium-3030]

So to carry off some of our velocity in LEO to help us get on the Hohman to wherever we want to go, we perform our burn-in LEO? What is the science behind taking some of your velocity in LEO and carrying it off? Is there any scientific explanation on why this happens?

[@GurraThePwner]

I was a little lost, but that pythagoras triangle really helped forming the picture :)

[@klikkolee]

you do a great job explaining the calculations done in the examples. however, the camera doesn't show a large enough section of the paper to follow along easily. you are constantly referencing values that have been off screen for the majority of the video that have long since been forgotten by us viewers. it would be greatly helpful if you could possible hold the camera further back, use a wider angle lens, or pan over to values that haven't been on screen for a while when you use them. thank you!

[@Virtuitus]

Scott, you should discuss the landing legs on the Falcon 9 that will land the rocket on land instead of at sea.

[@MillerNate]

It's strange, easy math like geometry and algebra I have a hard time with, but this comes so easy to me!

[@seadart07]

If you have two circular orbits in the same plane, one with a larger radius than the other, the potential energy of the higher orbit is greater than the potential energy of the lower orbit. But because the orbital velocity of the higher orbit is less than the orbital velocity of the lower orbit. the kinetic energy of the higher orbit is less than the kinetic energy of the lower orbit. Is that correct? If so, does the drop in kinetic energy exactly match the increase in potential energy when comparing a lower to a higher orbit? If the answer is yes, then this suggests that all co-planar earth orbits have the same total energy (potential + kinetic). Is that correct?

[@modakshantanu]

Please do more! Teach us how to calculate transfer orbits and phase angles stuff.where did you learn this ?

[@silvercomic]

Will we still get the episode on how to get the 3 km/s dV to get from Earth to Mars?

[@rocktheworld2k6]

Why is he using d to represent delta? dV makes me think of the derivative of V not delta v, shouldn't he be using the triangle that's the Greek letter for delta instead?

[@knallbi]

to be honest, i was wondering the same thing as well... i thought that maybe he was doing so to make in easier for poeple ho have no background in these kinds of mathematics... but then he uses µ...

[@rocktheworld2k6]

DJKnallbi If they have no background on these kind of mathematics, they'd probably get lost waaay before deltas haha. And I'm in physics 1, so we're just doing basic 2D and some 3D motion, haven't gotten into any orbital mechanics yet, we get into that kind of thing a little bit at the end of the semester, but right now µ to me means the coefficient of friction xD

[@Archgeek0]

He's probably just doing what I generally do and going lazily with 'd's instead of the more annoying to draw capital deltas, which if done correctly require you to make sure the upper left leg is thicker. It's my favourite Greek letter, but it's annoying to draw over and over.

[@bwjclego]

Daniel Wilson Yeah but nobody does that, its always just a triangle :P.

[@lolmandos]

rocktheworld2k6 When i studied electrophysics i guess some letters had like 3 meanings in a single semester :p

[@hereforfunandgames]

Scott is a good teacher

[@JimmyHoffa24]

Thanks for exploding my brain scott!

[@samueltass8157]

Hey Scott , fantastic physics class , reminds me of high school...Please make more videos... Have a question though...During Falcon Heavy's last flight the upper stage was travelling at 28800 Km/H ... Isn't that speed higher than the escape velocity from earth ? Silly question but it's been bugging me ...

[@jeffvader811]

How does he get the 3 at 6:20?

[@Raffo42]

Can I ask for a part 4, explaining the Delta-V calculations for landing on or starting from planets? The 4500m/s you need to start from Kerbin are hugely different from the orbital speed around 2300m/s, so where does the difference come from, and how do you calculate that?

[@moonasha]

is that the sound of a beer opening at the very beginning, as the music ends?

[@scottmanley]

It's the sound of a pen clicking.

[@zuinfibe947]

Hey Scott! Thanks for these vides I was struggling to find some information for my KoS programming adventure. I'm currently trying to use the V0=Mu*sqrt(2/r - 1/a) formula, to make a ship in a known orbit, to re entry (crash) back to the planet on a specific given spot. Let's assume that the orbit is with zero inclination, and our selected spot is on the equator at a given LONGITUDE and no atmosphere on the planet. How can I calculate the DeltaV , so that when the `r` of my orbit is equal to the radius of the planet, (so basically on the ground) it will collide at my given longitude? And when to burn? I was trying to solve a system of two equations using `a` that should be the only common value in the two equations. No luck for now. I beg your help! :-) Thanks!

[@OracleWest]

Very nice video. My only point of confusion is with the V_infinity. How can you have an r = infinity but your a term not equal to infinity? I know you mentioned you are now on a hyperbolic orbit there, is that where the discrepancy is? Should an alternate equation be used there?

[@benForDestruCtION94]

I wish Scot taught me physics in school

[@GermanBiking]

I think Scotts didactic abilitys leave something to be desired :D Giving a lecture for a bigger audience you first have to create some sort of foundation so everyone has a chance to understand it. You can derive all of this pretty lucidly.

[@4IN14094]

Divided by infinity!? Get ready Scott, KZbin mathematicians is going to make you their prime target now! D:

[@ThaFuzzwood]

1 / (bigass number) approaches 0. Happy? :)

[@Sean_735]

ThaFuzzwood (lim y -> infinity) of x / y = 0 for any real number x.

[@Sean_735]

MrSplodgeySplodge Mathematical operations like that can not be performed on infinity in our number system because it lacks the concept of infinitesimals. 1/∞ = 0.000... ...000 1, where the number of zeros in infinite. In our number system, that can't happen, so 1/∞ is undefined with a near-perfect approximation of 0. There are some advanced mathematical systems which allow the existence of infinitesimals though.

[@fiveoneecho]

I know this thread is old at this point, but I feel the need to make a correction: You can’t do mathematical operations on infinity because it is not a number. Infinitesimals are a whole different can of worms. Infinities are constructs that we use to talk about things with no end, and infinitesimals are constructs we use to talk about things that are so small, they have no value, but they are still bigger than zero. With this in mind, he was effectively cutting a corner around talking about limits by just saying “diving by infinity”.

[@RedsBoneStuff]

Is it better to do interplanetary burns just above the atmosphere or inside the top layers of the atmosphere?

[@kubixus]

Dunno what is going on but it's still interesting.

[@foefighter30]

I understood most of it, but I can't seem to wrap my head around the infinite velocity. It's difficult to conceptualize but mainly I have no idea where the 3km/s came from that we plugged in for Vinfinity. Is it just the dv we got earlier to change to a martian orbit that we're plugging in as the excess energy at Vinfinity?

[@scottmanley]

3km/s is an approximate value for the velocity at which you have to leave Earth's SOI to be able to reach Mars, you can compute it using the equations in part 1 and 2. Vinfinity is the desired velocity when leaving Earth's SOI

[@kalebbruwer]

+Scott Manley so the infinity in Vinfinity stands for the height? Also, did you just say Earth's SOI? sorry for being 2.5 years late...

[@thepiratejosh]

Mic volume sounds kind of weird, but interesting video.

[@endfinum]

that may be caused by the fact that he is behind the camera (and the mic)

[@jkuzem96]

Scott can you explain how to calculate synchronous orbit in another episode?

[@ThaFuzzwood]

I honestly had a blast deriving these myself just now. Thank you for this mental exercise. They are a bit different than how Wikipedia displays them, but I wanted to use already-Manleyed equations. A geostationary orbit is a circular orbit which has an orbital period equal to a full “day” of the body it is orbiting. For Kerbin, this is 6 hours or 21600 seconds. But let’s call this time T so that we can use the equation for all bodies. The orbit has a certain circumference which equals 2π * R meters, where R is the radius from the center of mass. For Kerbin, this equals to 600000 m + whatever the altitude gauge is indicating. The velocity V, which is nothing more than the traveled distance divided by the time it traveled, is equal to (2π * r)/t. After rewriting this for a circular orbit: Rcirc = (Vcirc * T)/(2π) From Scott’s first vid, we saw that the velocity squared of a circular orbit (Vcirc²) with radius Rcirc equals μ/Rcirc, where μ is equal to the Gravitational parameter (μ for Kerbin = 3.5316 * 10^12 m³/s²). After rewriting: Rcirc = μ/Vcirc² For any circular orbit, these 2 equations are equal to each other: (Vcirc * T)/(2π) = μ/Vcirc². Couple of rewrites: (Vcirc³ * T)/(2π) = μ. Vcirc³ = (μ * 2π)/T. Vcirc = cubic root((μ * 2π)/T). Plugging in the data for Kerbin (T = 21600 s and μ = 3.5316 * 10^12 m³/s²), this equals 1009 m/s. If we plug this velocity in the first equation, we get the radius of our geosynchronous orbit: 1009 * 21600 / (2π) = 3468750 m, or 2868750 m above Kerbin surface. Exactly what all those sites out there tell you. Test it out by plugging in values of other celestial bodies.

[@HALLish-bl3bm]

Could we see the maths for this done for ksp's body's and then you try and match it?

[@DavidHaverson]

Well the math is the same, and all the value you need are in the kerbal wiki ;)

[@DoNeeh]

Magic.

[@ruiningwang1644]

Applied to KSP, would it be more efficient to put the periapsis below 70'000 before escape? Something like 55'000 maybe so that the atmosphere is not too thick?

[@ruiningwang1644]

actually nvm asking, I'm going to try that myself now :D

[@Raethen4e]

The mathematics are a little too advanced for me, but am i correct in the understanding that the lower the orbit, the closer you are to the parent body, the less delta v you need to spend when transferring to other planets?

[@JosheyG34]

Holey fuck I love you. 3rd year physics we haven't learned shit about this.

[@ATGEC]

The one thing i'm struggling with is landing on a planet with atmosphere. I pick out a good location and can't seem to land anywhere near it. Is there somewhat basic method for calculating this?

[@ruzeenfarsad367]

I remember Scott did an episode about doing a munar rescue and how you calculate your landing place using Pythagoras so I imagine it'd be similar to that except for counting in the aerobraking from the atmosphere

[@impguardwarhamer]

you lost me at 9:37

[@KarlDutschke]

He lost me at 0:00

[@chrismusix5669]

He lost me at Infinity.

[@TheHypno15]

So the idea is that the closer you are to the body the most burning you should do because that's when the change in velocity is highest? Or can someone correct me if that's wrong?

[@ThaFuzzwood]

This is easy to explain. The energy of an orbiting body (Etot) is equal to Ekinetic + Epotential. Ekinetic is equal to 0.5 * m * v² where m is the mass and v the velocity of the craft. Epotential equals m * g * h where g is the surface gravity (9.81 m/s²) and h the distance from the center of mass of the body you are orbiting. You can see that only Ekinetic has anything to do with velocity. At any point during your orbit, there is an exchange between Ekinetic and Epotential; as long as you aren't firing your engines, you aren't changing the total energy of the system and thus Etot is a constant. At apoapsis, Epotential is at its maximal value as h is at its maximal value and Ekinetic will be at its minimal value. At periapsis this is the other way around, with Ekinetic and thus the velocity being at its maximal value. Now the Oberth effect which you are referring to, can be explained at its most basic level if you consider that squaring a small number (3² = 9, increase of 300%) is not as significant as squaring a large number (1000² = 1000000, increase of 100000%). Since your specific impulse does not change wherever you are on the orbit, you expend the least energy gaining a specific velocity if you are at your maximal velocity already. It does not matter if you wish to slow down or speed up.

[@markwilkinson7798]

How do you calculate how much Delta V you'll need to slow down to get into an orbit around mars?

[@adebirkful]

I'd like to see how you calculated Vem = 3km. Any word on that one?

[@bartonpaullevenson3427]

What is a for the hyperbolic orbit and how do you find it? You never explained that.

[@ancapftw9113]

If you did the transfer burn to Mars in low earth orbit (7.7 km/s), wouldn't that be v^2 = 7.7^2 + 3^2 or roughly 8.3? Why do you need to do it at escape velocity? Isn't escape velocity essentially an infinite distance orbit?

[@Cat8fish8squid]

So.... In ksp with one of the FAR jets i have reached mach 11 but i cant go faster because i go into space and when i look at the map im on an escape trajectory, my question is, is this possible in real life?

[@HerraTohtori]

Yes and no. As far as orbital mechanics are concerned, yes, you can be on an escape trajectory while in the atmosphere. Earth's escape velocity is much higher than Kerbin's though, so you would need to reach very very high velocity for this to happen, and because of the limitations of materials, it is rather unfeasible thing to do. Rail guns may be used to accelerate projectiles to very high velocities, but again there are problems with the durability of the projectiles - they just basically vapourize or explode on short order. Your vehicle would likely have to be rocket-propelled as well, because very high velocities are really problematic for air-breathing engines (although recent advances in intake air cooling technologies may make this possible some day). In other words, it's technically possible to go past escape velocity in atmosphere, you just shouldn't expect to be able to use the vehicle again, at least with current tech. It would also be horrendously inefficient and fun way to do space travel, so I expect Jebediah and his real world brethren would approve. However, doing things the other way is very much possible - that is, entering atmosphere at speeds higher than escape velocity. The trajectory might intersect with ground to begin with, but it's certainly possible for objects to fall into atmosphere at higher speed than escape velocity. It doesn't usually end well for those objects - just look at Chelyabinsk or Tunguska for reference - but for example the Apollo capsules did pretty high-speed atmospheric re-entries. The Apollo 13 capsule, due to their direct return orbit, had the highest speed at re-entry, at just over 11 km/s - which is just slightly less than escape velocity at sea level. With a proper heat shield, the kinetic energy is rapidly dumped into the atmosphere, though, so you wouldn't be going at that speed for very long.

[@MWSin1]

There was a steel plate ejected in one of the Operation Plumbob nuclear tests that was determined to probably have achieved escape velocity, but it is believed to have vaporized due to frictional heating prior to actually leaving the atmosphere.

[@lolmandos]

MWSin1 Was about to mention that.

[@physicshuman9808]

I designate semi major axis with SMA so that is not confuse for acceleration since acceleration is also designated with an a

[@TicTacMentheDouce]

Not sure if I got what meant that "Vinfinity²" velocity. Explain, EXPLAIN

[@ThaFuzzwood]

Vinfinity is the velocity when semimajoraxis a equals infinity and you just end up with 2μ/r. This occurs at the moment you reach escape velocity where your apoapsis is considered infinitely large.

[@TicTacMentheDouce]

That's what I thought, but wasn't sure, thanks !

[@MuhsinFatih]

make more of this!

[@juliansuse1]

Thank you, this helped me with my high school graduation exams

[@unpaidcomintern]

Could someone please explain what he means by velocity at infinity?

[@VentantoGame]

Can you make video about anomalies?

[@NovardNoodle]

Awesomeee!

[@ekkotutorials9643]

So... 1 potato + 1 potato = 2 potatos?

[@GenestealerUK]

No actually.... 1 potato + 1 potato = 2 potatoEs

[@ekkotutorials9643]

Ahh....

[@GeekFurious]

I'm not an engineer but it made sense to me. Granted, I did take multiple graduate level math courses.

[@adsfsd1]

Is it correct that the V-infinity is the same as the V-earth.mars??

[@kalebbruwer]

can someone please point me in a direction to find more advanced orbital mechanics? I am struggling to find anything on it because, you know, internet and math...

[@masondaub9201]

Had the same problem while trying to research general relativity.I eventually gave up on trying to use the internet and decided to take a look at a university library. You can find a lot more information about more advanced math and physics there if you have a university near where you live

[@kalebbruwer]

Mason Daub It's gonna take a lot of effort for me to reach a university.... any plan b to find these books?

[@masondaub9201]

Well you can look for orbital mechanics textbooks on the internet and then look for a pdf you can download

[@kalebbruwer]

Mason Daub I've seen textbooks I can BUY on amazon, but no pdf and I'd probably have to torrent it if I found one. Not a lot of people seeds such things. But I'll give it a try

[@brian554xx]

so confused what the difference is between escape velocity and velocity at infinity.

[@TheMajorpickle01]

Oh god he divided by infinity

[@remavas5470]

or limits :D

[@TheDantheprogamer]

I would say in general scott that , Leaving from a moon to intercept a planet , using a parent body is (when looking at dV budget ) Acceptable, since you get pulled alot in most cases, (however this does not include the gas giants ofcourse since their sMa is so vast ), But escaping from a parent body , slingshotting around a lower gravitational moon , is ofcourse , a bit insane , and very fuel ineffecient as you pointed out :p. It is interesting to see however how you can manipulate the dV budget , by doing Hohmann transfer , From lets say in ksp, Minmus, to earth, sling shot to the planet of choice , how low you dV requirement is in comparison to a traditional trans marsian or duna burn from LEO (LKO), and also see where these things do not add up. for example at Jool, or from gilly by Eve to where ever. I would find that hugely interesting , to see how KSP models these orbital mechanics (if you can even do that with the inaccurate manuevre nodes i dont know )

[@Rocketyfox]

Well you could just use these equations and a stopwatch instead of maneuver nodes and possibly get better accuracy, if you have the patience of course

[@bwjclego]

I feel like if I understood what the dV from earth to mars was (dVem) then I would get the last bit. I understand the maths, just not the physics. If you are on an escape velocity aren't you already able to be headed to mars? What is the dVem doing?

[@ThaFuzzwood]

At Vesc you only have enough velocity to escape Earth's gravity pull. However, you are on the same orbit, or just a bit bigger, than Earth's orbit around the sun. Since Mars is on an even bigger orbit, you need to compensate that by burning even more.

[@pyr0b1rd]

Escape velocity just means you are leaving (in this case) Earth's gravity well. All it really does if you don't add any extra velocity is put you in a slightly different orbit. If you want to head to Mars for example you need to put in additional thrust to raise your apoapse until your orbital path will pass close to Mars. Works the same if you want to travel to Venus, just you would burn on the opposite side of the planet so your periapse gets lower instead.

[@bwjclego]

ThaFuzzwood Ah, that makes sense. Of course.

[@MWSin1]

dVem would be the dV needed to raise your solar orbit's apoapsis from the radius of Earth's orbit to that of Mars'. If you don't apply that, all you'll be doing is popping yourself out onto a solar orbit pretty much identical to that of the Earth.

[@ThePred2009]

have you studied astrophysics scott?

[@iTzCharmander]

if you were to gravity assist at just the right angle you could in theory save massive amounts of Dv required couldn't you?

[@scottmanley]

Not for travelling from Earth to Mars.

[@Starius2]

How do I get to the real life Mun?!

[@kenanjabr1992]

So if μ=mu, then 2μ=Mewtwo?

[@Elzebub666]

Yay, more Manleymatics =D