The Vis-Viva Equation | KSP Let's Do The Math 

He says "highschool math." They dont teach you this until like year 3 in college.

Thanks. These videos are mostly a repackage of my tutorial series with only the math left in. goo.gl/UCFuHa There's a lot of stuff already there, but I'll definitely extend past that. What are you taking in college?

@@MikeAben Can you explain just a bit how you got the ∆v2. I'm a bit confused

The average cellphone is an order of magnitude less powerful than the "super computer," used to put men on the moon. You could literaly just plug the real world numbers into KSP and end up with an almost exactly correct flight trajectory provided you had the rocket.

@@JohnDoe-zu2cm No you wouldn't. In the video he even said that the moon and the earth wave such similar masses that Kepler's law doesn't work as well anymore. Have you ever used the principia mod? It simulates realistic gravity in a multiple body system and it is entirely different to stock ksp.

It's so funny when seemingly obvious things can slip right under the radar. Thanks.

Cool! Thanks. Don't hesitate to drop a comment, observation, or criticism. My background is more math than physics.

And from there, of course, use g = GM/r² = 9.8 m/s² at the surface to get the standard gravitational parameter, right?

@@hopeg97 yes actually.

You are getting a negative number, but there are no negative numbers in the question, so something is up there. 10^(-11) is not negative, but rather 0.000 000 000 01. So 6.6741X10^(-11) = 0.000 000 000 066 741. Similarly 5.2916X10^(22) = 52 916 000 000 000 000 000 000. The way I've written these numbers in the video are examples of "scientific notation". As you can see, it is more compact than writing all the zeros. Perhaps you haven't gotten to this yet in your math studies. I hope this helps. Let me know if it doesn't. Edited to add: It is very possible that scientific notation will first appear in your science class rather than you math class.

Thanks, and yes, for the math videos I write a script and read it.

Thanks. Linear momentum isn't conserved because the velocity is constantly changing while the mass would remain constant.

No problem. Here's some more detail. By 10:06 I have developed two formulas. Typing formulas always looks terrible, so you may want to write these out on a piece of paper. v_1^2/2 - v_2^2/2 = mu(1/r_1 - 1/r_2) and r_1v_1 = r_2v_2 I want to figure out v_1 (my required velocity at periapsis) so I rearranged the second formulas for v_2 and substituted into the first. I do all this in the video and end up with the formula v_1 = sqrt(2mu*r_1/(r_1(r_1+r_2))). Okay, that was for the burn at periapsis, for the burn at apoapsis I would need v_2, my velocity at apoapsis. So I would take my second formula above and arrange for v_1 getting v_1 = r_2v_2 / r_1. I would sub that into the first formula and solve using the same steps as in the video starting at 10:15, but this time solving for v_2. This would get the formula v_2 = sqrt(2mu*r_2/(r_2(r_1+r_2))). I next need to work out that velocity I would need for a circular orbit at r_2. This velocity would be v = sqrt(mu/r_2). So the delta-v of the burn would be ∆v2 = sqrt(mu/r_2) - sqrt(2mu*r_2/(r_2(r_1+r_2))). I can take out the common factor of sqrt(mu/r_2) (like I did at 12:40) to get the second delta-v formula. I hope this helps. To keep these videos from getting too long and tedious, I do leave stuff to be completed by the viewer.

@@MikeAben Thanks,It really helps. Didn't expect you to give such an elaborate explanation. 😁 Really a fan now.

@@truebark3329 No problem.

I think my presentation was better in the Rocket Equation video. I should have presented the formulas first and shown how to use them, and then derived them afterwards.

Thanks. There will be more episodes in this series so keep an eye out.

Yes, they do give different values, because the r_1 and r_2 switch positions, though your way will work if you define r_1 and r_2 differently from how I did. I defined r_1 as the lower radius and r_2 as the higher one, but if you define r_1 to be the radius at which you perform the burn then your way works if you ignore the negative sign you get. Be careful interpreting a negative delta-v as retrograde. These formulas don't consider direction. For example, the delta-v_2 that is calculated at the end of the video would come out negative using the technique I just described, but it is a prograde burn for the situation described in the video.

Thanks for the explanation! I will definitely use the way you present in this video then, haha.

Nice. If it helps, towards the end of this series I explicitly talk about how to calculate interplanetary transfers.

You're very welcome. I'm glad you're enjoying them.

That's a really cool idea, though I think it would be best as a separate series. I can't say when i would get to it. Lots of ideas. Need more time. I've got a circAt function myself that is part of my kOS launch script that's based around stuff in this video. It uses the Kepler's 2nd Law formula to predict velocity at apoapsis and then Vis-Viva to calculate the burn. Works great.

@@MikeAben that is exactly what I'm looking to do with my launch script. Right now I just point prograde and burn until the periapsis is within 3% of where the apaopsis started. I'd like to use the equation to get an exact speed to burn to and I'd also like to calculate an approximation of the length of time so I can know when to start burning. I've only started with kOS last week, but this is what I have so far github.com/FlexibleToast/kOS/blob/master/launchSat.ks

​@@FlexibleToast You certainly write more legible code than me. By the way, this is how I calculate burn time. dV is the delta-v of the burn. Let me know if any of the other variables aren't clear. finalMass = m / (e^(dV/(ISP*g_0)) startAcceleration = thrust / m finalAcceleration = thrust / finalMass burnTime = 2*dV / (startAcceleration + finalAcceleration) I would also use the maneuver node maker built into kOS. The structure is NODE(time to burn, radial, normal, prograde). Once you know the delta-v of the burn, then it's just NODE(TIME:SECONDS+ETA:APOAPSIS, 0, 0, dV). You can then use kOS to lock to the maneuver node rather than just locking to prograde.

Wtf is kOS

@@alstillplays it's a mod for KSP that allows you to write scripts for automation.

Is there a specific timestamp in the video you need help understanding? Or just the general concepts? (And even if it's the general concepts, a timestamp with a specific example you'd like help on would still be a good idea.)

I suspect you're over thinking it. According to Newton, the force of gravity is proportional to thet the masses and inversely proportional to the square of the distance separating them. For example, if you double one of the masses, you double the force of gravity. If you double the distance, gravitational force is a quarter of what it was before. Proportionalities like these are great, but to turn them into equations, you need to introduce a constant. For example, if I'm buying eggs, the number of eggs I get is proportional to the number of cartons I buy, but you actually don't know how many eggs I've bought until I tell you how many eggs are in each carton. That is the constant if I were to make an equation for this. That's all G is, a constant. It is calculated by literally taking two objects of known mass, putting them a known distance apart, and measuring the force of gravity between them. You can look up the Cavendish Experiment if you want more details. There's nothing more to a constant than that. It's the number measured to make the equation work. You may ask why it has the value that it does. Einstein's Field Equations can be used to derive Newton's Gravitational Law, but those equations have their own gravitational constant. Why does that constant have the value it does? No one knows! There are several constants in physics like this that are very precisely measured but no one knows why they have the values that they do.

Yoiks! Thanks.

Great thought. I started to play with it and, as is often the case, the devil's coming out in the details. You can get a formula for total energy per kg by combining epsilon = v^2/2 - mu/r with v = sqrt(mu/r). This gets epsilon = -mu/(2r)

Yeah, I see the problem now. I'll need to chew on this a bit...

Mass is irrelevant to these calculations and would just end up cancelling out anyway. Objects of different masses behave the same under a gravitational field, so you don't need to worry about it.

It's scientific notation. 2.24X10^6 is the same as 2,240,000. The exponent tells you how many places to move the decimal place. It's very simple once you're used to it, but can look confusing. On a scientific calculator there is usually a button like EE or EXP to let you enter the notation.

You're right. I just rounded to the nearest 10.

@@MikeAben Ah okay good to know I didn't mess up anything!

I'm lost at 10:47 by the way but I'm gonna understand it! I'll make sure of it. If I got to understand the light cones in Einsteins equations (or rather Swarzchild) with relativity, I can understand this for sure

@@Belfor09 Yeah. I threw a lot out there in this one.

As I taught high school math for over 30 years, I'm rather familiar with what is, and is not, in the curriculums.

@@MikeAben Must be in a different country/ state/ province.

Sorry, I'm not in the American system, so I'm not sure what's in Algebra I. In Canada, we would be hitting this in grade 10 and 11, though what I'm doing here is a smidge nastier and I blast through it pretty quick. If you have any question, don't hesitate to drop them here.

I don't think you really get comfortable with this math until around Intermediate Algebra or College Algebra. This type of math will be second nature by time you're through Calc I. The hardest part of Calculus tends to be the algebra required, but that's another topic. It's been a long time since I was in high school though, over 16 years ago. I did recently go back and get my bachelor and minored in math so I'm judging this off of my university experience and College Algebra I took there.

@@FlexibleToast Thanks. My previous reply was likely too brief. You're right, a student not up to at least senior high math or early college is likely to get a bit lost in the algebra shuffling starting around 10:30, though it's likely not because the topics haven't been introduced. One thing is my speed (I don't want a video to get bogged down), but the other is they are likely to get bogged down in all the variables. I am fairly convinced the latter is a brain development thing, though I've seen previous little literature on the topic. For example, 15 year-old math student is likely to know how to factor a difference of squares. That x^2 - 9 = (x - 3)(x + 3), but give them r_1^2 - r_2^2 and they'll likely see it as a completely different animal even though it is the exact same pattern. Somewhere in the late teens or very early twenties, this confusion disappears. I don't think it is due to exposure, but rather some developmental wetware wiring that has just occured.

Calculators are different in how they work, but a scientific calculator will have the ability to enter numbers in scientific notation. The button is usually EE or EXP.

​@@MikeAben I see, thanks very much for the quick response by the way, even after all these years since you uploaded the video

You could try looking for a video with the word Duna in the title rather than the word Math.

Hey, you've probably got it by now but here's an explanation anyway: The two values in the LHS were being subtracted, but they couldn't be subtracted as the denominators were unequal. In order to make them equal, the denominator of the first term was multiplied by r2^2. But, in order for this new value to be equal to the old value, the numerator also had to be multiplied by r2^2. Since v1^2 was taken common, only 1 was left as the numerator. This one was multiplied by r2^2, which obviously gave r2^2.

It got multiplied against the 2444.8^2.

Unfortunately, calculators are not consistent in how they work. Is the calculator scientific? If yes, there should be a button that says EXP or EE for entering numbers in scientific notation. That'll help. If your using a phone calculator it should either have a scientific mode or you can download a free scientific calculator.

No worries. This is a pretty dense video.

Yeah, it's kinda odd to think about that way at first. At infinity, the potential energy is zero, which makes that maths Kinga convenient. It also means that if the total energy is negative, you're dealing with a bound orbit.

Yes, this is a simpler version. I introduce the semi-major axis a little later in this series. I figured there was enough being thrown at you already in this one.

@@MikeAben Then I'm sad to say I'm just not doing it right. I'm either plugging values in wrong or I'm not understanding properly. I need a way to test myself.

@@rereturn The formulas presented here are definitely correct. If you are trying to replicate my calculations but getting different answers, then you are putting something into your calculator incorrectly. Unfortunately, calculators work differently from each other. So you'll have to find out how to do it with the one you're using.

@@MikeAben i definitely believe that so let me make sure I'm doing this correctly. We use this formula to change altitude of the ship, correct? So, say if I want to increase my altitude IRL from 150km to 250km i convert them to meters and use Earth's mass to get a new myu, right?

@@rereturn Yes. Don't forget the distances have to be measured from the Earth's center, so you have to add on Earth's radius.

It's about getting directions right, which I'll admit I only explained through one casual comment. Gravitational potential energy is relative, which means we can decide where E_p should equal zero, but it's important that as altitude increases, gravitational potential also increases. If E_p = +GMm/r, then as r gets bigger, E_p gets smaller. That doesn't model what really happens, but making it negative fixes this. If E_p = -GMm/r, then as r gets bigger, E_p becomes less negative which means it's going up. (-5 is greater than -10). It does mean E_p is always negative and approaches a maximum of zero as r goes to infinity, but that's okay. Again, where the zero point is doesn't matter, but that the numbers move in the right direction is.

@@MikeAben Didn't expect such a detailed answer if any, thank you very much sir

@@notintetesting You bet.

Actually, dividing by m gets rid of the the m in both terms. Here's an example with just numbers to help you see. I'll start by making an arithmetic statement that we can both agree is true. 20 = 7X5 - 3X5 Now I'm going to divide away the 5s getting 20/5 = 7 - 3 or 4 = 7 - 3 Notice how both 5s have to go for the statement to remain true. This is because division is "distributive" across subtraction and addition (so is multiplication). Let me know whether this helps clear things up. Don't be afraid to ask more questions.

@@MikeAben thanks for the quick and detailed answer. I have another question actually, when looking up the vis-viva equation in the wiki, it shows a formula where semi-major axis is used. Can you show how to derive the equation we got at the end of this vid to the equation in the wikipedia? Thanks a lot! :D

@@acr_-kj8gd I do exactly that in a couple of videos. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ZZvvj3i2eZk.html

@@MikeAben Fantastic!