Nice actually seeing this on paper. Gonna have to try some of these equations playing KSP. 🙃awesome video

Great video, Ryan! Looking forward to Part II

You have explained this amazingly, thank you!

This is interesting and useful! Thank you for preparing such a comprehensive lecture!

Great video, simple and effective

Loved the video bro, I look forward to watching some of your other stuff!

@martiancolonist

7 жыл бұрын

Thanks!

Great video, super helpful. Hope you’re doing well!

You forgot to point out that the total energy is the sum of kinetic and potential, but potential is negative because the G has a minus sign in front of it, and that's why you've got a minus sign in the equation despite having said "plus."

excellent video...thanks so much...

This is an amazing video!

3:47 _"That's not flying, it's falling with style!"_

Many thanks... more please 😊

I like your videos. You're full of energy and able to condense the information while keeping it simple. If you're not already a professor, I would recommend you try teaching. You seem to have a knack for it :) You helped me a lot, Thanks!

@martiancolonist

7 жыл бұрын

I do some teaching of Advanced Quantum Mechanics at the University of Cambridge, but still a decade or two away from having a chance at being a professor!

@LouiseBongala

7 ай бұрын

Watching this in 2023! I hope you're a professor now and if not, you are to me!

@martiancolonist

7 ай бұрын

@@LouiseBongala I'm applying and interviewing for professor jobs now!

realy cool. I learned a lot, thanks!

@martiancolonist

9 жыл бұрын

Glad you liked it!

Great video! Can't wait to learn more because I'm planning on being a test pilot and hopefully an astronaut so videos like this really help me with school!😁

@martiancolonist

7 жыл бұрын

Glad you found this helpful! Best of luck with your aspirations :-)

@classonbread5757

9 ай бұрын

How's it going, if I may ask?

Nice work

Thank you. Excellent

Excellent explanation. Thanks very much. I wish you were my Professor.

WOW! I really enjoyed this video. I'd like to make those calculations to see how big of a rocket do I need (almost 0 payload and by big I mean mass) to put it in orbit. I learned a lot. Thanks

@martiancolonist

7 жыл бұрын

Glad you enjoyed the video!

Thank You!!! much LOVE ♥️♥️♥️♥️♥️♥️♥️

Excellent explanation and details math on orbital logics…one simple recommendation is to improve lighting to improve visualization ok board …keep bringing this marvelous knowledge. Thanks

Thank you so much.

NOW i get it! (at 48 years old ugh)

@martiancolonist

8 жыл бұрын

+Pete Kuhns Excellent! Be sure to check out the second part where I use the ideas to derive the Hohmann transfer Delta-V.

@luneakvic

5 жыл бұрын

we're never too late to learn something new :3

@judgeericsmith997

4 жыл бұрын

lol, i just watched it.. so age is relative.. if you got it now, you would have anywhere from 16 onward. This is one of those things you get or you dont. But yeah, I lost my Oribital Virginity today.. i almost made it to 40 virgin status :((( 39 lol

@jonahlewismusic

3 жыл бұрын

I'm 15, and I first I was kind of confused, then you explained the material well. Thanks

thanks alot buddy

Great enthusiasm. I am an Aerospace Engineer with MS. I belong to aerodynamics and propulsion but I trying to switch to space craft dynamics. Let me know if could connect. I am currently working for Indian Space Research Organization as research fellow.

@martiancolonist

7 жыл бұрын

Feel free to add me on LinkedIn: uk.linkedin.com/in/ryan-macdonald-84bba57b

Hi, excellent video. In minute 17:35 you compare the circular velocity against the escape velocity. That confused me, because escape velocity is the launching velocity of the rocket, but when you compare it against circular velocity does it mean that circular velocity is also a launching velocity?? But the graph shows circular velocity tangent to the orbit so it is not a launching velocity, why do you compare them?

@martiancolonist

8 жыл бұрын

In an idealised case of a spherical planet with no atmosphere you could launch the rocket tangent to the surface of the Earth. This results in a circular orbit with the radius of the Earth, differing from the escape velocity by a factor of sqrt (2). I compared them to demonstrate that it takes less energy to get into a free-fall orbit than to radially escape. In the real world however, the launch velocity does not equal the circular velocity because you must first launch vertically to escape the atmosphere before turning to begin the circularisation burn.

Thanks! can you recommend any text books from where one can learn the orbital mechanics?

@martiancolonist

5 жыл бұрын

Two recommendations come to mind: 1) Analytical Mechanics by Fowles & Cassiday. 2) Orbital Mechanics for Engineering Students by Curtis.

@josephpentony4804

4 жыл бұрын

@@martiancolonist thanks

Anyone else 14 or younger?

@spaced___x

4 ай бұрын

I was younger than that when you left this comment 😂

@user93683

3 ай бұрын

Nice bait bro

Hallo teatcher, at you calculation you did not considered distance in square between boths mass at gravitation law equation... could you please explain why? Thanks.

10:15 just let ya know your comet's tail should be pointing away from the sun hehehe ;P p.s love the vid, nice work

Wow, this is a really great video! Just one question, in the conservation of energy, why do we subtract the potential energy to get the total energy instead of adding it? Also, I didnt fully understand why potential energy increases as you get closer to the planet, shouldnt it be the other way? thanks!

@martiancolonist

2 жыл бұрын

Great question! Potential energy is actually negative, so you are adding kinetic + potential, it's just that the potential term has a built in negative sign. To understand why potential is negative, consider an object at infinite distance from the Earth. That object has zero potential energy, since it isn't influenced by the Earth's gravity. When the object moves closer to Earth, it gains kinetic energy. So for energy to be conserved, it must have lost potential energy. Hence the potential energy is always negative for any finite distance.

@aryanthacker

2 жыл бұрын

@@martiancolonist Oh that makes a lot of sense, thanks a lot!

14:38 can we plug in any given variables from different planets?

This is wizardry

great video but why do you subtract the potential energy to get the total energy. I thought in mechanics PE + KE is constant?

@martiancolonist

6 жыл бұрын

They are indeed added, but gravitational potential energy is a negative number (-GMm/r). By convention, the GPE is 0 at infinity.

@S24W2

6 жыл бұрын

+Martian Colonist thank you for your reply, I understand now

erinvr sent me here

@martiancolonist

2 жыл бұрын

Cool, welcome!

I m not an engineer But i dont know how this information is useful... But i like the way youre explaining it... I also understood it well

Your powerful cannon analogy is incorrect. If a super powerful cannon fired a round into orbit it will still return to the surface and will not continue to orbit.

Nice video but re escape velocity you said "sitting on the surface" (meaning v=0 ?) but then talk about velocity (at launch? So the very first velocity at liftoff ? ) but then equate the energy to the energy when it escapes, but the rockets are firing all the way from liftoff to escaping (or at least most of the way) , putting in more energy, so how can these energies be equal to each other ? And how can v mean first the velocity when sitting on Earth but then mean escape velocity ? This section could be explained better. Otherwise great video and very helpful thank you.

I was watching your interview with Bas Lansdorp and Alan Titchmarsh recently. When Maggie Lieu mentioned that there could be dead bodies beneath the surface of Mars, do you think that she meant it?

@martiancolonist

9 жыл бұрын

Don't take it too literally, she was joking that we don't really know what we might find on Mars.

@Togwak1

9 жыл бұрын

As impossibly ambitious as it all is, it's good to watch as it develops. I'd love to go on such an adventure, to Mars and back, I'm not sure id survive the trip there though. I'd do anything to go to the Moon, or even low earth orbit, but to go to Mars with no intention of even trying to get back. After the initial adrenaline and euphoria of getting there wears off after weeks, months or years. After you're used to the 40% gravity and silent dust storms. It's an amazing thought and it'd be incredible to watch from 50 million miles away but I think I'm with Alan Titchmarsh.

@martiancolonist

9 жыл бұрын

Togwak1 You don't have to go yourself in order to share in the experience. Bringing those stories back and inspiring the world is one of the primary motivating reasons to go.

@martiancolonist

9 жыл бұрын

gandharzero It's so exciting to think about, it's the things we really don't expect to find that I can't wait for :)

@robertlatta2019

9 жыл бұрын

Martian Colonist Of course the whole mars one thing came about because humans are naturally curious and want to explore new and exciting places, so i'm most excited about the colonists going on drives on the surface, when you know they have more than just a few days to explore.

I learnt more in 8min 30sec than in a whole year of physics at school and first year uni. What a shame I left it because I could not understand physics.

Sir, How can the last two "orbits" be "orbits" if they never return. I thought Orbit meant a guaranteed return no matter how long or short in time it takes. Halley's Comet etc. If it only gets a gravity turn once forever how can the last two be "orbits"?

@mliuzzolino

7 жыл бұрын

There are a few things going on here with the term orbit. First, colloquially we understand an orbit as a necessarily closed loop - this implication doesn't exactly transfer over to the vernacular of astrodynamics. That is, an orbit with an eccentricity equal to or greater than 1 (parabolic and hyperbolic, respectively) will not be a closed loop orbit, with respect to the Earth. Secondly, notice the qualification, "with respect to the Earth." This is critical. The conic equations you use to describe an orbit depend entirely upon your inertial frame of reference. The concepts in this video set the Earth as the frame of reference from which the relationships are derived. To demonstrate, let's turn to the case of a rocket in a closed, elliptical orbit around the Earth. Here, eccentricity lies between 0 and 1. Now, let's imagine that this rocket fires its boosters, introduced some energy into the system resulting in an escape velocity such that the eccentricity is altered to exactly 1 - i.e., a parabolic orbit. Now, the rocket will no longer be in a closed orbit around the Earth - it has successfully escaped the gravity well. However, in the inertial frame of the Sun it will be in an elliptical orbit. Depending on when, where, etc. the boosters are fired, the rocket can be trailing the Earth, leading the Earth, etc. in an elliptical orbit about the Sun. Further, if the rocket fires the boosters to achieve a velocity further such that the eccentricity exceeds 1, we enter a hyperbolic orbit. The extra velocity is what will enable the rocket to reach out towards the gravity wells of other celestial bodies, all whilst remaining in an elliptical orbit with reference to the Sun, a hyperbolic orbit with reference to the Earth, and depending on the parameters, it will lock into either a hyperbolic orbit with reference to another planet (i.e., a gravity sling shot or fly-by) or elliptical orbit (such as with the Apollo missions or when we eventually get our butts to Mars). The upshot being that you can be in multiple types of orbits at once, depending on your frames of reference. For any of the Mars satellites, they are currently in an elliptical orbit around Mars, an elliptical orbit around the Sun, and a hyperbolic orbit around the Earth (in some sense - but maybe an astrophysicist can correct me here). Hope that long winded explanation helps and didn't just make it worse...

I've always wondered the math needed to go to Mars. This is truly amazing. Thank you for the video! A short question though..You said that E_infinite=0. But how is that also equal to E_surface? The velocity at surface is 0 for sure, but the potential energy part is non-zero. So shouldn't E_surface be a negative value?

@martiancolonist

3 жыл бұрын

Sure, the potential energy on a surface is negative. Which particular part of the video are you referring to?

@palindrome06

3 жыл бұрын

@@martiancolonist Around 13:20. E_infinite=0. That I understand. But E_surface turns out to be negative. Shouldn't they be equal to each other due to conservation of energy?

@martiancolonist

3 жыл бұрын

@@palindrome06 The escape velocity is defined such that you have just enough speed to have exactly zero velocity by the time you reach infinite distance. From energy conservation, you have the sum of kinetic and potential energy. E_infinite is actually the sum of the kinetic energy at infinite distance (zero by definition) and the potential energy (zero from r being infinite). E_surface (the sum of kinetic and potential energy) will then also equal zero *only* when we are calculating the escape velocity. We are essentially searching for the value of 'v' for which the kinetic energy exactly cancels the potential energy, allowing you to ascend to infinite distance.

@palindrome06

3 жыл бұрын

@@martiancolonist Thank you for the explanation. I see, so E_surface doesn't refer to the energy of the state where the rocket just stands still on the surface indefinitely.

@martiancolonist

3 жыл бұрын

@@palindrome06 Correct. E_surface is the energy at the moment just after the engine has fired. If you consider the state of rest beforehand, conservation of energy only applies if you account for the energy released from the chemical reaction during fuel consumption.

Idk if this is because it's a game but in KSP I just do the transfer orbit in a way that your transfer orbit gets bent by the moon creating a "pass-by" (idk the real name) orbit where you have to slow down at the apoapsis of this orbit to create an orbit around the moon.

@jetpond7904

2 жыл бұрын

That “passby” is called either parabolic or hyperbolic orbit

5:44 it would've been nice to lay out the full 3rd law equation

Good day Sir Why is the velocity required to keep a satellite orbiting a planet much lower than the escape velocity? For the escape velocity, the distance r increases therefore the velocity should decrease until zero, while for the circular velocity, the satellite is much closer to the earth, r is smaller therefore a greater velocity should be required Why is it the opposite in this case?

@umuraydin4337

5 жыл бұрын

the escape velocity in this case is velocity that you have to have when you just lifted off

You've also got your Moon rocket going in the direction the engine is firing.

Please would be better you put subtitles all vídeos, because a lot of people can't to follow your teaching. thank you for you to put this kind of vídeos to us watching them.

@martiancolonist

9 жыл бұрын

Edmilson Oliveira I may be able to do this in the summer, in many cases it takes longer to add the subtitles than editing the video itself! :-)

For some reason.. I understand this meanwhile I cant do shit for school math and have to put in extra hours

Hi Sheldon! 😁

@martiancolonist

4 жыл бұрын

Bazinga!

@vamp366

4 жыл бұрын

Martian Colonist 😂😂😂

nice hamster ball, I have the same one

Try an app called simple rockets. You can build simple rockets and play with them in space, and see the orbits.

For the last part, wouldn’t the moon’s sphere of gravitational influence just capture you is your velocity is slow enough? That is how I’ve learned it and known it, unless your case was different and I didn’t notice

@iamdogefied

Жыл бұрын

Well you wouldn't be able to reach the moon if your velocity was that low, if that's what I think you're asking.

@jetpond7904

Жыл бұрын

@@iamdogefied that was 9 months ago and I don’t know what I meant. But thanks for answering, I’ll rewatch and take a look.

@nickjones1422

8 ай бұрын

He means retrograde burn to slow down as you approach moon so you can use its gravity to create a moon orbit.

just one question , if two satellites one in low earth orbit and one in high earth orbit , which one will be faster ?

@martiancolonist

7 жыл бұрын

The low orbit is faster.

@seifraslan8186

7 жыл бұрын

Martian Colonist thank you so much that means that the iss is travelling faster the the moon

@martiancolonist

7 жыл бұрын

Yes, the ISS is about 7x faster than the Moon's orbit around Earth!

@ 5:31: How can a period (= time-unit) be equal to a distance (= length-unit)?

@martiancolonist

4 жыл бұрын

The constant of proportionality has units of time^2 / length^3.

@jacobvandijk6525

4 жыл бұрын

@@martiancolonist That explains a lot! So the =-sign is not quite right ;-)

@martiancolonist

4 жыл бұрын

@@jacobvandijk6525 It is correct in the specified unit system (years and AU) only, for which the constant of proportionality is 1 by definition ;)

@jacobvandijk6525

4 жыл бұрын

@@martiancolonist Explaining simple things in a difficult and confusing way helps nobody, mr Einstein.

Hold tha phone!

Hi, AFAIK - G in formula fo F(orce) is not quite correct. Standard constant is kappa which is 6 and something *10^ -11. If you use result from M(Earth´s),r*r and kappa you will get "g" Yes in other countries it could be different, but here is question how here can be seen difference between kappa(6.67*10^ -11) and G(9.98m/s/s) if both are G ;) - periapsis and appoapsis are general terms. In case of sun are called perihelium and appohelium and in case of Earth perigeum and appogeum. - I have a question. Why we need to accelerate to velocity which equals Moon´s. As far as I know it is enought to reach L1 point, where ship is accelerated by gravity of second object instead of its engines. Apollo 8 had such trajectory as mission sign. I think that it was based on reality. Your example is good if we would set something on the same orbit as Moon has, but in case of reaching Moon we can use its gravity to complete manoeuvre. (man is learning whole life ;) )

@martiancolonist

9 жыл бұрын

Newton's Gravitational constant G=6.67*10^-11, the acceleration of gravity at Earth's surface is g=9.81 m/s^2, so the equation for force is indeed correct. Correct, the general terms are also true for the sun, so I chose to use them instead of the sun specific term. But you are right, I should have said 'closest point to any gravitating object', but I hope the meaning was clear. This video focused on calculations involving just two objects (the rocket + Earth), but in the real world with 3 or more objects things become more complicated. There is no exact solution to the 3 body problem - which is where the Lagrangian points come from. My value for the escape velocity will be larger than what you need to get to the Moon (as you don't need to get to infinite distance away), and the Moon's gravity dominates once you get close enough.

@stanislavbandur7355

9 жыл бұрын

yes in general, it is much more complicated to make calculations on three and more objects with gravity. But in case of two, here is possibility to make gravity as function of distance (from one centre of gravity to another) and split it to two distinct equations. One from first gravity centre to zero gravity point (L1) and second from second centre of gravity to zero gravity point (L1) Centrifugal forces are included in v*v/r part of equation and ship gravity does not need to be considered. Yes of course, if we want to make superprecise calculations, we need to apply all forces (gravity and centrifugal), but it is much simpler to use correction blasts during the fly, because it is not impossible (by relativity theory I mean, not Newton´s mechanics), but so complicated that it not worth to do it.

Why would the object in orbit not eventually fall though?

In the escape velocity equation, why does he drop the mass of our rocket from the whole energy equation (E= 1/2mv*2 - GmM/r)? Be nice I'm not mathsy

@martiancolonist

6 жыл бұрын

The mass of the rocket (m) cancels out when you use conservation of energy to set the energy at two different points in the orbit equal to one another.

Do you guys think you could make 4x90 minutes math lessons for ~17 years old people of this topic?

@martiancolonist

4 жыл бұрын

I recommend you check out the Khan Academy's mathematics video series.

you remind me of the trio gang from buffy.

@martiancolonist

7 жыл бұрын

Wow, that's the first time I've been compared to them, bravo. (Most people go for Sheldon).

Shouldn't E be the sum of kinetic energy + Potential Grav Energy? Instead of minus?

@martiancolonist

3 жыл бұрын

They are added, but gravitational potential energy is a negative quantity (it is defined as zero at infinite distance, so when objects get closer and the potential energy drops it becomes negative).

@peksn

3 жыл бұрын

@@martiancolonist Ah right!

Nice video, but KEPLERs 3rd law does not state, that P^2 = a^3. Years can not equal km (or AU). It states, that P^2 is proportional to a^3. In other words, P^2/a^3 is a constant. Greetings from Germany

@martiancolonist

6 жыл бұрын

Indeed, P^2 = k a^3, where k = 1 (Yr^2/AU^3) by the definition of the Astronomical Unit. Physicists are lazy with dimensions (e.g., we routinely set the speed of light to c=1 in QFT), so I omitted them for expressional clarity. Further note that this also only holds in the solar system. More generally, P^2 = (4 pi^2/G(M_star + M_satelitte)) a^3.

@ganymedkallisto5561

6 жыл бұрын

+Martian Colonist well, I just thought, this is a little bit too lazy. Doing so, you could also say, that Ohm's law states, that current equals voltage ;-)

@martiancolonist

6 жыл бұрын

We only do that for constant quantities. 4pi^2 / GM_Sun is a well-established constant, whereas R in Ohm's law is a function of length, area, material, temperature etc - so not quite analogous :-)

@Dieterbe

6 жыл бұрын

Ganymed Kallisto yes. Units are a beautiful thing. I love how by applying them correctly you have an extra safety check to validate an equation. But I can see how it becomes boring for physicists who do the math so frequently they want to take shortcuts

I challenge you to do inclinations? Because anyone can do simple calculations like these.

@martiancolonist

7 жыл бұрын

I'm also tempted to do fully general relativistic orbits (derived from the principle of stationary action) and apply them to Mercury's orbit and black hole accretion. Ultimately this video is as far as someone can go without knowing calculus, which is the level I was aiming at in this presentation.

@wildmonkiesJR

7 жыл бұрын

when is that comin?

@martiancolonist

7 жыл бұрын

Not for a while, since I'm quite occupied with my own research (exoplanet atmospheres) at the moment. In the meantime, I do also have a more advanced video on Kepler's laws that you may find at a more appropriate level ;-) kzread.info/dash/bejne/dqmmrriBgMaXedo.html

@wildmonkiesJR

7 жыл бұрын

Had any success?

@martiancolonist

7 жыл бұрын

wildmonkies Jr Yes, writing a paper at the moment.

So, in a way, one could say an orbit is falling horizontally.

Yes i know1

What the hell is "angular mentum"?

@martiancolonist

3 жыл бұрын

en.m.wikipedia.org/wiki/Angular_momentum

This is useful and real basic material, but it is not complete. The demonstration does not take into account the influence of the sun over the spacecraft elliptical orbit and the influence of the moon over the the same orbit. The spacecraft is always under the influence of the earth, sun and moon, though at the beginning the earth influence is the most important, then the suns's influence increases or decreases and finallly the as the spacecraft is getting closer to the moon its influence is getting proeminent. So the energy equation must take into account the earth, the sun and the moon at all the time. So it makes the equation much more difficult to resolve analitically , it must be solved numerically in solving a set of differential equations that relates distance and speed together. The Orbital Mechanics 101 is just a crude and fatal approximation to get a rough estimate of the problem to be solved but should not be construed to describe the entirety of the vast and intricate problem to be solved by mathematicians, engeneers and programmers. It is the 3 body problem, never resolved analitically, but numerically solved via infinite series.

@martiancolonist

3 жыл бұрын

Indeed. The point of this video is to give people physical intuition for the key principles driving orbital mechanics: energy and angular momentum conservation. Jumping straight into numerical solutions of the N-body problem is not helpful for a first introduction to the subject.

@jeanbossy6950

3 жыл бұрын

@@martiancolonist Please, if you ever update your presentation, clearly indicates at the beginning that the calculations is a first approximation to the real problem, to show the basic principle of a spacecraft trajectory using the concept of orbit, or ellipse. You have rightly mentioned the influence of Jupiter over the trajectory which is far away from the Earth and Mars while the Earth influence over a mars mission trajectory is much more important that the influence of Jupiter. Also, stressed the important fact this is a three dimentional problem, plus the time dimention of course, in showing the different orbital planes involved from the spacecraft launch to Mars arrival.

@martiancolonist

3 жыл бұрын

Certainly. And if high accuracy is required you must also consider post-Newtonian series approximations of GR. The point being that we are always forced to make approximations to model any phenomena in the real world, everything is an infinite Taylor series 😉

nice video, but you have a strange way of holding a pen.

@ 14:52: Have you ever seen a rocket being launched at a speed of 11.2 km/sec.? ;-) No, me neither. Although I like your video, I'm getting a little itchy ... .

@willbrother9246

4 жыл бұрын

Jacob van Dijk That’s unfairly pedantic comment. The rocket only needs to reach that velocity once it is in an orbit. If you even attempted to do just ‘fly off the surface of the planet’ you would most probably crash back down.

@jacobvandijk6525

4 жыл бұрын

@@willbrother9246 Pedantic? Really? In my opinion, physics tries to describe reality! Taking off with 11.2 km/sec. is just ridiculous. Why didn't he tell me that (in reality) rockets get pushed away from the surface of the Earth, and that therefore such an unrealistic speed isn't needed?

poor sound quality

Your speech is fast, correct it

Orbital mechanics lol. It's a science fiction story

@thebibleman1349

5 жыл бұрын

what?

@ozzypigg7351

5 жыл бұрын

what?

Mission to Mars ...Yea right..Mission to Devon Island.. Just like trying to tell us we have a Molding magnetic earths core..and how has that ever been proven ... Looking forward to all the warped indoctrinational thought you all would just love to share that knowledge for us who may be a little baffled...When the deepest hole ever dug was shorter than 8 miles in a joint effort by the United States and Russia..Make it good please..

@ngc-fo5te

4 ай бұрын

This is all too much for you, isn't it.

I am sorry to say that, but you have a terrible mistake in this video: p² = a³ is nonsense! Kepler's 3dr law say that p² is PROPORTIONAL to a³, but NOT EQUAL! A time to the power of two can NEVER be the same as a distance to the power of three. That is simply not possible.

@martiancolonist

4 жыл бұрын

This expression has P in units of years, and a in units of astronomical units. By the definition of the astronomical unit, the constant of proportionality = 1 (yr^2 / au^3). For brevity, the units of the constant of proportionality are not written, as the constant is equal to unity.

@ZaHaDum1984

4 жыл бұрын

Martian Colonist The true relation between the orbital period p (usually written as T) and the semi-major axis a is T = 2*%pi*sqrt(a^3/(G*M)), with M being the mass of the orbited body. If T^2 = a^3 would be true, a circular orbit of a fixed height would always have the same velocity and period, independant of the mass of the central body that is orbited. Besides you can not just ignore units in science. Years are a multiple of seconds and AU is a multiple of meters. So your equation T^2 = a^3 is still nonsense. Proportional and equal are NOT the same.

@martiancolonist

4 жыл бұрын

*sighs* If you want to be pedantic, also note that 'M' in your expression should actually be M_object + M_star in the general formulation of Kepler's 3rd law. This only approximates to M_star for M_object

@ZaHaDum1984

4 жыл бұрын

Martian Colonist You are absolutely right about M. Mostly it is accurate enough to use only the mass of the orbited object, for the reasons you mentioned. Nevertheless Kepler‘s 3rd law doesn‘t say T^2 = a^3. It says (T^2 / a^3) = const. (if M does not change). Again, proportional and equal are NOT the same.

@martiancolonist

4 жыл бұрын

Sure, we are in agreement there. The constant happens to have a magnitude of 1 with the units chosen in the video. This is perhaps an abuse of notation, but was intended to avoid going off on a tangent about G, M etc. Reminds me that I never got around to making a video where I derive Kepler's 3rd law from first principles!