DERIVING the Orbit of Our Home Planet from Newton's Law of Gravity: Physics Mini Lesson

Поделиться
HTML-код
  • Опубликовано: 10 дек 2024

Комментарии • 134

  • @dwaynep6174
    @dwaynep6174 3 года назад +71

    I’m a freshman physics major and this channel gives great insight into things I may be learning in the next few years. Love it!

  • @williamsutter2152
    @williamsutter2152 2 года назад +19

    I'm a maths and stats major, haven't done any physics courses, but for a C++ computing course I took I created C++ programs for modelling the trajectory of the Moon and Earth. I derived my equations of motion using the Euler-Lagrange equations, which I taught myself in high school. Back then the great resource of your videos didn't exist, the best available were MIT open courses and Susskind's lectures at Stanford.

  • @bastianherrera6408
    @bastianherrera6408 3 года назад +47

    The clarity with which you explain these concepts is outstanding. Your videos are awesome, hope more people appreciates the quality of your content. Keep it up!

  • @vassilioskaxiras5724
    @vassilioskaxiras5724 3 года назад +20

    I never knew physics could be so "grounded". Keep up the good work!

  • @TIO540S1
    @TIO540S1 3 года назад +9

    The work you put into these is amazing. That you add notes and problems makes it even more so. I hope you can keep it up! Thank you!

  • @javierrucal4813
    @javierrucal4813 3 года назад +2

    As a first year physics major, this is hugely inspiring. Your channel is fantastic! Please keep it up! Love from Guatemala.

  • @sasamladenovic6120
    @sasamladenovic6120 3 года назад +4

    These videos are fantastic. Explanations are very clear and understandable, but without the loose of generality. I wish that I had youtube and this channel when I take a physics course in university! Keep going, great job!

  • @informalchipmunk5775
    @informalchipmunk5775 Год назад

    Wow! Just found this now, I love how you explain everything clearly whilst also forcing me to do some work as I watch

  • @Darthvanger
    @Darthvanger Год назад +2

    Just how genius Newton was to come up with all of these by himself in 17th century.
    Today ppl take the orbits for granted, as "obvious". Only a few realise how actually hard it is to solve this problem.
    Personally I was astonished at the difficulty level after solving simple stuff with no integrals in high school :)

  • @andreaskleinert7756
    @andreaskleinert7756 2 года назад +1

    It was incredible deep and fun. Guys like Ptolemy spent their lifes thinking in this kind of problems. I want more of this types of videos!

  • @chrisr9320
    @chrisr9320 2 года назад +5

    Great video, just an addition/clarification to what you said about the perihelion precession of Mercury: the precession itself can be derived from Newtonian gravity if you take into account there are other planets that influence Mercury's orbit, plus other smaller perturbations. You will need General Relativity to get the right amount of arcseconds of precession, though (575"/century instead of the 531"/century predicted by Newtonian perturbation calculus, the famous missing 43")

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +6

      Thanks Chris! Yes, though what I was trying to say here is that even for a single planet orbiting a star the orbit will precess in a way that's not predicted by Newton's law of gravity

  • @agrajyadav2951
    @agrajyadav2951 2 года назад +1

    best physcis channel!

  • @press2701
    @press2701 8 месяцев назад

    A very clear, clean and thoughtful lecture, sir. Well done.

  • @kmahadevan6110
    @kmahadevan6110 Год назад

    Thank You! Today I learnt how a topic can presented in its simplest but rigorous form.

  • @AffaAu
    @AffaAu 6 месяцев назад

    Brilliant. Have been looking for such mathematical explanation!

  • @anishsharma6702
    @anishsharma6702 2 года назад

    this videos are like gold , waiting to be explored , by a curious mind,

  • @chrispw144
    @chrispw144 3 года назад +4

    Awesome video! You could say it was a "stellar" explanation :)

    • @PhysicswithElliot
      @PhysicswithElliot  3 года назад +7

      Stellar? I have no "Earthly" idea what you're talking about Chris

  • @gaeb-hd4lf
    @gaeb-hd4lf 3 года назад +2

    Awesome channel bro, glad i found it!

  • @alejandroalmarza8447
    @alejandroalmarza8447 Год назад

    You are excellent profesor …keep this great physic videos coming

  • @varunahlawat9013
    @varunahlawat9013 2 года назад +1

    Really really helpful!
    Please more lectures like this :)

  • @JasonAStillman
    @JasonAStillman 10 месяцев назад

    That was awesome, really enjoyed that, thanks E.

  • @TeamHarrisonMachine
    @TeamHarrisonMachine 6 месяцев назад

    Thank you!!! I was just thinking about this kind of tutorial!! I learned that Uranus was discovered through mathematics before it was located visually and I wondered how. How did the math look? There’s others too. For example, Kepler realized the orbits were elliptical… how? How, by only looking trough a telescope were they able to figure out the precession of Mercury? I was thinking if I had enough money someday I would pay a physics instructor to make a tutorial about those things. This is a step in the right direction. Thanks for this!

  • @過客-c5u
    @過客-c5u Год назад

    Very clear and concise explanation.

  • @truthprevail2742
    @truthprevail2742 Год назад

    I really thank you very much, i was looking forward this concept mathematically for so long🙏

  • @sergiolucas38
    @sergiolucas38 2 года назад

    Great video, man, so much quality :)

  • @macrosense
    @macrosense Год назад +1

    How would you know the mass of the sun, the earth, or the ratio between them?

  • @zack_120
    @zack_120 Год назад

    So cool! Literally you pictured the pathes of planets by actual numbers. In some cultures you would be painted as a supernatrural figure, fortune teller or magic.

  • @peterhall6656
    @peterhall6656 2 года назад +1

    Feynman did an even more basic version of the path derivation in Volume 1 of his Lectures on Physics pages 9-6 to 9-9. That derivation did not involve all the calculus employed here.

  • @paulboro5278
    @paulboro5278 3 года назад +2

    Awesome explanation.

  • @JulieanGalak
    @JulieanGalak 11 месяцев назад +1

    11:37 - am I missing something here? Every calc book I've ever read emphasizes that "d/dt" (or whatever variable) is not really a fraction, it's an operator. So how can you "cancel" the "dt"s here? Am I missing something fundamental about the math involved?

    • @gmncnr
      @gmncnr 3 месяца назад +1

      Welcome to physics

    • @EvilDudeLOL
      @EvilDudeLOL Месяц назад

      When they say, "df/dt is not a fraction", they mean to say that df/dt is a ratio.
      It is not like saying "df things out of dt things".

  • @steffenleo5997
    @steffenleo5997 2 года назад +1

    Great Videos..... 👍👍..... Thanks a lot..... Have a nice weekend

  • @aoooriel2361
    @aoooriel2361 2 месяца назад

    wow enjoyed the video so much.

  • @ravikantpatil3398
    @ravikantpatil3398 7 месяцев назад

    Excellent very exciting .. thanks Eliot

  • @tgylfason
    @tgylfason 2 года назад

    Brilliantly lucid lectures, thank you.

  • @astridlovespie
    @astridlovespie Год назад

    Amazing video and great explanation! I am so glad I came across your channel while studying this topic.
    I have a question regarding the video though, how do we know for sure that the integration constant from 14:33 is indeed the eccentricity of the orbit?
    Thank you for the video and keep going!

  • @sairithvickg6663
    @sairithvickg6663 2 года назад +1

    I am amazed ! Thanks a Lot for this!!!

  • @pedrojuan5550
    @pedrojuan5550 2 года назад

    Very well put. Thank you.

  • @kennerlyjomorala1658
    @kennerlyjomorala1658 Месяц назад

    Absolutely amazing👏👏👏👏

  • @sairithvickg6663
    @sairithvickg6663 2 года назад +1

    2:32 couldn't we just observe planets motion in Sun's frame, since in that case the modification we need to do would be converting k to k' and then we are set
    By the way great video

  • @maxdemuynck9850
    @maxdemuynck9850 2 года назад +1

    Hi Elliot, love your videos, there just great:) was just wondering if maybe you could do a series on a second year E&M course?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +2

      Thanks Max! I certainly hope to make more E&M videos. It's possible I'll make a full E&M course down the line. Right now I'm working on creating a course on Lagrangian mechanics.

  • @andreaq6529
    @andreaq6529 Год назад

    Thank you, I was trying to derive the ellipse equation by kinematics only and totally forgot about using energy.

  • @owen7185
    @owen7185 2 года назад

    Another awesome lecture

  • @jdalton4552
    @jdalton4552 Год назад

    These formulae could be more easily solved using the formulae for p, the semi-latis rectum of an ellipse: r=p/(1+e cos theta) and p=a(1-e squared). However the beauty of his calculus is worth the extra effort.

  • @normdesbiens726
    @normdesbiens726 2 года назад +1

    Great derivation, but can you spend a bit more time with the differential equatiron? Won't u(theta) = epsilon sin(theta) also solve the equation? Why do you pick cos? Also I'm not sure how you handle the + 1, since the differential of 1 with repsect ot theta is 0 so it would drop out with the first differentiation.

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +2

      It's the simple harmonic oscillator equation, but for u-1 instead of u. You could define a new variable z = u - 1, and then the equation is z'' = -z. Solve for z, and then get u back from u = z + 1.
      Yep, you could also write down a sine solution. Or equivalently, you could write the solution as cos(\theta - \theta_0) with another parameter \theta_0. Different choices for \theta_0 just amount to different choices of how you set up your coordinates. I set \theta_0 = 0, which means that the planet's point of closest approach to the star happens at \theta = 0.

  • @sarthaksamal182
    @sarthaksamal182 3 года назад +1

    Great explanation man! Where can I get the animation that you showed in the end?

    • @sarthaksamal182
      @sarthaksamal182 3 года назад +2

      Actually I want to present a ppt on the runge-lenz vector so it would really be a help if I get the similar kind of simulation, that you showed, somewhere

    • @PhysicswithElliot
      @PhysicswithElliot  3 года назад

      Thanks Sarthak! You can find all the links here: www.physicswithelliot.com/orbits-mini

  • @yosemiguelsotomayorcarnead9909
    @yosemiguelsotomayorcarnead9909 2 года назад

    you are amazing man

  • @jordanfreidel1751
    @jordanfreidel1751 Год назад

    Why at 5:15 do you draw the perpendicular component to the velocity? I don't see why.

    • @josemanuel9305
      @josemanuel9305 10 месяцев назад

      Cross product by definition needs two perpendicular vectors.

  • @dr20231may
    @dr20231may Год назад +1

    It would be better if the explanation done here is either done entirely on a paper and just record that paper work or use a neat font for computer explanation and just don't use digital pens for illustration , use pre-sets like LINE CURVE SHAPES SURFACES VOLUMES etc ,

  • @Ninjamanguy732
    @Ninjamanguy732 Год назад

    Thanks for the great video! Can i ask why you aren’t factoring the Earth’s spin into the angular momentum and kinetic energy calculations? Thanks!

    • @Jackie-wn5hx
      @Jackie-wn5hx Год назад

      I'm definitely not Elliot, but I was curious as well. He used point mechanics to derive the orbits and celestial mechanics.
      Were you asking whether the rotational motion and kinetic energy of the earth will have a significant
      impact on the shape of its orbits around the sun?
      I've seen gravitational problems in general relativity that have to factor in the effects of frame dragging and spinning black holes and neutron stars.
      For most orbital problems in Newtonian mechanics and relativistic physics, point masses are used.

  • @briannguyen6994
    @briannguyen6994 3 года назад

    I don't understand the step at 11:50 when you divide by (dtheta/dt)^2 to get the green equation. more specifically, where did the (l/mr)^2 term came from as a multiplication. Could you care to explain in a little more detail for me please? :D

    • @PhysicswithElliot
      @PhysicswithElliot  3 года назад

      I'm dividing (dr/dt)^2 by (d\theta/dt)^2 there to get (dr/d\theta)^2. From the angular momentum equation d\theta/dt = L/(mr^2). I skipped a step where I multiplied that (L/(mr^2))^2 in the denominator on the right-hand-side over to the numerator on the left-hand-side of the equation.

  • @palfers1
    @palfers1 Год назад

    Your value for the eccentricity is >1, which (according to you) is hyperbolic if E>0. I assume that one can prove that E

    • @EvilDudeLOL
      @EvilDudeLOL Месяц назад

      Potential energy is always negative

  • @frankreashore
    @frankreashore 2 года назад

    Really nice!

  • @joaco4ever11
    @joaco4ever11 Год назад +1

    is there a way you can solve the equations with respect to time ?

    • @itoshiigrauben
      @itoshiigrauben 5 месяцев назад +1

      Actually, by combining some geometry with the fact that the area-sweeping speed is constant , one is able to solve the time respective to the position of the planet t(x,y). However, the equations of x(t) and y(t) for the inverse-square law orbit are non-elementary. You can look up Keplers' Equations for elliptical orbits.

  • @jmmahony
    @jmmahony 2 года назад +2

    Twice you said hyperbolic orbits were like a comet going by the sun. This is generally not true: comets are generally in elliptical orbits, but often in extremely eccentric orbits, where the part of the path we can measure (when it's close enough to the inner solar system to be observed from earth) may be indistinguishable from a parabola (the borderline case between a closed elliptical orbit and an open hyperbolic trajectory). But still these are thought to generally be on elliptical orbits, ie, still bound to the sun, in a closed orbit. Only very recently did we finally spot an object "'Oumuamua" that was clearly an extrasolar object on an open hyperbolic trajectory, veering by the sun on its way through the solar system. And then we found a second, but that's been it, so far.

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      Very cool I hadn't heard of that object! Indeed comets can certainly follow elliptical orbits. I think I mentioned Halley's Comet in this video

    • @carultch
      @carultch Год назад +1

      A hyperbolic "orbit" would be like a rogue comet that isn't bound to the sun, that finds its way into the solar system, only to escape and for us to never see it again.

  • @r2k314
    @r2k314 Год назад

    I always think when I see this kind of derivation, how do you know that you haven't irretrievably lost a variable or constant or some other relationship. I wish there was a course on the limits of substitutions, differentiation, etc. Is there a reference which deals with all the legitimate tricks in one place?

  • @MarceloRobertoJimenez
    @MarceloRobertoJimenez 2 года назад

    Congratulations for the excelent content of your videos.
    Can I suggest you a video? Use the data Newton had available to derive the orbits of the planets.
    Best regards.

  • @Daniel-c3l1k
    @Daniel-c3l1k Год назад

    For the first time I have come across what I wanted to learn most: how Newtonian laws apply

  • @anthonyorf8827
    @anthonyorf8827 10 месяцев назад

    so how do I calculate dr/dt and dθ/dt when given the radius vector and the velocity vector

  • @briangoad8016
    @briangoad8016 2 года назад

    Can you use this equation to tell where a planet will be in the night sky? Does time T represent calendar time?

    • @carultch
      @carultch Год назад

      Yes. This is exactly what they use to predict the positions of planets.
      The t doesn't necessarily represent calendar time, but rather just time relative to an arbitrary point we chose to call t=0. Usually lowercase t represents a specific point in time, while in this context, capital T represents the period of orbit.

  • @raulcantu6998
    @raulcantu6998 Год назад

    Is there a position vs time formula, asside from aproximations, and if so, could you give the solution :o

  • @asifshahriyar6382
    @asifshahriyar6382 Год назад

    How amazing!

  • @Structure-Dubai
    @Structure-Dubai 2 года назад

    hi sir, what is the physical meaning of a position vector? is it some kind of position coordinates or direction of position or what?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      It's an arrow pointing to the location of a particle from whatever origin you've chosen

  • @jordanfreidel1751
    @jordanfreidel1751 7 месяцев назад

    Why can’t I use sin theta in the final equation? Or does it not make a difference, or am I just blind and short circuiting when doing the second derivative?

    • @robert.ehrlich8942
      @robert.ehrlich8942 7 месяцев назад

      There is something missing at this stage in the video. After saying that cos theta or sin theta are solution of the equation without the "1", he keeps only the cos variant without further explanation. For such linear equations, any linear combination of solutions is also a solution,
      so the general solution would be A cos theta + B sin theta. Not surprising, a second order differential equation usually depends on 2 integration constants. This can be rewritten as C cos (theta - phi) with the two new constants C = sqrt(A²+B²) and phi = atan(B/A). In the following he assumes that phi = 0, which can always be obtained by rotating the x,y axes by phi, so eliminating one of the two constants. This is why all the following trajectories are symetrical relatively to the x axis, which is not implied by the initial description of the problem.

    • @jordanfreidel1751
      @jordanfreidel1751 7 месяцев назад

      @@robert.ehrlich8942 then that would mean that using sin theta would be an acceptable answer correct?

  • @alfredkokou2013
    @alfredkokou2013 2 года назад

    Great video. But how does the value of epsilon get lower than 1 given that you derived it as epsilon = sqrt ( 1 + x) where x is always positive ( x = 2EL^2 / mk^2) ?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      E is negative for circular and elliptical orbits, so no x doesn't need to be positive!

  • @monawykayla2156
    @monawykayla2156 5 месяцев назад

    Very well explained. However, I didn't quite get the final conclusion. If the equation applies to circles, ellipses, parabolas and hyperbolas, why are the planets orbiting in ellipses? He mentioned the angular speed, but can someone explain it further? I think I didn't get the "that's why earth is orbiting in ellipses".

    • @mikev4621
      @mikev4621 5 месяцев назад

      The type of conic section followed by any solar satellite is dependant on the total energy the satellite has. They all 'sling-shot ' around the sun, but not all have enough energy to escape

    • @EvilDudeLOL
      @EvilDudeLOL Месяц назад

      ...if the planet's trajectory was a parabola/hyperbola, then it wouldn't really be "orbit"-ing the star, would it?

  • @mandystarr404
    @mandystarr404 Год назад

    Does anyone know if this equation could be written as a vector expression with i^ and j^

    • @AwfulnewsFM
      @AwfulnewsFM Год назад +1

      Yes, you can use the simple transform equation
      r_vector = r*cos(theta)*i^ + r*sin(theta)*j^
      Just put in the expansion for r

    • @mandystarr404
      @mandystarr404 Год назад

      @@AwfulnewsFM thanks 👏🏼

  • @Paolo-sy7wf
    @Paolo-sy7wf 2 года назад

    Hi, your videos are really amazing for those who want to go deeper in the theories and know where the laws come from. But i had an issue with the solution of the differential equation r(theta). As you explained (for example) in the pendulum video the solution for that DE is: a sin(x)+b cos(x) and i don't get how did you ended up choosing b cos(x). Did you consider the initial conditions to determine that? I tried also to figure it out with another method (taking square root of dr/dt and separating variables after the division of the 2 equations) but i can't solve the integral which should result in theta+theta0 = arccos((b^2 - ar)/(sqrt(a^2 - b^2))), with a and b being some rearrangements of G, M, m, L (this method is presented in a book where it misses the solving part).

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +2

      Thanks Paolo! Picking the cosine solution just means that I set up the coordinates so that theta = 0 when the planet is at its point of closet approach to the star. You could also add a sine term-it would just amount to rotating the ellipse around.

  • @kaypee9187
    @kaypee9187 3 года назад

    Sir, A very good video. One question though, would one have a Hyperbola in an attractive central force such as gravity. Even comets have highly eccentric orbits. The voyager spacecraft would be a parabola of course its angular momentum might be changed by the rocket thrust at the appropriate moment. Also the derivation of the locus was an exercise in mathematics!. Congratulations!

    • @PhysicswithElliot
      @PhysicswithElliot  3 года назад

      Thanks Kaypee! The trajectory is an ellipse when then eccentricity is less then 1 and a hyperbola when it's greater than 1.

    • @Salien1999
      @Salien1999 3 года назад +3

      Or in physical terms, if the object has reached escape velocity, it will follow a hyperbolic path. Otherwise, it will follow an elliptical path (at least according to classical mechanics)

  • @arkopro30
    @arkopro30 2 года назад

    Why we failed to find celestial objects moving in parabolic or hyperbolic curves ?

    • @carultch
      @carultch Год назад +3

      Because those aren't the orbits that stand the test of time. By definition, those trajectories are escape paths of bodies that don't stay in the solar system. A parabolic "orbit" is a special case of orbit theory, where an object has exactly enough energy to escape the gravitational field. If you had a spacecraft that you put in motion exactly at escape velocity, and then continue to coast, it would follow a parabolic "orbit" as it escapes its home world. A hyperbolic "orbit" is what you get when you have more than enough energy to escape.

  • @glennrubel4326
    @glennrubel4326 Год назад

    So how do you get an elliptical orbit from an eccentricity that is always greater than one?

    • @itoshiigrauben
      @itoshiigrauben 5 месяцев назад

      Total mechanical energy E is a negative quantity for elliptical orbits

  • @maxst6647
    @maxst6647 2 года назад +3

    My dear sir,
    Right at 13:00 you leave out the crucial step in how to proceed from a DE in r'(\theta) to one in u'(\theta). You are therefore doing the same thing that articles and texts in mathematics do: "Clearly it follows..." or "We will leave it to the reader to show...". This is not pedagogically sound nor is it convincing.
    Without justifying the step at 13:00, you have proved exactly nothing. It's not that your conclusion is wrong, but rather that you have not proved it. Furthermore, you cannot treat dr/dt (or any other derivative) as a fraction and claim that the "dt's cancel"; this is just pure rubbish.

    • @EvilDudeLOL
      @EvilDudeLOL Месяц назад

      I'm sorry, but you *absolutely can* cancel dt's. I don't speak alone here, take a quick search on RUclips or Google: not only is it perfectly okay to cancel the differential under all conditions, the differential is, in contrast to popular belief, a tiny but absolutely *REAL* number, not an infinitesimal. Thus it makes perfect sense to cancel out the dt's.

    • @EvilDudeLOL
      @EvilDudeLOL Месяц назад

      Furthermore, I would like to note that your use of the term "proof" is lossy. I think the term you meant to refer to was "derive", not "prove". In which case, I agree, this isn't really a "derivation" as promised by the video's title, but it is certainly a "proof".

    • @maxst6647
      @maxst6647 Месяц назад

      @@EvilDudeLOL No, you cannot cancel differentials. Go back to school.

  • @chritophergaafele8922
    @chritophergaafele8922 3 года назад

    Can you explain the mecury perihelion shift solution to einstein's field equations

    • @PhysicswithElliot
      @PhysicswithElliot  3 года назад +1

      Possibly! I was thinking about whether I could cover it in an accessible way

  • @chritophergaafele8922
    @chritophergaafele8922 3 года назад

    I like orbital mechanics

  • @mcalkis5771
    @mcalkis5771 2 года назад

    So the final equation still requires us to know a planet's total energy and angular momentum? How is that done?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +2

      Those just amount to the initial conditions, like when you throw a ball you need to tell me what it was doing at t = 0 to be able to write down the trajectory after that.

    • @mcalkis5771
      @mcalkis5771 2 года назад

      @@PhysicswithElliot So how would we know the angular momentum of the earth if we were just standing on its surface? How did Kepler and Newton do it?

    • @Tobi21089
      @Tobi21089 2 года назад +1

      @@mcalkis5771 I guess by watching the skye and looking how long it takes for one orbit and maybe they did calculate R by triangulation of the planet but that's just a guess for their velocities around the sun

    • @carultch
      @carultch Год назад +2

      @@mcalkis5771 At the time of Kepler and Newton, we couldn't know what the momentum of the Earth is, in kg-m/s. We could only get the relative masses of the planets from extrapolating the observational data, and comparing it to the theory behind orbital mechanics. For instance, we could know the Earth is 81 times the mass of the moon, and that the sun is 333000 times the mass of the Earth, but we couldn't know how many kilograms each one of them were. The same is also true with knowing how many meters or kilometers away the sun is. We could know the relative scale of the solar system, but not the actual distance
      It wasn't until Cavendish, whose work was in the century after Newton's life, that we could know the actual mass of the Earth and celestial bodies. Cavendish "weighed" lead spheres in each others' gravitational fields, in order to determine the universal gravitational constant, and isolate the G from the GM product of astronomical bodies. From that information, we could solve for M to know their masses, as an application of this knowledge.
      Similarly, it also wasn't until Captain James Cook, and Lewis Swift, until we could know the actual scale of the solar system, and know that the sun is 150 million km away. They observed the transit of Venus from a known distance apart on opposite sides of the world, to determine this.

  • @bygabop9368
    @bygabop9368 9 месяцев назад

    A straight line must also be a solution, i.e. initial angular speed equal to zero, I guess.

  • @realcygnus
    @realcygnus 2 года назад +1

    Nifty

  • @jcffilms256
    @jcffilms256 2 года назад

    GOAT

  • @observer137
    @observer137 Год назад

    Why is there no solution where object falls into the sun?

  • @veronicanoordzee6440
    @veronicanoordzee6440 Месяц назад

    @ 7:21 Should be r x F.

  • @billthomas7644
    @billthomas7644 Год назад

    A bit of a sleight of hand here. Makes the assumption that the m in Newton's gravitational force equation has anything to do with the "m" used in the momentum and energy equations. Using the same symbol doesn't justify it.

    • @JulieanGalak
      @JulieanGalak 11 месяцев назад +1

      Huh? Mass is mass, it's not just the same symbol, it's the same quantity....

    • @billthomas7644
      @billthomas7644 11 месяцев назад

      @@JulieanGalak Nope. You have to prove that inertial mass and gravitational mass are the same. Look up the principle of equivalence. Hundreds of experiments have been carried out attempting to either prove or disprove this.

    • @JulieanGalak
      @JulieanGalak 11 месяцев назад

      @@billthomas7644 - interesting, I'll have to look at that.

    • @JeffreyHoeper
      @JeffreyHoeper 8 месяцев назад +1

      @@billthomas7644 Great point. Part of the genius of Einstein was that he made it a fundamental law of nature that inertial mass and gravitational mass are the same. Thus, there is no difference in nature (and physics) between a phone booth that is accelerating in space and a phone booth that is suspended in a gravitational field. That leads to the complex transformation equations of non-Euclidian curved space. And that leaves me very, very confused!

  • @davidrandell2224
    @davidrandell2224 3 года назад

    Galilean relative motion has the earth approaching the released object. D=1/2 at^2.The earth is expanding at 16 feet per second per second constant acceleration: gravity. Or 1/770,000th its size. Where are the lower case “ a” and “ b” in Newton’s first Proposition? The Final Theory: Rethinking Our Scientific Legacy “, Mark McCutcheon for gravity facts.

    • @carultch
      @carultch Год назад

      The earth is not expanding toward objects that fall toward it. We would notice that by now.

  • @agrajyadav2951
    @agrajyadav2951 Год назад

    0:50 well not really, pretty sophisticated models, even though not as good as newton's, but still pretty good existed before that, for example, surya siddhanta en.wikipedia.org/wiki/Surya_Siddhanta

  • @asifshahriyar6382
    @asifshahriyar6382 Год назад

    You deserve something really really great❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️