Gravitational Attraction: Time it takes for 2 objects to collide in free space

[@ethan_martin]

2 and a half minutes of algebraic manipulation just to avoid using the product/quotient rule? Exactly what I would have done xD

[@YitzharVered]

*O O F*

[@nyanbrox5418]

It did make it very easy to follow, I think I needed that kind of assistance because I'm not that great at this. As a result I found it pretty easy to see all the steps, which was cool

[@Larry640]

lmao I'm in calc physics 1 right now and the hardest integrals we've ever done in the class are reverse chain rule related

[@prydin]

Dude, you hit all the marks for me. Starting out with a seemingly simple expression, taking it through some insane algebraic witchcraft and ending up with pi at the end. This is why I watch this channel!

[@mipmip4575]

Its really satisfiying when a large term at the beginning turns into a small expression at the end :)

[@konstantinub]

If someone had told me I'd enjoy watching a derivation of this formula back when I had it on my Physics I exam in freshman year of uni, I never would've believed them. But it's true - I really enjoyed watching and following through all of that.

[@zakoxy8088]

Sorry Flammy, but that's not how a physicist would do it. A physicist would recognise the straight line as a degenerate ellipse, apply Kepler's third law and be done. Or at least skip the first integration and just use energy conservation and dt=ds/v. TL;DR: 9/10, nice maths, but too rigorous for god tier physics.

[@JoseWui]

Zako XY holy shit you’re right

[@viliml2763]

>A physicist would recognise the straight line as a degenerate ellipse, apply Kepler's third law and be done. oh fuck, 300 IQ move right there

[@morbidmanatee5550]

Right, but this is Flammable Maths, not Flammable Physics

[@ammarali5007]

can you explain for me how can i do that?

[@cerwe8861]

I like how just blindly applying Keplers 3rd Law skips all the Integration and is way more beautiful. But it is also beautiful to see how all this rigorous Math leads to the same result (what it should). But treating a line as a really flat Ellipse is clever.

[@srikanthtupurani6316]

life is so difficult. just two objects so much math.

[@neonblack211]

Srikanth Tupurani and it’s only describing one out of an infinite amount of variations, that doesn’t mean it’s difficult, life can be as simple or as complex as you want to make it, I would substitute difficult for amazing/beautiful

[@ianmathwiz7]

Wait until you get to three objects.

[@joelhaggis5054]

Good Luck separating 2 point masses once they've collided.

[@koenth2359]

I'm sure papa Schwarzschild will help out disecting your black hole

[@anirbanroychowdhury5080]

Einstein won't like that.

[@thephysicistcuber175]

PAPA HAWKING TO THE RESCUE!!!!!

[@u.v.s.5583]

You just need to catch the kinetic energy of the collision in a flux capacitor, then use it to separate the masses again. Or leave the masses together as they are and enjoy your UNLIMITED POWER (and unlimited energy too)!

[@nyanbrox5418]

use a 2d knife, duh

[@Pianothegamer]

YES. I had a problem similar to this (but of a much smaller calibre) in my high school physics exam (no calculus) where we had to find the velocity of a meteorite very far away from a planet when it hit the planet. The meteorite is initially at rest. I love how at 9:15, the equation of kinetic energy pops up along with the gravitational potential energy at a point in a planet's potential well. Why kinetic energy even though there is no mass there? BECAUSE THE MASS CANCELLED OUT FROM BOTH SIDES. That's exactly what we had to get and realise that the mass of the meteorite is irrelevant. That problem took me the last half an hour of the exam to solve along with another subsection. Wow, this was an amazing representation in calculus form. Thank you so much.

[@michaelg5932]

at 13:35 you can put R=Ri*cos^2(theta) for an easier substitution, but make sure the lower bound is pi and not zero (cos^2(theta)=1 means cos(theta) is -1 or 1, so theta is pi or 0. For theta zero we get the negative absolute value for the time at the end)

[@marcelopradocionek6259]

Last year I was trying to figure out the distance r as a function of time for an object falling on a much more massive object and realized it's impossible to obtain a r(t) function, but instead I got the t(r) you just derived. The full expression would be t(r) = (R³/2GM)^½arctan((R/r-1)^½)+(rR(R-r)/2GM)^½. When you take the limit of t(r) with r→0, you get the same result for total time. I was very happy that I could verify the formula using "experimental data" from simulations of Universe Sandbox². Still, I wanted the damn r(t) and spent days trying to figure out a way to approximate the function. The graph looks a lot like a shifted and tilted ellipse of the form r(t) = At+b+sqrt(Ct²+Dt+E), and so I tried applying the four initial conditions (one of which was the total time T = π(R³/8GM)^½) and left the fifth constant to determine numerically. After days trying to manipulate everything, I finally found the correct equations for the four constants as functions of the fifth. After I compared the flipped t(r) (which is the r(t)) with my hypothesis, I realized the functions are ALMOST identical, but they differ in growth in the middle of the trajectory. I should have realized since the beginning that if my function was correct, then its inverse would also be a shifted and tilted ellipse, but that couldn't be true since I already knew the weird form of t(r). Anyway, thanks for the video! It took me a long time to find someone trying to solve that same physics problem.

[@skylardeslypere9909]

So were you able to find the actual function r(t) then? Or did you stop trying?

[@nyanbrox5418]

@@skylardeslypere9909 It looks like he found an approximation but couldn't figure out a general solution mathematically

[@edmundsmaths3980]

It is possible to solve for r(t) exactly for both particles, using incomplete beta functions. Check out my playlist to see how! kzbin.info/aero/PLJpdwO0Y3erGc9DX3ZcBFdYFrf1I0Gr1Z

[@marcelopradocionek6259]

I didn't know about Beta functions back then, that solution looks very beautiful! I actually managed to solve for r(t) using power series. Start with t(r) = (R³/2GM)^½arctan((R-r/r)^½)+(rR(R-r)/2GM)^½ Let k = (R³/2GM)^½ and tan(x) = (R/(R-r))^½ You get that r(x) = R/(tan²(x)+1) = R cos²(x). Now you plug in that substitution in the t(r) to make it t(x) and a lot of things will cancel out, so you end up with 2kt = 2x + sin(2x). that's a Kepler Equation of the form M = E - e sin(E), with M = 2kt, E = 2x and e = -1. Somehow an equation related to angular motion shows up in the unidimensional case! That equation cannot be solved directly, but I've searched for solutions in power series and found out about Lagrange-Bürmann's formula of inversion used to solve Kepler's Equation. There are at least two forms of the solution for x(t): 1 - A Fourier series whose coefficients are Bessel functions: r(t) = R cos² [ kt + sum{1 to infinity} J_n(n) sin(2 n kt)/n ]. 2 - A power series with coefficients defined by insane L'Hospital limits of the n-1-th derivative of (θ/(θ-sin(θ))^n with θ approaching 0. Someone in Wikipedia calculated the first coefficients, but it seems there is no pattern that can be expressed as a simple formula, otherwise this would be just a simple Taylor series. I believe the vertical slope / infinite derivative at the end of the trajectory makes it difficult for the solution to converge with few terms, so neither of them are very practical to use.

[@Gameboygenius]

When Papa Flammy tells me to keep something in mind, I keep it in mind!

[@o_-_o]

Oh, boi! Du are gut! Finally, I can see some practical value of calculus, not just pure, abstract delights.

[@AndDiracisHisProphet]

Maximum respect. This explains why I never saw anyone do this :D Also Walter Lewin just recently had a much easier version of this problem (already relatively hard) in his bi-weekly physics problems

[@AndDiracisHisProphet]

That is, probably, where the Kepler's law trick suggestion came from

[@blackpenredpen]

I should learn more physics...

[@AndDiracisHisProphet]

yes

[@koenth2359]

blackpenredpen Need to hold your nose though ;)

[@koenth2359]

AndDiracisHisProphet In what respect was that problem much easier?

[@AlwinMao]

Astronomy/Physics intuition: Kepler's Third Law says the period for two particles to orbit each other is sqrt (R^3 / 8GM)*2*pi. In a "full orbit" between 2 particles, they come together and then come apart to end up where they started. The collision time is hence 1/2 of the total period. In a conceptual sense, v^2 ~ GM/R gives the typical velocity, R gives the distance, and sqrt(R^3 / GM) is the typical timescale. The other factors are geometric details and indeed pi comes from the periodicity of an elliptical orbit. Notes: a = R/2 for those familiar with T = (a^3 / GM) * 2 * pi, M = m1 + m2, and at 0 distance the particles would be travelling infinitely fast, allowing them to escape infinite force if you ignore relativity, but more realistically some other force other than gravity prevents 0 distance. Since the particles are moving fastest at smallest distance, they spend very little time there, so the final answer would only be wrong by epsilon / V ~ epsilon / sqrt(2GM/epsilon) ~ epsilon^(3/2) if epsilon is how close they actually get.

[@dox1755]

27:37 dont try to verify i tried for you in universe sandbox and its correct !

[@pablog.887]

Excellent video, reminds me of the method we had to use to calculate how much time a deposit would take to empty itself if only gravity were to act upon it.

[@skinsheroes]

8:52 “zero is just zero” new level of genius updated that’s a good shit over there

[@hOREP245]

I think these kind of videos are my favourite. The physics derivation ones are always great. The "abstract proof" kind of videos always feel like there is too much going on and I have no idea what the stuff is used for, or what any of it means.

[@colt4667]

I'll have to watch this about four times before I really truly understand every step. And I WILL watch it several times more.

[@Alessar30]

I m a physicist so I play *physicist* (nose sound)

[@rossetto23]

Great video! I always wanted to do this calculation, but I always got stuck in the dark magic part. Some remarks: - At 1:40, it's not 2nd law, it's 3rd law. - There's a simpler way to do the integral of sin^2(x). Use the trigonometric identity (that is really simple to derive from the double arc of cosine): sin^2(x)=(1-cos(2x))/2.

[@МаксимРусских-й1ь]

to count the time approximately. you can say that straight line is very thin ellips and use the third keplers law to find the "period". the time is a half of the "period"

[@HilbertXVI]

AndDiracisHisProphet What are you, 12?

[@AndDiracisHisProphet]

that's a very mature response to a "yo mamma so fat", joke. not

[@МаксимРусских-й1ь]

it works always

[@deeptochatterjee532]

AndDiracisHisProphet No it does work, you just have to use the concept of reduced mass, which actually was used in this video, but it was very subtle.

[@AndDiracisHisProphet]

yes, you are right. i already checked it myself. i stand by the mom thing, though

[@NebulaM42]

It makes me so happy when Physics and Calculus hug like that. I'm new here, and already love your channel! Keep up the great work, from an Engineering student (still obey you, Maths

[@samonellasgayclone1054]

Ah. Thank you, I needed this

[@samonellasgayclone1054]

Flammable Maths

[@abrarshaikh2254]

The easier way to do that intregal! U can express dr/√( 1/r + 1/R) = √r*dr/√(1- (r/R) ) Now put r/R=(sinX)^2 Simply,u will get intregal of (sinX)^2

[@DanieleBattesimoProvenzano]

And, if the force was repulsive, one should use the sinh!

[@frede1905]

Wow, incredible. The most incredible part though, is probably that pi is showing up here. It can be hard to understand how the circumference of a circle divided by its diameter has anything to do with the time it takes for two masses to hit by the influence of gravity lol... That just shows how amazing maths and science can be... Great video! 😄😄

[@neonblack211]

what does a circle have to do with physics and math? A lot, I do know what you mean though, I’m not at a high enough level to see these relationships unfold and derive/understand them myself, so for now I just accept it’s true, .. waiting till the PI cools enough for me to get cloe, close enough to reveal his sexy secrets.... then bam....

[@Max-bz8ev]

Depending on your perspective, it's just as easy to define pi as double the x value of the first root of cos(x) in the interval [0,2] and then wonder why it shows up in circles. Still, the connections are crazy.

[@crazyphil7782]

Pi will pop up any time you’re dealing with a central force. The r^2 in the denominator means that it’s spherically symmetric, because it’s divergence is uniform.

[@ДимитърПопчев-н2ч]

That means the collision time squared is proportional to the displacement between the two objects to the power of 3, e.g T^2 \prop R^3, noooow if there was any place I have seen this before.... Of course the third law of Kepler of orbital periods, very interesting...

[@discretelycontinuous2059]

These bodies are obeying keplers laws (it obeys the 1st law if you consider a straight line to be a special form of an ellipse (an ellipse of infonite eccentricity?))

[@douglasstrother6584]

The insidious "zero angular momentum" case!

[@alpha-gy1es]

At last guys, it will be( π-2) instead of only π, because at last he took that infinity assumption to make the 2nd part "0" but it could have saved by not substituting the( x-sinx.cox) back into the form of u, instead he should have changed the limits according to u=tanx that at u=infinite, X=π/2 and at u=0, X=0. Then put that in( x-sinx.cosx) then he could have obtained the final result without any assumption or error.

[@aryanjain9957]

Papa, I just realised that this video probably saved my ass. In the US, we have an exam called F=MA, which is basically a non-calculus Newtonian mechanics based test which will allow you to take the USA Physics Olympiad if you perform well enough. On this year's test, there were these weird questions about objects colliding under gravity, and I just happened to remember this formula which most likely got me the point that I needed to make it past the cutoff. If I do qualify for the national exam, I owe you big time.

[@nicholasleclerc1583]

Seriously, I’ve been trying to dot his on my own for a looooong time now, you don’t wanna know; true, I was taking long pauses, but I was stuck, bro; Thanks do much !!!

[@bobbicals]

Awesome! These mathematical physics videos are my favourite things on youtube

[@jonni2734]

just beautiful...

[@aleksandarprodanov4454]

Great video! It was very fun for me to watch it! I just wanted to say that that the time can be negative! And indeed, if the force was repulsive by nature, then the collision time you would get doing all the calculations would be negative.The physical meaning of this is that the collision happened before the moment of time we defined as t=0. The real reason for the minus sign in front of the square root thing is that dR/dt is in fact negative. That's because R is getting smaller as the time goes forward(R(t+dt)-R(t)0). I look forward to your reply!

[@philipphoehn3883]

27:26 epic chalk drop

[@KakoriGames]

The negative answer to the problem also makes sense physically, because given no external influence, in order for 2 bodies subject only to the gravitational force between them to have a initial velocity of 0, they must have been going away from each other previously, which implies a "collision" has happened in the past, given no other forces act on the system. In fact, given no other forces, if you allow both masses to go through each other, they would go back and forth in a Simple Harmonic Motion. EDIT: Ok, not sure if would qualify as a simple harmonic motion, but it would be a periodic motion of some sort.

[@balajisriram6363]

Hi I had attempted to solve the same problem 6 years ago. but I couldnt and as the time went i forgot about this problem. Now i'm happy to see the solution thanks very much!!!!

[@utkarshverma1]

9:45 It's quite sad that all the previous effort with integration only yielded the obvious energy conservation equation. 😅

[@abrarshaikh2254]

This question is as same as problem given by Lewin Walter !🔥

[@coleozaeta6344]

It’s nothing short of gold that such a serious subject is met with use of the word “boi”. 🔥

[@yurigouveawagner9432]

i'm a freshman at an undergrad physics course. i almost tried doing this on my own in 2 dimensions before searching this. I feel like the character from those stories where someone decides not to go on a car trip and the car falls off a cliff.

[@jasimmathsandphysics]

THANK YOU SO MUCH!!! I have been working for days trying to find the answer! Just two days ago I found the equation you got and thought it must be wrong so today I wasted 5 hours trying to do it another way and thanks to your video, I can stop wasting more of my time 😅. When I was integrating I did not use limits (constants I used were pretty much the same as inputting the limits) and I did not set the initial velocity equal to zero to further generalise it (I'm not sure if that makes it incorrect but it seems to make sense on desmos). If I set the initial velocity equal to zero then I get the equation in your video. Thanks again

[@danielortega5211]

Amazing video! I'm enjoying these last few a lot, thanks

[@peterclark5244]

For the first part you could also, without loss of generality, use coordinates such that m1 is at 0 and m2 at R, which would give you the differential equation for R without having to consider r2’’-r1’’

[@discretelycontinuous2059]

I think that I beg to differ. r1 may begin at 0, but mass 1 does not remain stationary (as it is acted upon by mass 2). There are no assumptions made here regarding the relative sizes of mass 1 and mass 2.

[@Newtonissac6]

I am a new subscriber here and I am already loving this community. Everyone seems to be respectful and learned and eager. And you are very good at keep my attention on you. I am really excited to see this channel growing because this has all the potential of becoming a math/physics giant on KZbin P.S. I absolutely love your accent. Makes learning really interesting.

[@MadSideburns]

Daaaaam I love the variety of topics you're bringin' us lately. Love U

[@rayniac211]

Holy shit this was magnificent! I've seen an example solution for this problem but it used some convoluted integration formula for the integral and the final solution was nowhere near as elegant! I am really in awe how you were able to arrive to this result step by step. The units check out too btw. The final formula truly does spit out seconds, which is a very good sign :)

[@Cosmalano]

I had always wondered about setting up a way to look at the acceleration value as a function of distance but I think the equation you ended at is a lot better than anything I was able to come up with. Thanks.

[@BogdanAlex]

A friend at work came up with the question - what if the earth stopped rotating around the sun, how long would it take until collision. I thought it was a really cool question, started working on it but got stuck soon enough. Many thanks for the video. Jeez this was painful. (looks like around 64 days or so, in case anyone was wondering :D).

[@dox1755]

Wow.How pi popped up in a linear system...

[@angelmendez-rivera351]

Alper YY It is explicitly not a linear system, as was described at the beginning of the video. It was a second-order non-linear differential equation. Also, the solution to the two-body problem is an ellipse, and this case is just when the initial frequency is zero.

[@lucastellez2558]

Was going to pull up a KZbin workout but this appeared on my feed. Can't miss out on papa, another great video congrats.

[@PapaFlammy69]

:)

[@yourdad9885]

In cold Russia we rush trough this kind of integral just using the formula sin^2(x)=1/2*(1-cos2x). Then use the linearity and that’s over.It’s simpler but your way much more intriguing, yeah

[@elijahl-s5184]

you are a natural teacher

[@srpenguinbr]

to solve that second order non linear differential equation, you can multiply both sides by dR/dt (i.e., R dot) and integrate. The problem is dealing with the complicated integral, but maybe by doing definite integrals it is more comfortable

[@thehashslingingslasher5336]

It also further simplifies to 37.7(Ri/M)^(1/2) if Tc is in hours :)

[@jahirbasha5318]

you may use binomial theorem at (14:16) that may short down process

[@xCorvus7x]

Quascience I fail to see how that would help. What do you have in mind here?

[@EmissaryOfSmeagol]

1:41 Newton's 3rd 2:03 That's Newton's 2nd

[@harrisidh]

EmissaryOfSmeagol smh what a fake physicist, right. I bet he can't even find the collision time between two masses in free space 😤😤😤

[@Drimypso]

Great video! Can you solve the problem for a general potential kr^n ? And try to see what happens for a different n? Maybe a hamiltonian formalism could be nice to watch!

[@deeptochatterjee532]

I think it was pretty clear that the plus or minus would turn out to be minus because the velocity had to be negative, since the distance between the two masses was decreasing over time.

[@elfaroukharb3979]

Love the Random Week Pappa flammy

[@alainbrizard4719]

At 11:40 a physicist introduces a trig substitution R = R_i cos^2(theta) and gets an integral (from 0 to pi/2) of 2 R_i cos^2(theta), which gives R_i pi/2. Done in less than one minute.

[@lagranginabile]

10:30 heyyy I rember that formula, I found it once using conservation of mechanical energy! I think it's interesting seeing how the same result can arrive from basically different methods. I think that's a prove that physics and mathematics works!

[@heylofellas]

Hey! Just a shorter way to do the last integral! Take R as (R_i*sin²(t)) and it'll give you the sin²t integral in a few easy steps! Which you can also do by Cos(2t) identity but it's Trivial post that point :).

[@MCLooyverse]

Wow. When I try this, I get to the second order differential equation, and find out that solving it is *literally impossible*. Thanks, internet, for lying to me. Great video, by the way. Holy cow.

[@alwinpriven2400]

4:38 Ah yes, like my linear algebra professor used to say. "Dark/black work with indexes" (he said it only once but still)

[@nicholasleclerc1583]

This is why I want him to tackle atmospheric friction ballistics !!!!

[@jackvernian7779]

Absolutely fascinating.

[@hydropage2855]

It made me mad when you back substituted into u instead of just figuring out the upper and lower bounds of the substitution lmao

[@waylonwilde4147]

This video was incredible. Integral time is over. Now, soup time.

[@ianluebbers5492]

My brain has melted. Thanks for the cool vid! This is why I want to pursue physics

[@MrJdcirbo]

Papa Flammy is doing physics????!!!!! 🙂🙂🙂🙂 I love your pure mathematics videos, but I have a special love for physics. Please do some qm or gr field equation solutions if you ever feel adventurous!!! 😊

[@PeterBaumgart1a]

Great work Jens! I felt exhausted at the end (and I can only imagine how exhausted you must have felt), but it was very enjoyable, and in a way absorbed me like a suspense story! Can you solve it for a case if initial speed is not zero, but perpendicular to R?

[@willful759]

he's too powerfull!

[@jehovah0121]

I have to get accustomed to the accent to watch and understand more of your fantastic videos, dude.

[@Just_a_user3]

Thank you for this great video. I am planning to learn german, wish me success papa flammy :)

[@tibortresla]

Now i get why uni teachers avoid explaining this proof in Physics 101 classes.

[@discretelycontinuous2059]

This is all 1st year calculus concepts. It ought to be understandable to a switched-on 1st year physics student

[@HaniSantosa]

I have done this peoblem in the past. I think it is easier if on 13:41, you define R = Ri sin^2 θ. The end result will involve arcsin function. (If I remember correctly.)

[@admink8662]

Crazy video

[@omarabu-khalaf6043]

I think your result is highly related to Kepler's third law of planetary motion. It can be thought of as a reduction model of planetary motion from 3D to a 1D space ( Planets orbiting each other in a straight line rather than in a 2D ellipse).

[@Musicrafter12]

Deceptively hard question: Imagine a christmas tree as a right circular cone with height H and base radius R. We have an idealized light string of length L. We want to wrap it around the tree so that it perfectly stretches from the top to the bottom with none left over, maintaining a constant angle to the horizontal while doing so. Given the dimensions of the tree and the length of available light string, what is the angle? I worked on this one for a while..... never got it.

[@ankushpatanwal6535]

You know you could have easily integrated that bad boi there by substituting R = R_i * Sin²θ

[@qbasic16]

legendary move ❤️

[@PapaFlammy69]

[@koenth2359]

That looks a lot like Walter Lewin's problem #36. I solved it like this (but I took the scenic route): First express the velocity as a function of position x, using conservation of energy, and the potential -MG/x: v(x) = -√(2MG(1/x-1/x0)). Then we can set t = ʃ dt= ʃ dx/v = ʃ -1/√(2MG(1/x-1/x0)) dx Substitute u = √(1/x-1/x0) so that du = 1/(2ux^2) dx and dx = -2u / (1/x0 + u^2)^2 du Now t = -1/√(2MG) ʃ 1/u dx = √(2/MG) ʃ 1/(1/x0 + u^2)^2 du Next substitute u = tan(v)/√x0 so that du = 1/(√x0 • cos^2 v) dv Now t = √(2/MG) ʃ 1/(1/x0 + tan^2(v)/x0)^2 / (√x0 • cos^2 v) dv = √(2 x0^3/MG) ʃ cos^4 v / cos^2 v dv = √(2 x0^3/MG) ʃ cos^2 v dv Substitute w=2v Now t = √(x0^3 / 8MG) ʃ cos(w)+1 dw We arrive at t = √(x0^3 / 8MG) (sin(w)+w) + C, where w = 2v = 2 arctan(u √x0) = 2 arctan √(x0/x - 1) We want t to be 0 for x=x0 => w=0, so that we can conveniently set our integration constant C = 0.

[@atrumluminarium]

Interesting that Tc(Ri) has the same functional dependence as Kepler's third law (i.e. T² ~ R³), even though we are analysing collisions not orbits

[@pavlopanasiuk7297]

This is actually an orbit. Think of it as of degenerate ellipse

[@luispuente1554]

It'd be nice that you make a video doing the experiment. Keep the good work!!

[@Don-ut6fs]

This is so cool and u cuda done sin^2 is nothing but (1-co2u)/2 which is much faster to integrate

[@yarooborkowski5999]

Flammy, now try to derive in this way the time that two bodies rotate around their center of mass in free space. You're doing good

[@BlueCollard-h2u]

You used Newton’s Laws incorrectly - Third law not Second 1:41 - other than that thoroughly enjoyed such a marvellous calculation

[@dylansbjpm]

This video also got me thinking: If we were to place these two objects on the x-axis at positions (r1, 0) and (r2, 0), could we find out the point where the two objects meet in terms of G, the initial positions and the masses of the two objects?

[@juliouribepadilla1240]

Very nice video. I'm gonna subscribe finally hahaha 8*pi*G also appears on Einstein's Field equations btw

[@PapaFlammy69]

:33

[@46pi26]

Just now got done with work and came home to the smell of fresh memes. I'm finna drink this video straight through my third eye

[@Mike_SC]

I have a simpler way , take it as an extreme ellipse orbit, use Kepler 3rd law. a³/T²=G（m/4）/4π², a=R/2, T=t/4, same result

[@harikishan5690]

what if the two objects were charged equally and opposite,how would we account for radiation,magnetic effects etc??I had a question that I was stuck on for quite a while,consider two oppositely charged objects whose mass is 'm'.and consider them to be falling under the influence of a uniform gravitational field,what would be the time taken to collide,and how much energy would be lost in radiation

[@TheMauror22]

Wow! This was amazing!!!

[@GrandMoffTarkinsTeaDispenser]

Hold those boys tight

[@nindocomic]

Oh man, I really love your videos. You're amazing boi, keep it up

[@jameswilson8270]

Excellent video my man! However, I believe I saw a loss of generality by assuming the initial velocity to be zero.

[@jameswilson8270]

I just got around to examining your claim. It's been a pretty lazy day for me. What if we let v_i < -sqrt(2GM/R_i) (say this corresponds to time t = t_i with R_i given)? If you consider this case, you should find it implies that the bodies could never have been released from rest.

[@jameswilson8270]

Flammable Maths As long as we know some generality was lost

[@ersin486]

Ich hab das so gemacht: Da auf beiden Objekte die selbe kraft wirkt mit aber unterschiedlichen massen müssen beide auch unterschiedlich beschleunigen (F=ma). Mit F=GMm/r^2 = Ma, bzw = ma hab ich jeweils a = GM/r^2 für den Körper mit Masse m raus und a = Gm/r^2 für den Körper mit der Masse M raus. Da beide die selbe Zeit brauchen bis zur Kollision und die Summe der jeweiligen zurückgelegten Strecken r ergeben, hab ich stehe r = s(m)+s(M) = [a(m)+a(M)]t^2/2. Umgeformt nach t also t = wurzel{2r/a(m)+a(M)} Die Gleichungen für a von oben hab ich einfach eingesetzt und vereinfach und bekam als Ergebnis t = wurzel{2r^3/G(m+M)} Da ich dich als Lord der Mathe memes nicht schlecht dar stehen lassen will frag ich mich, wo mein denk Fehler ist :( Ps: wenn ich deins mit mein Ergebnis gleich stelle kommt pi = 4 raus huch Entweder heißt das ich werde Ingenieur oder irgendwas ist faul

[@konradgebura3985]

My cousin looks like you and I called him Papa Flammy. -Great video by the way I enjoyed every second of it.

[@adandap]

All good, but a few numbers at the end would have been nice. I calculated how long it would take two 1kg masses initially separated by 1km to collide - it's interesting to take a guess first and then see what the answer is. (Left as an exercise for the reader. :D )