Kepler's Third Law of Planetary Motion Explained, Physics Problems, Period & Orbital Radius

This physics video tutorial explains kepler's third law of planetary motion. It provides physics problems where you have to calculate the period of Mars or the distance between Venus and the Sun using the orbital radius of earth and sun-earth mean distance. This video also provides the formula needed to calculate the mass of the sun.
Static Friction & Centripetal Force:
• Static Friction, Centr...
Normal Force on a Hill:
• Normal Force on a Hill...
Banked Turn Physics Problems:
• Banked turn Physics Pr...
Non-Uniform Circular Motion:
• Non-Uniform Circular M...
Gravitational Force:
• Gravity, Universal Gra...
______________________________
Universal Law of Gravitation:
• Gravity and the Univer...
Gravitational Acceleration:
• Gravitational Accelera...
Gravitational Field Strength:
• Gravitational Field St...
Satellite Speed:
• Speed of a Satellite i...
Circular Motion Review:
• Centripetal Accelerati...
____________________________________
Kinetic Energy and Potential Energy:
• Kinetic Energy and Pot...
Work, Energy, and Power:
• Work, Energy, and Powe...
Final Exams and Video Playlists:
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Full-Length Videos and Worksheets:
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Physics PDF Worksheets:
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[@TheOrganicChemistryTutor]

Physics PDF Worksheets: www.video-tutor.net/physics-basic-introduction.html Full-Length Videos & Exams: www.patreon.com/MathScienceTutor/collections Next Video: kzbin.info/www/bejne/laHWeoqqhad7sKM

[@darachristy2422]

Pls what can I do My calculator never seems to give me the same answer as yours

[@Kurto.]

5:11 mean distance of planets to find period of one around the sun 10:04 distance between two planets based off how long it takes to orbit 12:56 mass of the sun using using the distance of the earth

[@mdaffanpagdiwale8685]

👍

[@chemiflask7692]

I have been watching his videos for ages, he is so helpful! I will continue with his legacy as soon as I get my PhD.

[@kairu9373]

ah yes the classic "calculate the mass of the Sun" meme

[@sayanhaldervii.a.3203]

😂

[@aryansaishiv8461]

The explanation is fantastic and the applications are quite good. This video can make us understand better about Kepler third law so I recommend everybody who is watching this video to save it or download it

[@muhammadanas8213]

you are greatest tutor on youtube.I love the way you explain.

[@tormentedsunbeam]

Him and drphysicsa are my favorite

[@sebastianblandonlondono9658]

Thank you so much! I didn't find out a clear explanation in spanish.

[@raeelemental7084]

13:24 okay... you blew my mind honestly, that was amazing.

[@johnquinn8052]

You are the best - I too love the way you explain . Top guy

[@Happygirltales]

Thanq very much ....your video is really helpful ....cuz this online classes make no sense..

[@LONE-MUHAIB]

I want to become a physicist or I want to do physics engineering...I love your tutor . Thanks for giving such helpful videos on youtube..❤️💞💕💖💗

[@mynameishidden1025]

Thank you very much, clear and quick to the point. Effortless.

[@val-ub3nt]

thank u for this!!!!

[@rlynotfaye]

You are a life saver!!!!! Thank you T^T

[@North.X857]

Thank you for the explanation,,,, highly appreciated 🍃,,from Zambia 🇿🇲

[@nicotengco2775]

In the 2nd example In my calculator I always gets 12.82x10raised to 32 am I missing something or its my calculator incorrect?

[@memokingdom7590]

Superbbbb....

[@treymoulton4811]

This Man is the Math God... I want him.

[@takta3265]

when i first tried to understand keplers third law, i was completely lost... and then this guy makes it sound so easy and understandable - absolutely amazing, thank you

[@nestoons4539]

"I am the man who carried every math and science student" "Bullcrap, school taught them" "You sure?" ... "That's right... Now, say my name." "The Organic Chemistry Tutor" "You're goddamn right"

[@kaeez]

Haha, good one.

[@johnclydemontajes2472]

this video so nice

[@Dipjyoti143]

Thanks sir 👍👍

[@jherica4755]

Im inlove with you thank u 😭❤️

[@funkymonkey9011]

for question 3 i was wondering if you could use r^3/t^2 = GM/4pi^2 to find suns mass however im getting a different answer to your mass. Is this formula wrong?

[@Spectrojamz]

I've made a promise to myself that I must meet you in person😭. You've really made an impact in life. Thank u Respect🙇🙇

[@cedricmendoza8316]

I'm having trouble understanding this. Any recommendations on pre requisites? Topics to study before this?

[@asorockboxingclubAjayi-sq6og]

Newton law of universal gravitation

[@reggyreptinall9598]

How do you figure out the mass of the earth, or even better the sun?

[@fatimarana6319]

Any video on second and first laws?

[@joycemicah6893]

Seriously what would I do without you😫😫.

[@zeroz88]

Truly astonishing. Can’t believe Man is capable of doing this. Looks like we have only just begun to touch the foundation of the entire world

[@farhanurmiah2635]

thanks

[@alissanabila2439]

For question 3, instead of 2.01 x 10^30 kg, i got 2.01 x 10^8kg instead as my final answer. I cant recognize where did i went wrong since my calculation perfectly matched with yours. Am i missing something?

[@pol7332]

you literally saved my life i love you and you are welcome to my house god bless you

[@siyayas9134]

Idon't even speak english but i understand it ... But i'll never forget you Never sir

[@anilmawji]

How do you determine which planet uses the variables T1 and R1 and which uses T2 and R2 in the equation? Does the missing variable always take T2 or R2?

[@hitorilollipop7573]

it's relative so it doesn't matter which you make T1/R1 or T2/R2 but you just have to stay consistent example: if you make the period of the Earth T1, the distance from the Earth to the Sun has to be R1

[@lynette4975]

@@hitorilollipop7573 thank you!

[@kaeez]

Doesn't matter. You can take as you please for convenience but it won't matter much. For example when calculating by paper, for ease, you can take the top one (dividend) as the unknown one, so you can just take the other terms to the opposite side of the equation easily. Regardless, taking the reciprocal at the end would suffice too. But make sure you correspond both the T and R to be for the same object. TL;DR: If let's say Earth is taken as R1, then T1 should also be of the Earth, not the other way around. You would be fine.

[@eternalanimation1734]

If we are talking abou the centripetal force then does that mean we are saying that the planet is moving in a circular path but that contrdicts the 1st law of kepler that says that a planet moves in a elliptical path around sun

[@swapnilm9774]

Not bad for timing JG

[@aksamlaraksam4609]

AT LAST EXPLAINED!!

[@joshuaethanyap2657]

Legendary

[@yeswowferengee6140]

M.r organic chemistry you are our organ

[@rikipashaj7120]

How can we use 2πr if he said in the first law that the shape of the orbit is not a circle

[@joshuaethanyap2657]

Where can I find you mate

[@souljarohill8795]

If you can answer this question, why do we use seconds in the third problem instead of just the 365 days for T?

[@shushanksivana5286]

Because when we attempt a question given in metres we use seconds

[@kaeez]

To convert it to S.I unit so as to get the answer in meters.

[@bernardo6925]

Is there any book (or online article) of which you retrieved this content? It would help me a lot to reference this explanation in my article.

[@jessepinkman681]

Giancoli Physics.. almost same content, i think it would help.

[@bernardo6925]

@@jessepinkman681 Thank you!

[@stevedahl9503]

You start by showing: (T1/T2)^2 = (R1/R2)^3 But then you go to derrive that the equation is actually : (T2/T1)^2 = (R2/R1)^3 ...am I missing something?!

[@majncraftchlapik]

Both of those equations mean the exact same thing (because they're ratios)

[@tommax6884]

That is the same thing as raising both sides to the ^-1 power. Or taking the reciprocal on both sides. same thing

[@1Mayoii]

its basically the same thing but reciprocated. so not missin on anythin

[@redbean9410]

its the same thing lol

[@willswan8896]

it's just flipped same difference between the variables

[@abbybaer4074]

I made a mistake taking astronomy :(

[@carlosmunoz8789]

same girl

[@rionoma4630]

Really?🙁 I'm planning to take astronomy

[@abbybaer4074]

@@rionoma4630 if youre good at math youll be okay but im not

[@rionoma4630]

@@abbybaer4074 but how about now? Do u still u made a mistake?🥺

[@Editoryadnesh]

​@@rionoma4630hi...did you choose astronomy???

[@brightchirindo5393]

thank you kkkkkkk

[@KaranKumar-ey8ly]

You are great sir

[@reathpanom]

I got you but, how can I calculate speed of the sun

[@wlo23ex81]

At 14:30 can someone please explain to me why he uses seconds instead of years this time for the period?

[@wlo23ex81]

I mean can there be, for T2/T1=r2/r1, different uses of units? For example, can the units also be inputted as T= hours and r=kilometers? If so, why???

[@bartonpaullevenson3427]

@@wlo23ex81 When he was working with ratios, the units didn't matter. When he was working with absolute amounts, the units had to be consistent.

[@PRIMEPOWER13]

@@wlo23ex81 He uses seconds because the unit for G (that is the gravitational constant) is N•m²/kg² . Since N (that is Newton, standard unit for force) is equal to kg•m/s², then we will arrive to another unit for G that is m³/kg•s² As you can see in the his equation T is in squared thus the unit s would be squared as well. Now if we wish to multiply the units of G and unit of T² we can see that the unit s² cancelled. (m³/kg•s²)(s²) = m³/kg Make sense right?

[@comicaljedi2064]

If I'm using a standard calculator, how can I do a cubed root? I've tried pressing 3, and then the root key, and I've tried raising the number to the 1/3 power. No dice :(

[@chunkychops]

try raising to 1/3 power instead as it's the same thing

[@kaeez]

I know this is late, but make sure while raising to the 1/3 power, you put it in brackets. Eg: (3.14)^(1/3) otherwise if you would write like this :- 3.14^1/3, according to PEDMAS, it would compute the exponent first (i.e, 3.14^1) and follow that with division by 3. So essentially, it would do 3.14/3 which is not what you want.

[@kristoffereliasullumthores1609]

They really expect you to just understand this in school, meanwhile it takes a 20 minute video to actually understand.

[@harrisoncarpenter1038]

Jesus Christ, you made this seem like simple addition after you explained it.

[@fofoca1731]

could any one tell me what is the different between the velocity in circular orbit an in elliptical orbit?

[@gracepancalstatela4422]

Fofo Ca he didn’t say that his prove bases on that planet circulates sun in a circular mention .

[@bartonpaullevenson3427]

For the elliptical orbit you'd need the vis-visa equation.

[@SENsanzo]

i find this very difficult i want an explanation for ninth grade not sure if this is it but i watched it non the less its much beter than nothing

[@TheMacC117]

Are you not going to show how you arrived at 472183? Because no matter how many different calculators I use, none of them get me Nowhere close to that.

[@hannakennedy3720]

you must be forgetting a cube or a square exponent it works perfectly fine for me the way he has it.

[@marcinna8553]

I wonder how Kepler derived this equation. He did not have Newton's law of Gravity that is used in this analysis. I have also always been curious about the trigonometric methods Kepler used to deconvolute the respective orbits of the Earth and Mars using Tycho Brahe's data. It is a very complex problem -- how to calculate an orbit of another planet based on measurements taken when you yourself are also orbiting the sun and spinning on your own axis, which itself is tilted. I think this is the crux of Kepler's achievement: to find mathematical ways to factor out all of this motion and be left with the orbit of Mars. And to be so precise about it that the orbit could be shown to be slightly elliptical.

[@chadmc336]

I like dinosaurs

[@PhysicsMath]

They use some kind of indian trigonometry technique.. the circle geometry

[@chiomauche2138]

bro you’re deep .dint think about it too much..cuz then you’ll create your law and our kids have to study it in school

[@yuva-so2bp]

Today is my exam and I am watching your video

[@amnaalmansouri1621]

In the 3rd question, why do we convert T into seconds, why is it not in days, hours, minutes, or even milliseconds?

[@romielynpescuela4448]

Because seconds is the unit of T

[@PRIMEPOWER13]

He uses seconds because the unit for G (that is the gravitational constant) is N•m²/kg² . Since N (that is Newton, standard unit for force) is equal to kg•m/s², then we will arrive to another unit for G that is m³/kg•s² As you can see in the his equation T is in squared thus the unit s would be squared as well. Now if we wish to multiply the units of G and unit of T² we can see that the unit s² cancelled. (m³/kg•s²)(s²) = m³/kg Make sense right?

[@iflo8171]

Why do we assume constant radius when planets orbit is an ellipse and second their velocities are not constant. How is this derivation valid with those incorrect assumptions?

[@redbean9410]

the derivation is valid because newton found those equations with the correct assumptions. The reason we have velocities and radii constant for a classic high school physics class (even ap physics) is to keep the math within Algebra. Doing the problem realistically with changing velocities and changing radii requires differential equations and integral calculus, which newton actually invented so he could do these calculations in the first place. The assumptions in this video are there for simplicity.

[@laze_amvs424]

@@redbean9410 and from what I understand these assumptions lead us to answers pretty close to the actual values so I don't see the harm in them.

[@zubayrfarooq8851]

this derivation is valid only when a body is not accelerating (d=vt) which is in the above case. Secondly, with the passage of time radius changes but at any instant distance between sun and planet is fixed (constant)

[@kaeez]

What do you mean by the "radius" here? I assume the distance between the planet and the Sun? In that case, how can you say that this distance changes but the distance between sun and planet (which I suppose is same as the "radius") is constant at any instant? I'm sorry but sounds absurd to me, correct me if I'm wrong. @@zubayrfarooq8851

[@maddieek6865]

Wait so can we always use the T^2= 4(pi) ^2 or do we have to change it around

[@maddieek6865]

@@abdirahman8111 ty, managed to hopefully survive, waiting on my final score LOL.

[@Freddy18w]

When you present the initial law (T1/T2)squared=(R1/R2)cubed but when you do the problems You have (T2/T1)squared=(R2/R1)cubed why are you using the reciprocals of the initial equation?

[@kaeez]

Those are ratios, it doesn't matter how you take them as long as the T and R corresponds to the same object. Just like how 2/1 = 4/2 and 1/2 = 2/4 are same.

[@ederlinabano8311]

According to the Law of periods that the ratio of r3/T2 is the same for each planet, so what do you think is the period of revolution of an imaginary planet if its mean distance from the Sun is 337.9 x 106 km? Comparing its period of revolution to Earth, is the imaginary planet near or far from the Sun? Show your solution.

[@bartonpaullevenson3427]

Not bad. Some of the figures were a little off, but always by trivial amounts. A very nice presentation.

[@Lapston-iy9hp]

I'd say pretty good, not 'not bad'.

[@kaeez]

How are the figures off?

[@lll8112]

If an 8,700 kg satellite is orbiting a planet 8340 km away from the midpoint of the planet has a period of 6.8 hrs, what is the mass of the planet?

[@aksavxavier3080]

IF Earth moves with constant Speed then, How can the 2nd law is Satisfied?

[@bowerow]

"use the period of the moon for tea" 14:00 ... nice.

[@pendonkenmark3824]

I think 10^-11 not positive 11

[@1Mayoii]

he wrote -11

[@BruceLee-lk8uc]

Lol

[@avylynnejadezy7972]

A small planet was discovered 15 times farther than earth from the sun. Earth is about 1.5 × 10⁸ km away from the sun. How long does it take for this small planet to orbit around the sun compared to earth which takes an average of 365 day?

[@kaeez]

So, let's take the unknown planet's parameters as T2 and R2, and Earth's as T1 and R1. Given, R1 = 1.5*10^8 km R2 = 15*R1 (Small planet's distance from the sun is 15 times that of the Earth and the Sun) T2 = ? T1 = 365 Days According to Kepler's Third Law, (T2/T1)^2 = (R2/R1)^3 (T2/T1)^2 = ((15*R1))/R1)^3 T2^2 = T1^2*3375 T2 = sqrt(365*365*3375) = 21204.5838 Earth Days ≈ 58 Earth Years. So the time it takes for this unknown planet which is 15x farther than Earth from the Sun would be equal to around 58 years. Which is 58x more than the time it takes for the Earth to orbit the Sun.

[@mathsforclass6505]

But Newton himself derived his law of gravitation through Kepler's third law so how can you use newton's law to prove something that proved newton's law?

[@TeddysReadingNook]

Keplers laws were founded based on observations. Newtons laws were made to explain them, so you can derive keplers laws from newtond laws

[@valentinadegennaro7361]

u make me hate physics a little less

[@puregamingpg5086]

My question is why there is only T²/R³

[@dylanbreedveld_5280]

Huh the internet says the distance sun-earth = 1,5 x 10^8

[@dylanbreedveld_5280]

Never mind it was measured in km

[@ahmadmustafa899]

im finished

[@sanjanagupta9013]

✖

[@tharios4289]

1:44 Earth is NOT moving with a constant speed around the sun, the amount of the speed earth has while orbitting changes due to its distance to the sun. Earth orbits not in a perfect circle shape but in a eliptical shape around the sun, so its distance to sun changes time to time. If earth is closer to sun, it will move faster, and the more earth is away from the sun, it will move slower due to gravitational force changing by distance's square. Since the Earth is not orbitting with a constant speed around the sun, therefore you cant use the d=vt constant speed equation. I understand that you are trying to prove it simply by accepting that earth orbits in a perfect circle shape around the sun and there is no problem with that BUT you should state the "reality" too. You are saying "earth orbits around the sun with a consant speed" like a fact. You should say something like "guys, earth actually does not orbit with a constant speed around the sun but for the simplicity sake this time we wil accept that it does." Let people know the truth.

[@PoisonNemesis]

this is different from what they teached us😭

[@7mda17]

I should’ve dropped outta high school because this is not normal why is it so hard I still don’t get it

[@monero.jeanniton]

Did you learn to demonstrate Kepler's third law in high school?

[@Marccccc09]

Baro naya

[@Migsb1379]

dlsksissksk LMAO

[@skellingtonmeteoryballoon]

i call the mean the nice now. the mean between them is getting boring.

[@HanHazRosChi]

Wadada

[@hiramani399]

Yaar kash Newton, kepler aur Galileo merai hath lagtai mai unka hashar nashar kar deti.....😔😔😔

[@elmerfudd5650]

You make this so much harder than it needs to be. I don"t need lessons in algebra. I want to see how the harmonics are derived. Get over the algebra and use known values for orbits and such. I want to see music!

[@redbean9410]

then you're in the wrong channel

[@mollerbt]

You made a mistake at 14:22. 6.67? That G is supposed to represent Earth gravity. It should be 9.8 shouldn't it? You don't explain where that number came from. I hope your not just picking random numbers...

[@nagamanivanagundi8860]

The capital G is the gravitational force of Earth on another object which is 6.67x10^ -11 and small g is the acceleration due to gravity which is 9.8N

[@abdmalikyusuf2799]

nagamani vanagundi 9.8m/s²

[@dylanbreedveld_5280]

Malik Yusuf 9,81N\kg

[@abdmalikyusuf2799]

Dylanbreedveld _ 😁 another angle

[@bartonpaullevenson3427]

G is the gravitational constant and applies to any body. g is a particular planet's gravity.

[@lyra7660]

could u make this a little less boring. like maybe talk like u actually care and make the visuals more exciting

[@kaeez]

This is not a complete beginner's guide. There are a plethora of other videos to be found on KZbin with colors and rainbows to keep you at ease. You ought to have some pre-requisite knowledge to follow up with this. After all, he is working out problems related to this concept, what are you doing here if you don't even have mere conceptual understanding of how all of this works coherently?

[@nolanyoung4480]

Don’t take the Lords name in vain please.

[@TheBlueBaronMoto]

Sooo bad at explaining what is what and why

[@Luxedecor.online]

Man I love you I swear I just watched 4 videos and you are a legend ❤❤❤

[@ayandangesi7743]

And 5 years later this video helps me with introduction to astronomy!🤍🤍