 # Kepler’s Second Law

Kepler’s second law states that planets sweep out equal areas in equal times. Let’s take a look So if we have this planet here closer to the sun than it is when it’s over here Because it’s closer the Force due to the sun is going to be greater, and so it’s going to be moving with a faster velocity Right here a faster. Speed [okay], so it’s moving quick right here It’s zipping around right here, but when it gets over here because it’s so far away from the from the sun it’s not going to be moving as fast this the sun is going to is going to be pulling on it, less and So it’s going to be moving slower over here as a result and so the the trajectory might go something like this in time-lapse fashion Whenever it’s over here at aphelion. It’s moving slow Just snail’s pace crawling along, but then it’s speeding up because it’s getting closer and closer to the sun And it’s moving the fastest when it’s right here at perihelion, and then it’s you know it sits around and then it slows down It’s moving slower and slower until it reaches a billion and it’s moving at its lowest and then so on Okay, something like that. You can look for some animations [online] to see a you know a better description of that or more more accurate time lapse fashion okay, so KEv will argue that the time rate of change of the Planet right of the area that the planet sweeps out must be equal to a constant value because look whenever the planet is here It’s pole in the mail, it’s zipping along So in some time interval [Delta-t]. It’s going to cover some arc length right and so a line connecting from the sun to the planet is going to sweep Out a big area, right? Okay, but whenever the planet gets over here It’s moving slower, and so it covers a smaller Arc length and so Delta theta is small But the the radius from here all the way to here is compensates for that Such that it sweeps out an equal amount of area okay, and so this is the law equal areas in equal times Okay, so we can use that [to] deduce something very useful for solving problems Let’s for the first thing we’re going to do that Let’s find the angular momentum of that planet over [there] at this You know distance away here, so the angular momentum is equal to R cross p Right where R [is] the distance from the axis to the particle in question? And it’s the vector from from the sun to the particle and then p is the momentum right the linear momentum So we know that only the perpendicular component [of] the linear momentum will contribute in the cross product We’re going to label this or right here, and then let’s draw the linear momentum It has a velocity that’s all flike that write which results You know in V roll results in some linear momentum like this and if we find its components it has components Like this this is p Radial we’ll call it and also. It has [ap] perpendicular, right? [and] the perpendicular means that it’s perpendicular to that moment arm. So this is P perpendicular Ok well, it’s only this perpendicular component of the momentum that contributes So this is R times p Perpendicular so now we’re just going to drop the vectors and just realize of course that the right hand rule says that it’s moving you know of course it’s moving around this way and the right hand rule says [that] the Angular Momentum will be out of the board [okay]? So then we can substitute for P perpendicular we can put in MV perpendicular right because this is just MV perpendicular So put em V perpendicular but then for V perpendicular We can substitute in omega [R] because we know that whenever something is out some distance R And it’s moving with some angular velocity that the perpendicular velocity to that moment arm, right is given by Omega R So we can put r Times M times omega R like this so let’s rearrange things a little bit We’ll pull the m out front and then we’ll say this is r-Squared Emedia like this Okay Now let’s consider that will be our first equation, and yeah Let me just separate it [like] this now. Let’s consider the area of this triangle here Okay for the non calculus folks fret not it’s not you know a huge leap But we’re going to consider a small change in Arc length, right and we’re going to call [that] d s [d] s is an infant testable amount it’s just a small amount of width okay, and then we know that that is equal to R [dtHeta] Right from the classic equation you know s equals r theta for a circle, okay? We’re just going to extend it to this ellipse because it’s such a small distance and if we want to calculate the [area] of this triangle the area of a triangle We’re going to call it da because it’s a small amount of area is It [go] [to] [1/2] the [base] times the height well the base of that triangle is the DS, right? so already theta so we can put that in for the base of the triangle and What’s the height of the triangle the Height is just going to be equal [to] R? so we put that in for R like this and so this is equal to 1/2 [R] squared D theta combining the ours [ok] so this occurs over some you know amount of time, right so we can divide by Dt Over both sides get the time rate of change for the area and the time rate of change for the angle and then this Notice this is just once d theta dt. That’s just equal to omega So this is 1/2 [or] squared omega All right Now from this equation We can eliminate R. Squared omega between both of these two and it will yield the powerful result so from this equation We can say that r squared omega is equal to look we just divide by m so it’s L over L Like this, and so we’ll make that substitution in for this thing Okay, so then this is equal to and then 1/2 times L over m is L over 2m, right? And so this is equal [to] Da Dt but kepler stated previously That the time rate of change for the areas that planets sweep out equal areas in equal times [write] that the time rate of change? Of the area swept out is equal to a constant value so you concluded that the angular [momentum] for A planet L must be equal to a constant value also so the conclusion Was that L naught must always be able to L final? Unless there’s some outside force acting which there’s not okay. There’s only the force of the planet acting Essentially, it’s a central force problem [ok] and this result will be useful Because you can say with this with this result you can say m R naught V naught is equal to M our Final V final Right and so you’ll be given you know you’ll begin some information It’s at this distance moving this fast or whatever you know or it has this angular velocity at this moment and so you know that will give you enough information to get this stuff over here because V is equal to omega or not and Then you know then you be asked to find the final velocity later or something you know Underline what you know is sort of what you don’t solve Okay, see you next time

## 33 thoughts on “Kepler’s Second Law”

• ### Saul Rémi Post author

@kingplatoedwinkaku There are 2 answers really: 1. So I exaggerated the change in velocity as I was going around the ellipse for demonstration purposes. The orbit of the Earth is very nearly circular and so our speed is very nearly constant (66,000mph) throughout our orbit around the sun. See "Kepler's First Law" for for a discussion on eccentricity of orbits. 2. Einstein proposed that space is curved and if we fall freely then we don't feel our own weight. See "Principle of Equivalence"

• ### Saul Rémi Post author

@kingplatoedwinkaku I was running short on characters. Good question and I hope this helps. Let me know if you need any further clarification but be sure to check out those other two vids either on our YouTube channel or on our website. BestDamnTutoringCom.

• ### Noesgaard Post author

thx m8. you actually saved my exam in physics. you did it in a total different way than my book. thx a lot

• ### TaiMoa Ula Post author

Wish you were my physics teacher.

• ### Aui Ken Hoong Post author

My physics class should just play videos of you teaching .

• ### Saket Dongre Post author

if we put our earth in place of sun and moon in place of earth, then this law is applicable??

• ### tomatomaniaman Post author

Thanks so much! Really helpful, well explained!

• ### jaan baz Post author

thank you very much sir. it was very helpfull for my seminar. god bless you.

• ### PLANTLOVEAFRIKA Post author

I WISH I COULD SHOW U MY LECTURE NOTES FROM CLASS, TOTAL RUBBISH. YOU MADE IT SO SIMPLE.

• ### Sam Richardson-May Post author

Cheers dude, you make it so simple and explain it so well

• Thank you so much this will be on my physics test tuesday!

• I fully understand this thank you so much!

• ### mohamed medhat Post author

why we cancelled the "P" and took its perpendicular component . r ??

• ### Fzyon Zarzuela Fernandez Post author

It is also a video lecture for my upcoming first Astronomy exam and last workbook h.w. this Thurs., March 13 & I hope it will be helpful hint

• ### Fzyon Zarzuela Fernandez Post author

,so thanks !

• ### Julius Mboli Post author

love this man, its pretty brief and concise. many thanks.

• ### uchi hata Post author

does the moon obet kepler's seconds law？

• ### Layla Abdon Post author

helped a lot, thanks!

• ### eli alfa Post author

thanx. ..شكرا 🙂

• ### Pulkit Midha Post author

The noise produced from the friction b/w marker and the board is too disturbing. plz have a solution for that.

• That's more than awesome…
but I'm really confused that how equal momentum can lead to equal area ?? I mean can we prove law rightly by proving equal momentum??

• ### Archana Gowda Post author

thank u so muchh
helped me lott

• ### Dean Pavlovič Post author

when you took deriavtive of dA=(dtheta*r^2)/2 how come you not differentiate r? we are on ellipse, not circle so its not constant or is it?

• how can l get in bangla plz tell me any one

• ### swapnil prithish Post author

thanks sir……understood great ly

• By The scientific evidence that Newton's law is error in calculating Centrifugal force C.F = MV2 / R
• And the correct the centrifugal law must be derived from the law:
• torque = force (mass x angle velocity) x arm.
• And The centrifugal law shall be
• C.F = M x ( V x 2.Π.R )
And the Earth is standing still

بالبرهان العلمي اثبت خطأ قانون نيوتن في حساب قوة الطرد المركزي
• C.F = MV2 / R
• قانون الطرد المركزي مشتق من
• قانون العزم = القوة ( الكتلة x السرعة الزاوية ) x الذراع
• ويكون قانون الطرد المركزي كما يلي :
• C.F = M x ( V x 2.Π.R )
والارض ثابته لا تدور حول نفسها ولا حول الشمس
International Arbitration Consultant in Aviation Disputes and Accidents
Specialized in the Hijri calendar, Gregorian calendar and prayer times
: ATP + F/E + F/D = F.A.A
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• ### Era Smith Post author

Great explanation understood precisely…..

• ### Jim Keller Post author

Now there's an intuitive concept for you: " the right-hand rule says that the angular momentum will be out of the board." I'll bet not one person can explain what that in fact means.

• ### Adharsh T S Post author

thankz

• ### A L. Post author

sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss

• ### Tlhomotse Moteme Post author

Oh my God you just made it simple! Thank you.

• • 