Kepler's Second Law: Explanation + Derivation!

After some time, we'll take up the space to discuss Kepler's Third Law, but first let's give equal area to Kepler's Second Law
Sources:
-www.cantorsparadise.com/keple...
-burro.case.edu/Academics/Astr2...
-hypertextbook.com/facts/2000/...
Intro and Outro:
"Astronaut in the Ocean" by Masked Wolf, beats by FaMusic

[@SambridhTheZenith]

thanks for simple explanation ma'am.

[@spreckrosekrans667]

Nice elegant explanation, thank you.

[@tasnimchowdhury7609]

Love the outro ❤

[@madhavmittal1560]

Superb and concise explanation💯

[@hadialsayed2992]

lovely explanation!

[@hichamessa-kiy9575]

very informative thanks

[@RM-zx5hi]

This video really helped me for my exam
This Channel deserves to be no1 education channel

[@astronaughtpov]

So glad it helped with your exam!

[@Planck944]

thanks

[@mauriciopetersen2622]

sorry but I have a doubt... angular momentum (L) is equal to m.r.v only if the velocity vector is perpendicular to the radius vector... otherwise you write as L = m.r.v.sin(tetha), and that is the case of an elliptical orbit... the angle tetha is constantly changing, in fact the velocity vector will be perpendicular to the radius vector only at the aphelion and perihelion, so the assumption of L=m.r.v at any moment wouldn't be wrong?

[@mauriciopetersen2622]

oh... I think I get it... you wrote Vtetha, not just V... so I presume that you've considered Vtetha only as the velocity perpendicular to R, not the hole velocity of the body. Wich makes sense considering that Vtetha is R.dtetha...
Ok, I think I get it hahah lol

[@astronaughtpov]

@@mauriciopetersen2622 No need to apologize for doubting; it is always good to question! You are correct in both of your comments. L=m(r x v), which in general is not the same as saying L=mrv. In this case, v_θ represents the tangential velocity, so L becomes mrv_θ.

[@resmirk1011]

Thank you

[@userutoob7]

Your videos are very instructive. This analysis definitely works if we assume the orbit of the planet is circular. We all know that Kepler’s laws work for all planets in all star systems, and that the orbits of the planets in our solar system are “very” circular. But strictly speaking, does this proof really work for an elliptical orbit with a “high” eccentricity, one much higher than the eccentricity of the planets in our neighborhood?
OBTW, your presentation is excellent. What software do you use for your videos? If you use a writing tablet, which one do you use? AB

[@astronaughtpov]

Hello, thanks for the comment! The derivation for Kepler's Second Law is independent of eccentricity. That said, the shape of the ellipse should not matter.
The tech portion of these videos is really quite simple. I record myself drawing on an online sketchpad and then speed up, pause, or slow down portions of the visual elements to correspond with the voiceover. I try to speak slower or add an extra delay for concepts that take a bit longer to process.

[@sussybaka3420]

It doesn’t matter because if you consider it in time t to t+dt and let dt->0 then you can regard earth as it is having circular motion in that time so the area dA is like that of a circle. Sorry for bad English

[@madhavmittal1560]

Can you please make a video on -"the effect of rotation of earth on g"

[@astronaughtpov]

I’ll keep it in mind!

[@tasnimchowdhury7609]

Love the intro ❤

[@astronaughtpov]

Thanks haha. This song came out when I first made the channel and it felt fitting

[@francisenemali7786]

Nice

[@Jacob11227]

Very useful thank you. But how can you take the arc length to be rd(theta)?

[@astronaughtpov]

Thanks for the question! Arc length can be found by multiplying the angle by the radius. This is basically a percentage of the total circumference of the circle. Arc length = (angle/360)*2*pi*r when the angle is in degrees. When converting to radians, this simplifies to theta (the angle in radians)*r. We use d(theta) in this case because we are looking at a very small angle.

[@PTNLemay]

3:00
When we move dt to the other side and try to multiply L/(2m) by time, the result would be square meters? As in, we don't need to do something special with the units of area or of time? I tested it quickly using the Earth and the Sun's parameters and it doesn't seem to work. For a full circle (time being 60*60*24*365) I get an area that's much smaller than what you would actually expect for a disc of radius 149 million km.

[@user-yz9nm9yi5v]

did u find the ans now?

[@kareem6148]

Why thank you good sir your assistance is has greatly helped me in my physics endeavor .I will remember you when I become famous. Yours truly future Albert Einstein

[@kareem6148]

Everyone seeing this message subscribe and I will mention ur name when I get famous

[@xyz_124]

Thx

[@deathonmondayyt7192]

I had a question. where did the other r go?

[@astronaughtpov]

The tangential velocity is equal to angular velocity (dθ/dr) times r, so the r is included in vθ.

[@krupt5995]

NOT THE ASTRONAUT IN THE OCEAN LMAO. Nice video though

[@astronaughtpov]

Haha I had the song stuck in my head for a while when I first made this channel

[@contagiousscorpio7048]

Okay the integration part was more understandable. Lol

[@kareem6148]

The outro could use some work

[@astronaughtpov]

Fair enough haha. I intend to eventually change both the outro and into

[@Cooososoo]

Are you astronaut

[@neverarguewithan18wheeler10]

Very educational but STOP USING METRICS!!!

[@madiaw5553]

Cry harder

[@divaykey]

nietzche what