Inverse Square Law - Physics

[@TheOrganicChemistryTutor]

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Ameen

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[@Mattropolis97]

You’ve made me understand something very important about myself today… So I’ve always SUCKED at math. In high school I’d always apply formulas to problems but I’d still always somehow get the answer wrong. Even knowing my mistakes I never got why I made them. I paused and gave your final problem a whirl on my own and just thought about how I’D solve it logically. I continued watching and you started going off on all this crazy ratio and fraction stuff which I had no idea how you decided was the appropriate step to take, etc. and I was like “ok…I suck” but in the end I ended up getting the right answer, and I didn’t even need to do all the stuff you showed. I figured it out starting from the last step you showed. So what you’ve made me realize is that for all those years in school, I was too hung up on formulas. I was so busy trying to make sure I remembered what the next step was that I completely isolated the parts of my brain that use context and logic to solve problems, and as a result I never had that fail-safe voice in my head telling me “wait that wouldn’t make sense, don’t do that”. Everyone’s brains work different I guess

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[@LOVE123_LOL]

6:12 There is an easier way to do this figure: We know that we have divided the original intensity by 1/16 to get 800W/m^2 ,so we multiply by 16 to get the original one(12800)...we complete multiplying 12800 by 1/9 to get 1422.2 repeating...

[@oliverjantrembos4854]

I used that as well. It is an easier way, but those formulas are useful too, in more complicated problems. When I saw that written problem I directly calculated the number 1422,2 (periodic) the way you wrote above. Then I just switched to the end of the video!

[@eZaFJDUBB]

It's amazing a simple formula has given us insight into our universe huge distances

[@davedeatherage4902]

Thank you teacher; sir, until 2:43 you explained crystal clear the Inverse Square Law. Thank-you. I wil continue to review latter half of film until I get the formula down you presented. Again thanks! ( your instruction is simple and precise in your administration of the formulas,and very understandable to me )👍

[@TheH8redd]

You made it very easy for me to understand it, great job sir!

[@jassonsmall]

Freshman year got a whole lot easier thank you

[@alexisubaldo3724]

My teacher literally took your example on my exam😂

[@ButterflySimmer]

Oh wow XD, lazy bum

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You’re teachers smart 😂 if you studied then you get the answers

[@Repz98]

All videos on hes channel have been used by teachers around the world, literally.

[@geetikagautam3730]

Lucky

[@deansawich6250]

Thanks. Sound also uses the inverse square law.

[@daffaasyhada4248]

Really good explanation. Thank you

[@bobobobobobobobobo9393]

you can make it more simple by taking the square of the inverse, 3m/4m -> 4m/3m -> (4m/3m)^2 -> 16m/9m then multiply 800 by 16m/9m

[@bobobobobobobobobo9393]

NY can that too

[@dear_imran]

@@nycan7725 isn't it same?

[@antisymmetric237]

You are a great teacher. Thanks

[@ButterflySimmer]

What I did is 800/ 1/16 to get 12800, then i divide it by 9 (as that is 3squared) to get 1422.2. i always seem to do calculation my way and that is different from other people's methods. But good method.

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Thanks so much :)

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Very very thank you🙏🙏🙏😊

[@fruffers]

I don't understand 5:11 at all. How are we getting those numbers by * R_1^2 by the two fractions? :0

[@buer_reub]

Wait so you plug in the two because its double the distance right? If it was triple the distance would that have been a 3?

[@mdobaidullahshake5186]

Nice video I see from Bangladesh

[@paulmobleyscience]

Hello sir, I have an honest question for you that needs an honest answer and is to me one of the most important questions of our time. In the Bureau of Standards volume 3 on pages 81-82 published in 1907 it states that the inverse square law does not apply to extended sources of radiation because the center could not be known therefore the intensity could not be calculated and that a more sophisticated law and formula must be used on extended sources of radiation. Can you tell me what that more sophisticated formula is because surely in over 100 years there should be a formula used for extended sources of radiation. I ask this question because this law and formula is used to not only calibrate Geiger Muller tubes and various other radiation detection monitors but also for taking measurements to calculate the intensity of ones exposure from a source of radiation. If this law does not apply to extended sources of radiation then how could we not only calibrate a Geiger and take measurements but also how could our sources of radiation samples of known radioactivity (Cesium, Cobalt etc.) be correct at that point? Not to mention the inability to ever have a true zero to begin with. When talking about the intensity of light from a lightbulb, the sun or any other light source an approximation may work but when we move into the realm of extended sources of radiation especially here on the surface of our planet with an atmosphere and a water cycle to move it further then how could we ever know the intensity of exposure using the simple inverse square law and formula? Thank you for you time

[@bharathbecrux4761]

Well umm if you have 2 charged cylinders of length 6 cm with uniform charge all over them and they are separated by 10cm can you help me find the potential at a distance of 5cm from both cylinders?

[@Gismotronics]

Ok, I tried but couldn't work it out... If we have 1 lumen of light intensity from the full moon and the moon is 238,855 miles, what is the lumens intensity on the surface of the moon? Use Km perhaps... 384,400 Km avg. distance from Earth to Moon. Let's also simplify it by asking the question like this... What is the Lumen light energy at 1 Km from the moon. Zero is tricky number! So, using the formula in the video, what if we want to know I2 when R2 = 1? Makes it easy, right!? However, it's so easy that it's confusing because it simplifies to I2 = R1 squared. The issue is, if you use meters instead of Km (384,400,00), if you square that, you just end up with a way bigger number for I2, which can't be right. I'm not a mathematician! help?

[@alirahimkhan3741]

Thank you sir

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I'm doing my post-grad yet I'm still here lol thank you so much

[@jeffreychavey4161]

Great explanation. BTW - you sound just like Bobby Flay the chef guy

[@RicardoFerreira-et1ez]

Question. The moon is 0.1 lux from earth. So how bright would it be if I was really close to the moon? The earth is 384,000 Kim’s from earth. Thank you

[@myztazynizta]

Is there an easier... or at least "different" way of answering the world problem that I might actualy be able to understand?

[@jonashammar2099]

42

[@AaroneousMaximus]

Before continuing all of these steps, I took a crack at it. Did 800w x 16 = 12,800w / 9 = 1422.2222etc. There’s always a lazier way, my friends

[@John-100]

Here is a real world survival use of the inverse square law. A 500 KT Russian nuclear weapon detenates over your town, at 0.4 miles the pressure wave measures 30 PSI, at 2.2 miles the pressure wave measures 10 PSI, what will the pressure wave measurement be at 10 miles from ground zero?

[@raviiyer6874]

thankyou!

[@Fatima-zf5ym]

I’m trying to understand from an hour … I just wanna sayyy thankkkkk you!!!!!

[@serpentzachary1340]

I= p/4π r^2

[@John-100]

Not sure about you physics guys, but the answer can be found just knowing 7th grade math, example X(1/4^2)=800 Solve for X, X =12800 You now know the watts at the source. So: 12800(1/3^2)=1422.2 watts/m^2

[@drewlappin2203]

Or a far more simple 800 x 16/9

[@nutcrackerreal100]

Einstein would love to be your friend

[@Seedavis397]

So how could we see stars at millions of light years away??? Doesn’t seem possible…. Either the formula isn’t correct or the stars are not that far away

[@authenticNL2]

Because our sun is insanely bright even at 93 million miles. When you find the original brightness by dividing how bright sunlight is on Earth (about 120000 lumens) by Inverse Square Law (1/93 million miles, in meters), you get 1.80180223*10^16. Earth Atmosphere deflects about 30% of light, and another 23% scatters/diffuses across the atmosphere. So, 0.30*0.23=0.069. Finally, 1.80180223*10^16/0.069=2.61130758*10^17. Unfortunately, some variable I don't know is throwing my maths completely off. The sun is about 35.73*10^27 lumens, but my point still stands.

[@devanggupta9007]

i just did (800*16)/ 9 [cuz 4^2=16 nd 3^2=9]

[@shanejoy7491]

why don't we take 4pi into account cuz when u have 2m and square it won't u have to times by 4 so you get power/16pi so basically Intensity is proportional to 1/16

[@Zreknarf]

you also have to multiply the 1m by 4 which gives you 1/4pi which is 1/4 of 1/16pi

[@andrejrockshox]

you sure do have some really strong bulbs...

[@johnnyreno7200]

So...basically..if we look at our solar system using the sun as the light source and use the planets Mercury which is the closest to the sun and Pluto which is farthest from the sun?.. is that the idea?

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I stopped crying because of you lol thanks

[@kevinkall8547]

If light diminishes over distance and the sun is supposedly 93 million miles away, how bright would it need to be one mile away?

[@anshpatel1927]

1/scale factor^2

[@ethenq2002]

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[@emperorpalpatine3125]

Wouldnt this equation make it impossible for the intensity to ever be zero? Because no matter how far away you were you would always have a number. Albeit a very small number, but a number.

[@emperorpalpatine3125]

@@DixieDee I'm saying that no matter how far away you are you would always experience some intensity, which doesn't make sense because light bulbs do not have infinite power

[@emperorpalpatine3125]

@@DixieDee All good, there is probably some equation to calculate the limit of its intensity, but i obviously dont know it

[@h2w25]

It wouldn’t be Perceptible

[@paulmobleyscience]

@@emperorpalpatine3125 Great point...I think the answer lies in the law itself that does has parameters and a single stipulation of the use of the inverse square law. The inverse square law does not apply to extended sources of radiation written in the Bureau of Standards volume 3 pages 81-82 in 1907.

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My favorite tutor! 😘😘😘 Second one to comment here!

[@wakiwakr]

Light-get-more-dim-when-it-more-far-away

[@h2w25]

Really shoots a hole in the heliocentric model considering the nearest star is supposed to be 4.3 light years away…

[@Keloot]

No it dosent. Héliocentrism is just the sun being at the center of our universe and earth or eating around it. BTW the closest star is our own sun

[@025.kalpeshmahajan4]

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[@justin1995so]

Well where the 4pi goes?..can't understand why there's no one ever asked this

[@christians2898]

4pi cancels, likewise with the p variable

[@justin1995so]

@@christians2898 I've some confusion about this, so in the case mentioned in the video P=power=360w/m2, when R=distance=2m the equation should be I=360/4pi2^2. but instead, it takes away the 4pi to get 360x1/4 to get I=90. why is that?

[@balancius8381]

What if I want to move it from 2m to 3m. Can it be done without going back to 1m. So the distance is not moved by double or triple, but only half distance

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Mark Wahlberg?

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I got a number way too large that is all

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face reveal?

[@danimorcos7226]

guys i got a question cuz im kinda confused about this topic. Why doesn't starlight follow the inverse square law? im probably just an idiot xD

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First one to make a comment

[@garyryan7852WR]

Proving the sun is not 93 million miles away . ty

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