Physics 18 Gravity (16 of 20) Gravitational Potential Energy

[@abdelmannan2332]

I don't understand how you can fully explain a seemingly counter-intuitive topic so well in a short video. I wish all teachers were like you. You sir are a legend !!

[@jacobthomas6908]

WOOOOW. I am blown away by how perfect this explanation was.

[@BoZhaoengineering]

This concept is tricky. I came across some videos on KZbin, your video is one of the clearest and most comprehensive interpretations. Using dot product alongside the radius orbit from the center to outer space is a smart way to proceed with vector product. Thank you for your excellent explanation.

[@MichelvanBiezen]

Glad you found us and the videos are helping. All the best.

[@frankdimeglio8216]

​@@MichelvanBiezen WHY AND HOW FRANK MARTIN DIMEGLIO HAS FUNDAMENTALLY REVOLUTIONIZED PHYSICS: TIME is FULLY consistent WITH WHAT IS E=MC2, AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). Consider TIME AND time dilation ON BALANCE. What is E=MC2 IS WHAT IS GRAVITY. Consider what is the orange (AND setting) Sun ON BALANCE. It IS the SAME SIZE as what is THE EYE. NOW, consider what is the fully illuminated (AND setting/WHITE) MOON !!!! The rotation of WHAT IS THE MOON matches the revolution. The stars AND PLANETS are POINTS in the night sky. c squared CLEARLY (AND NECESSARILY) does represent a dimension of SPACE ON BALANCE !!! INDEED, consider what is THE EYE !!! Magnificent. NOW, consider why and how it is (ON BALANCE) that there is something instead of nothing. Great. The bulk density of WHAT IS THE MOON IS comparable to that of (volcanic) basaltic lavas ON THE EARTH. The density of lava IS about THREE times that of what is water ON BALANCE. LOOK directly overhead at what is the TRANSLUCENT AND BLUE sky. Don't forget about THE EYE ON BALANCE. (The Earth is ALSO BLUE.) The density of pure water IS HALF of that of what is packed sand/wet packed sand ON BALANCE !!!! The diameter of WHAT IS THE MOON IS about ONE QUARTER (at 27 percent) of that of what is THE EARTH ON BALANCE !! Great. ACCORDINGLY, ON BALANCE, WE MULTIPLY ONE HALF TIMES ONE THIRD in order to obtain the surface gravity of WHAT IS THE MOON in DIRECT comparison WITH WHAT IS THE EARTH/ground. (The maria then occupy, predictably, one third of the near side of what is THE MOON ON BALANCE.) Notice that the blue Moon IS INVISIBLE ON BALANCE. GREAT. So, consider what constitutes (or is) the orange Sun, ON BALANCE, AS TWO THIRDS TIMES ONE QUARTER IS ALSO ONE SIXTH !!! GREAT. AGAIN, consider what is THE EYE ON BALANCE; AS E=MC2 IS GRAVITY. BALANCE AND COMPLETENESS GO HAND IN HAND. IT ALL CLEARLY makes perfect sense ON BALANCE. By Frank Martin DiMeglio

[@SH-wn1mq]

Thank you! This was exactly what a part of my homework was on, but my professor hasn't given a lecture about this yet. This was really helpful

[@MustafaYlmaz-ri6qv]

I think direction of F would be inwards and so it would be F.dr.cos180 and then last formula then consists of minus sign. So at the end there is no need to change the sign manually.

[@that_guy1573]

Wow... thank you so much for this video. I was confused as to how you find the equation for gravitational potential energy, as we were taught it in a different, more confusing way, but using integrals makes a lot more sense.

[@MichelvanBiezen]

Glad it was helpful!

[@dddfanqi]

THANK YOU VERY VERY MUCH. This is much better than the $** 4-month subscription we are forced to pay for.

[@anzatzi]

Finally get this--sort of. The negative sign seems very semantic and unintuitive

[@AB-ts7gi]

why is the force in this case equal to Fg? like if that would be the case wouldn't it point backwards?, I know a force needs to be applied to move the object, but why is this force equal to Fg?

[@MichelvanBiezen]

The weight is pointing towards the Earth, but the force required to push the satellite higher points outward.

[@hussainbohuliga3613]

What will happend to the potential energy if we have R3?

[@carultch]

Good question. The way you can find it out, is to use the relationship between force (F) and potential energy (U), which is dU/dr = -F. Integrate both sides and get U = - integral F dr. F is given as -K/r^3 in this hypothetical example, where K is a positive constant, and r is the distance from the origin, which would mean U would equal -1/2*K/r^2 + C. Simple application of the power rule, where integral x^n dx = 1/(n+1) * x^(n+1) + C. Set n = -3, and the exponent on the integral of r^n dr becomes -3+1 = -2, and the leading constant also gets divided by -2. The C is an arbitrary constant, which we can define to equal zero as we ordinarily do with GPE already. So we can say that U = -1/2*K/r^2, when F is given as F=-K/r. One place we run in to problems with using the power rule to do this integral, is when our equation for F is F=K/r. Using the power rule would give us a denominator of zero, and an exponent of zero, which would be big problems for evaluating it in practice. Fortunately, Calculus has an answer, which is that integral 1/x dx = ln(x) + c. This would mean the potential energy function would be given as U = -K*ln(r) + C. We can ignore the absolute value, since r is always positive.

[@kalopsia4124]

Explanation so clean it squeaks

[@MichelvanBiezen]

Thanks. 🙂

[@huzaimkhan9269]

Wow thank you for Good explanation about negative sign in gravitational potential energy. I have never seen a teacher like you. GOD bless you!

[@MichelvanBiezen]

Thank you.

[@zakirhussain-js9ku]

When an astroid enters gravitational field it gains kinetic energy. This energy comes from potential energy of gravitational field. Loss of gravitational potential energy should weaken gravitational field. This cannot happen since mass remains constant. Question is what is source of energy gained by the astroid if energy of gravitational field does not diminish.

[@MichelvanBiezen]

When an asteroid enters a gravitational field, it does not weaken the field.

[@cheezecats4085]

Amazing explanation. Thank you very much!

[@ironuranium3927]

what is the physical meaning of minus sign in gravitational potential energy ?

[@MichelvanBiezen]

It is a matter of a reference point. On the surface of the earth we use PE = mgh as the potential energy equation. If you go below the surface like in a subway, the potential energy would be negative. In space the zero reference point is at an infinite distance away from the gravitational source like a planet or a star and thus as you get closer to the planet or star the PE is negative.

[@pahulgill6114]

I love you

[@albertmendoza1468]

I have a question Sir Michel. At 1:41, why is the force postive? Shouldn't it be negative since gravity is acting in the opposite way relative to the direction of motion?

[@MichelvanBiezen]

Which force is doing the work pushing the object up? (It is not the force of gravity). Thus the force doing the pushing is in the same direction as the displacement.

[@albertmendoza1468]

Isn't that an external force? If there is also a force (gravity) acting on the opposite direction as that external force which is also equal in magnitude, shouldn't those two forces cancel leading to no acceleration or movement at all? I'm sorry for my naivety.

[@Ahmed-vs1ui]

Can you explain why exactly we chose to integrate here sir??

[@MichelvanBiezen]

When the force varies over distance, integration is necessary

[@Ahmed-vs1ui]

Michel van Biezen okay, im really trying to understand it but i think its a problem in my understanding of integration cuz this equation has always confused me and still is unfortunately

[@carultch]

@@Ahmed-vs1ui If we had a force that was uniform at every position along the way, and aligned with the direction of motion, and wanted to calculate the work done by the force as an object moves across that distance, it would be a simple multiplication of force * distance. If there were an angle between the two vectors of force and displacement, the multiplication would become a dot product. When the force is NOT uniform with distance, that is when we need to integrate. What you ultimately are doing, is multiplying infinitesimal displacements with each force at every point along the way, and adding them up. Except integration allows us to be continuous in our approach, and less computationally intensive. When the force is neither uniform with distance, nor aligned with motion, that is when our integral becomes a vector field line integral.

[@Ahmed-vs1ui]

@@carultch i really appreciate you answering even after all this timehas passed Thank you so much

[@sakshinaikwade1720]

That was really helpful sir....! Ur explanation is perfect. 👍👍

[@MsLoversky]

Thank you very much Sir Mike, It's helpful for me.

[@MichelvanBiezen]

Glad to hear that

[@dawn-of-newday]

I really like the video and it covered everything but I didn't get the integration R raise to minus 2 divided by dr

[@MichelvanBiezen]

The rule for integration like that is that you add 1 to the exponent (-2 +1 = -1) and then you divide by the new exponent (divide by -1).

[@nandini.sreddy5845]

the best teacher

[@memyselfiyo_its_me_lmaooo8769]

Nim ammin na kaiya

[@ardajakuzi2081]

you are an ataturk to me, everytime saving me from the enemies i can't defeat on my own thanks a lot

[@MichelvanBiezen]

Happy to help. You still have to do the heavy lifting.

[@kishorenavale8246]

Thank you sir,love from India

[@CatsBirds2010]

Very clearly explained and thanks.

[@thinkmatelearningsolutions5415]

Force vector is Inward (F) and Position vector (dR) is outward to move towards infinity angle should be cos 180 and not cos 0. Cos 180 is -1 ? anyone

[@davehumphreys1725]

The quantity mg, when m=mass of the earth, is the force of gravity of the earth, isn't it? But this force of gravity is exerted on every other massive particle in the universe? I just cant understand why mgh is invalid when h gets large?

[@MichelvanBiezen]

Because g is not constant when h is large. The equation PE = mgh only works if g is constant over the change in height.

[@davehumphreys1725]

Thanks for your reply. OK how about this one. PE = mgh is positive. GPE = -GMm/r is negative, yet both equations describe the same quantity don't they? So why the different signs?

[@MichelvanBiezen]

The reference point is different. On the surface of the Earth (g = constant), the reference point is PE = 0 at h = 0. In space where g is not constant. PE = 0 at h = infinity (and negative everywhere else).

[@dawn-of-newday]

I hav a doubt about binding energy of a satellite on the surface at rest . it uses the same formula u derive in the video for gravitational potential energy but on earth surface g is constant so why not mgh ?

[@MichelvanBiezen]

When you use PE on the surface, the reference point is the surface (PE=0) But in space the reference point is at x = infinity and that is where the PE is zero. So it depends on where you want to place your reference point and how far above the surface you want to compare it to. (g is not constant when leaving the surface of the Earth).

[@josephwheelerton]

Why is the potential energy zero at r = infinity? I agree that the force would be zero but the area under an F(r) function from r=0 to r=infinity is not zero.

[@MichelvanBiezen]

+Jose Molina Remember that the potential energy anywhere else is negative, which matches the negative area under the curve when you integrate.

[@carultch]

It's an arbitrary convention we assign to keep the math simple. When you took calculus, and had to write +C on every indefinite integral, you may have wondered why we do that, and why you can't just let C = 0. Well in many applications, you can let C=0 to keep it simple, which is what we do with universal GPE. Since we are ultimately interested in differences in GPE, and absolute GPE has very few (if any) applications, we don't really care what C equals, and set it equal to zero for simplicity. The consequence of doing this, is that we assign GPE to equal zero, infinitely far away. There are applications of keeping the +C constant of integration around, and ultimately solving for a value for it, but that is a topic for another day.

[@naeemghafori5046]

absolutely awesome Explanation.

[@shabbirgheewala14]

good explanation

[@jannatuladhaneva]

Thank you. For this explanation

[@DushyantDeshwal]

Great Video ! Completely understood that !

[@dishant8126]

Wow thank you so much

[@MichelvanBiezen]

You’re welcome 😊

[@pholosocalven6937]

wow thank you sir...that really helped much..

[@fahyen6557]

I wish I have learned integral, so that I can understand it even better. Good video still.

[@rummusLoL]

Very, very good video!

[@danv8718]

All these years I thought he said "Welcome to electron line"...

[@akromaansah2283]

why integrate the force

[@MichelvanBiezen]

If the force is not constant it must be integrated.

[@akromaansah2283]

Michel van Biezen thank you

[@dawn-of-newday]

thank you that's explain a lot

[@ahmedragon2903]

makes it look easy thanks

[@belle30222]

you are a god among men

[@MichelvanBiezen]

Not a god, just a person like everyone else. :)

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[@joshuaoriendo7781]

I didnt go to school so I need to study about our lessons even so im absent ! XD THen i find myself advancing to our lesson XD im just search for PE=mgh then i find it those not apply ! ^_^ eehehehe Quite nice learning to you Sir Thumps up :)

[@abdelalsnayyan959]

Thank you for the videos!!!!

[@Mustamaggara]

now I get it. PS the interface of ilectureonline is pretty poor, it should have some kind of menu-submenu structure so that finding specific topics were easier

[@MichelvanBiezen]

Are you referring to the ilectureonline web site or the youtube site. (We've been ignoring the site to spend more time on the youtube videos.)

[@Mustamaggara]

Michel van Biezen I meant the website

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[@aromelzeus6317]

nice vedio

[@afsaralabiba9461]

Dammmn you have explained it really well!Thank you so much *_*

[@snusinanus]

Amazing explanation thank you very much

[@kholoudkandil5878]

على الطلاق بالتلاتة اسيادنا راضين عليك p.s: don't try google translate . It simply means good job

[@hussainbohuliga3613]

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[@mo.elbanna]

حلوة

[@Ahmed-vs1ui]

ههههه

[@lisro21]

Thanks

[@SameerKhan-oc1dj]

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[@MichelvanBiezen]

Welcome to the channel!

[@mohammadmunzurulhaque3197]

7 people who unliked, dont understand English...

[@prashantnagre2798]

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[@oscarlin8450]

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[@MichelvanBiezen]

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[@trungac2414]

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