Gravitational potential energy derivation

GPE is a form of mechanical energy caused by the height of the object above the surface of the Earth (or indeed, any other source of a gravitational field). Any object that isn’t at the lowest-energy point in such a system has some gravitational potential energy, and if released (i.e., allowed to fall freely), it will accelerate towards the center of the gravitational field until something stops it. Although the process of finding the gravitational potential energy of an object is quite straightforward mathematically, the concept is extraordinarily useful when it comes to calculating other quantities. For example, learning about the concept of GPE makes it really easy to calculate the kinetic energy and the final speed of a falling object. Definition of Gravitational Potential Energy GPE depends on two key factors: the object’s position relative to a gravitational field and the mass of the object. The center of mass of the body creating the gravitational field (on Earth, the center of the planet) is the lowest-energy point in the field (although in practice the actual body will stop the falling before this point, as the Earth’s surface does), and the farther from this point an object is, the more stored energy it has due to its position. The amount of stored energy also increases if the object is more massive. You can understand the basic definition of gravitational potential energy if you think about a book resting on top of a bookshelf. The book has the potential to fall...

in the example of the pendulum, what force is doing the work?? the least i can make out is that it isnt gravitational force that is doing the work of making the pendulum move because the acceleration or displacement of the pendulum is sideways and not downward, am i correct and someone please answer my question It is the gravitational force only which does the work. I appreciate your assumption that it must not be gravitational force as the motion here is sidewards. However, the downward pull exerted by the gravitation here acts sidewards as the pendulum is connected to a rigid end and can't come downwards. What I say maybe doesn't make absolute sense, because the rigorous explanation of your question requires the mathematical understanding of vectors and vector components, which you will learn in the Class 11 physics course Hope that helped. Feel free to comment if you need further help :) - [Narrator] If I hold this bowling ball at some height, then we say that the ball has gravitational potential energy. And the reason we say that is because if I let go of that bowling ball then gravity can do work on that ball. And the question you want to try and answer in this video is exactly how much potential energy does this ball have? Meaning if I know the mass of this ball, let's say it's M, and I know exactly how high it is, what is the potential energy? That's what you want to try and figure out. All right, so how do we do this? I think we can do that if we understand exactly...

Discussion introduction Short bit of calculus. U g=− ⌠ ⌡ F· d s r U g=− ⌠ ⎮ ⌡ − Gm 1 m 2 dr r 2 ∞ U g=− Gm 1 m 2 ⎛ ⎜ ⎝ 1 − 1 ⎞ ⎟ ⎠ r ∞ and here it is… U g=− Gm 1 m 2 r where… U g= gravitational potential energy m 1 m 2= masses of any two objects r= separation between their centers G= universal gravitational constant (6.67×10 −11Nm 2/kg 2) Note that there is no ∆ in this expression. Discuss here. Discuss potential vs. potential energy somewhere. small distance approximation What about the old equation? ∆ U g= mg∆ h It's hidden in the new equation. U g=− Gm 1 m 2 r Let me show you. ∆ U g= U f − U i ∆ U g= U g( r+∆ h) − U g( r) ∆ U g= − Gm 1 m 2 + Gm 1 m 2 r+∆ h r Combine terms over a common denominator. ∆ U g= Gm 1 m 2(( r+∆ h)− r) r( r+∆ h) ∆ U g= Gm 1 m 2∆ h r( r+∆ h) Multiply by "one". ∆ U g= Gm 1 m 2∆ h r r( r+∆ h) r Swap terms in the denominators. ∆ U g= Gm 1 m 2∆ h r r 2 r+∆ h Factor some stuff out of the numerator. ∆ U g= m 2Δ h Gm 1 r r 2 r+∆ h Do you see it? If r is the radius of the Earth, m 1 is the mass of the Earth, and m 2 is the mass of something being lifted, then… g= Gm 1 r 2 is the acceleration due to gravity on the Earth's surface. Making this substitution (and dropping the subscript, since we only have one mass left), we get… ∆ U g= mg∆ h r r+∆ h The first part of this expression is our old friend, the original equation for gravitational potential energy. The second term is a correction factor. For ordinary heights, this term is essentially one. Let's con...

I was shown this derivation for the gravitational potential energy, and I'm not very happy about it assuming that $\frac You're just having an issue with $\lim_$$