What is the significance of U at infinity in relation to Earth?

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The discussion centers on the significance of potential energy (U) at infinity concerning Earth. It clarifies that potential energy is conventionally defined as zero at infinity, making U negative for distances less than infinity. The formula for potential energy, typically simplified as U = mgh, is acknowledged as a simplification for constant gravitational acceleration. For escape from Earth's gravity, a more complex integration approach is required to determine potential energy changes. Ultimately, potential energy is a relative concept with no absolute value, and its reference point can be chosen arbitrarily.
John Mcclane
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hello sir ...can anyone explain me the value of (U) at infinity with respect to Earth as reference point
 
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Potential energy is given by 'mgh' where the 'g' is acceleration due to the Earth's gravity. As soon as you escape the Earth's gravitational field, it stops affecting you. So the value of PE at infinity doesn't really come up.
 
U of what?

The absolute value is arbitrary, it is chosen to make the problem as simple as possible. For experiments in the lab, potential energy is typically zero at the floor, for experiments in space, it is more convenient to set the potential "at infinity" to zero - but you do not have to do this.
 
The usual formula for U uses infinity as the reference point. U of Earth referenced to infinity is the opposite of U of infinity referenced to earth. So just take the usual formula, find U of Earth referenced to infinity, and flip the sign.
 
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siddharth23 said:
Potential energy is given by 'mgh' where the 'g' is acceleration due to the Earth's gravity. As soon as you escape the Earth's gravitational field, it stops affecting you. So the value of PE at infinity doesn't really come up.
As the other answers implied, that isn't correct. Mgh is as typically used is a simplification for constant g. But for "escape", you'd combine with the equation for gravitational acceleration and integrate over the infinite distance to escape. That's how escape velocity is found and you can find the derivation on its wiki page.

And due to the continuous nature of the gravitational force equation, there is, of course, no distance where the force is exactly zero and Earth's gravity stops affecting you.
 
russ_watters said:
As the other answers implied, that isn't correct. Mgh is as typically used is a simplification for constant g. But for "escape", you'd combine with the equation for gravitational acceleration and integrate over the infinite distance to escape. That's how escape velocity is found and you can find the derivation on its wiki page.

And due to the continuous nature of the gravitational force equation, there is, of course, no distance where the force is exactly zero and Earth's gravity stops affecting you.
Ya I realized that. Do I feel silly!
 
John Mcclane said:
hello sir ...can anyone explain me the value of (U) at infinity with respect to Earth as reference point
As others have pointed out, typically the potential energy is conventionally defined as U = 0 when the distance is infinity, r = ∞. Following this convention, U is negative for values of r < ∞. In other words, most of the time U is negative when an object is near Earth.

This is merely a convention though. You can define U = 0 at any distance you wish, but it's usually chosen to be zero at r = ∞. There is no such thing as "absolute" potential energy.

Although this is not part of your original question, if you wanted to you can find the rest of the formula by evaluating the work done by slowly lifting a mass from the surface of the Earth R, up to infinity.
W = \int_R^{\infty} \vec F \cdot \vec {dr}
or more generally at an arbitrary radius r (such that r is greater than the radius of the Earth, of course),
W = \int_r^{\infty} \vec F \cdot \vec {dr&#039;}Then use the work-energy theorem to express that work in terms of potential energy.

I gather you know what the gravitational force, \vec F is, as a function of r?
 
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