- #1

- 1

- 0

## Main Question or Discussion Point

hello sir ....can any one explain me the value of (U) at infinity with respect to earth as reference point

- Thread starter John Mcclane
- Start date

- #1

- 1

- 0

hello sir ....can any one explain me the value of (U) at infinity with respect to earth as reference point

- #2

- 248

- 26

- #3

mfb

Mentor

- 34,055

- 9,918

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.

- #4

Dale

Mentor

- 29,066

- 5,328

- #5

russ_watters

Mentor

- 19,425

- 5,575

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.

- #6

- 248

- 26

Ya I realized that. Do I feel silly!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.

- #7

collinsmark

Homework Helper

Gold Member

- 2,890

- 1,213

As others have pointed out, typically the potential energy is conventionally defined ashello sir ....can any one explain me the value of (U) at infinity with respect to earth as reference point

This is merely a convention though. You can define

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

[tex] W = \int_R^{\infty} \vec F \cdot \vec {dr} [/tex]

or more generally at an arbitrary radius

[tex] W = \int_r^{\infty} \vec F \cdot \vec {dr'} [/tex]

Then use the work-energy theorem to express that work in terms of potential energy.

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

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 15

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 18

- Views
- 4K

- Replies
- 2

- Views
- 7K

- Last Post

- Replies
- 4

- Views
- 1K