# B Question about potential energy

1. Jun 3, 2016

### Xalkias

Let's say we have a mass M at the ground of the earth with speed U going upwards so its energy is E= kinetic. The speed is enough to surpass earth's gravitational field so now it has a speed U2<U and its energy now is E= kinetic+potential. So my question now is: Is this potential energy lost? I mean when it's in the gravitational field its kinetic energy converts to potential and potential to kinetic when back down... So when it exits the field potential can no longer convert to kinetic? If this is the case where does potential energy goes ?

2. Jun 3, 2016

### nasu

It never exits the field. The gravitational field of teh Earth is not limited to a finite region of space.

3. Jun 3, 2016

### Aniruddha@94

Like the post above mine says, the body never escapes the field. The potential energy is $U=-\frac{GM_em}{r}$ . As the distance increases, $U$ tends to zero.
For a body to escape the pull of earth its kinetic energy $K$ must be greater than $U$ i.e., the total energy $E=K+U$ must be positive ( $E$ is constant).
As the distance increases, $U$ goes to zero and $K$ slowly decreases. The total energy $E$ will always remain constant though.

4. Jun 3, 2016

### Xalkias

What about U=mgh ? As far as I know acording to this equation potential energy seems to be positive but also getting bigger as h growing. What's going on?

5. Jun 3, 2016

### Aniruddha@94

$U=mgh$ is only an approximate relation ( where g doesn't change appreciably with distance). For your case, where the body wants to escape the earth's attraction, you cannot assume $g$ to be constant. Also, in $U=mgh$, the potential at the earth's surface is zero and at a very large distance it's infinity. Compare this to $-\frac {GM_em}{r}$

6. Jun 3, 2016

### jbriggs444

Only differences in potential energy are physically meaningful. The actual value at a chosen reference point is irrelevant.

When using $U\, = \, mgh$, one is implicitly assuming a reference point at ground level, a height measured up from ground level and a potential energy of zero at the reference point.

When using $U \, = \, - \frac{GM_em}{r}$, one is assuming a reference point at infinity, a radius measured out from the center of the earth and a potential energy of zero at the reference point.

The difference between potential energy of a mass at a height of 10 meters above the earth's surface and the potential energy energy of a mass at the earth's surface is the same, either way. Both formulas give the same answer for the difference.

7. Jun 3, 2016

### sophiecentaur

What's going on is that the gravitational potential on the Earth's surface is Negative (with respect to infinity). Moving upwards is going in the Positive direction (doing work against gravity). It's like climbing up a mineshaft. The whole shaft has negative height but your height is increasing as you move up. I heard the term "Number Line" on the Radio, this morning. Moving right on the number line is always Increasing; going Up through the gravity field is Increasing your Potential.
-GmM/(Greater R) is more than -GmM/(Small R)

As far as gravity is concerned, the absolute potential Everywhere in the Universe is Negative (until they find something with Negative Mass).