# If g decreases as h increases, then why does gravitational potential energy increase?

1. Jun 6, 2005

### michaelw

EP = mgh
as an object moves further and further away, the value of g decreases and approaches 0.. if this is true, then wouldnt potential energy approach 0 as well?

According to my book, this is not true, the gravitational potential energy constantly rises the further 'up' you go

If thats true, then do we have MASSIVE amounts of gravitational potential energy when compared to a large star a few light years away? Where did this energy come from? How come it is not being translated to kinetic potential energy and we go flying towards that star, similarly if you raised a ball (gave it GPE) and dropped it, it would fall to the ground with final velocity of sqrt(gh)?

Im confused >_<

2. Jun 6, 2005

### gokugreene

Any specific height you are looking at? We don't go flying towards the star because the gravitational effects from a star light years away is negligible compared to earth's.

3. Jun 6, 2005

### dextercioby

That increase is applied to a negative number.It increases towards 0...

It all makes sense to consider a free body to have 0 gravitostatic potential energy.And that happens at an infinite distance from the source of the field...

Daniel.

4. Jun 7, 2005

### jdavel

michaelw,

Your equation Ep = mgh is an approximation and only works for constant g, that is, near the earth's surface. Also when this equation is used, potential energy is usually set to zero at the earth's surface. So the Ep is positive anywhere above the earth's surface.

For large distances from the earth, where g can't be considered constant, assigning the earth's surface as the location of zero potential energy seems sort of arbitrary, and isn't very convenient. As dextercioby said, it makes sense to assign zero potential at an infinite distance from the center of the the earth. Then objects at any finite distance have negative Ep.

5. Jun 7, 2005

### OlderDan

The fact that g decreases as you go higher does not mean that potential energy decreases. Where you choose to call the zero of potential energy, and whether it is positive or negative is not an issue. The issue is that the change in gravitational potential energy is equal to the work you would have to do to lift and object against the force of gravity. For convenience, let's say that an object at the earth's surface has zero potential energy. To lift that object without acceleration requires a nearly constant force of magnitude mg. If you move it a distance h, then the work you have done is mgh. That is why near the earth's surface, potential energy is PE = mgh relative to any point you choose to call h = 0.

Now suppose you keep lifting higher and higher. You still must apply a force in the direction the object is moving, so you are still doing work against the gravitational force. Therefore the potential energy keeps increasing. As g gets weaker and weaker, all that is happening is that the rate at which potential energy is increasing is getting smaller. In other words, the higher you go the greater the distance you must move the object to achieve the same change in potential energy. For the potential energy to start getting smaller as you go higher, g would have to reverse direction, and that never happens. What does happen is that as h approaches infinity, g approaches zero, and the potential energy approaches a limiting value, but that value is far from zero if the reference level where PE = 0 is on the earth's surface. It is equal to all the work you did to move the object from the surface out to h = infinity.

Because this limit exists, and because you are free to choose the height at which you call the potential energy zero, the zero level is sometimes chosen at h = infinity, or better yet r = infinity where r is the distance from the center of the earth. If you make that choice, at the surface of the earth the potential energy is a hugely negative number. No matter how you define the zero level, the potential energy always increases as you move higher.

From the definition of work it can be shown that for a gravitational field, (or any other potential energy situation) the force on an object depends on the rate of change of the potential energy with distance moved, not the potential energy itself. Near the surface of the earth the difference in potential energy between two elevations is proportional to the difference in height by PE = mgh. The force is the rate at which potential energy changes with height, or the slope of this equation, which is mg. If you want to call gravitational potential energy from earth's gravity zero at the earth's surface, and the gravitational potential energy from some distant star a gazillion, you are free to do so. But that does not change the fact that the rate of change of the earth's potential energy is mg, while that of the distant star is essentially zero. The force of the earth's gravity is far in excess of the force acting on you from any celestial body. You are stuck in what is sometimes called a "potential well" The total potential at the surface of the earth can be enormous if you choose to define it that way, but no matter what you do it is going to increase as you go higher.

A reasonable thing to do is to choose the zero of gravitational potential energy from all those distant objects to be zero at our place in the universe. A more natural and convenient mathematical choice is to call it zero at an infinite distance from an object. With that choice the potential energy is proportional to -1/r, where r is the distance from you to the center of the mass attracting you. That makes the PE from all those other masses out there essentially zero where we are in the universe.

6. Jun 7, 2005

### michaelw

ah
thanks a lot that makes sense :)