Why is the Gravitational Potential Energy formula negative?

Hey guys

So if I lift a book from the ground it gains potential energy. I do work on the book and when I release it the potential energy that I gave the book pulls the book back down to Earth.

However I was looking at a potential energy formula for planets which was

V = -GMm/r

But this doesn't make sense because a) why is potential energy negative and b) why does the absolute value of it get smaller with increasing r? The potential energy should increase with increasing r (it does, but because I don't understand the negative sign I'm thinking of its absolute value).

Cheers

the potential energy of electric charges works exactly as you think it should. This is because like charges repel.

the potential energy of gravitational fields is considered to be 'negative energy' because masses attract rather than repel each other. Otherwise the total energy of the system would not be constant.

Your choice of a zero really makes no difference during calculations since you take the change, the reason that the formula is negative is because we take the zero of potential energy at infinity. It might seem odd at first but potential energy is calculated as the work in going from your zero to your end point (the point who's potential you are trying to find). If you want the zero to be at the centre of the object you're measuring against (planet, star etc) you get a division by zero, so instead we take the zero as infinity as a matter of convenience :)

When the acceleration due to gravity is approximately constant, we take any convenient zero reference for the gravitational PE and we can use mgh for the change in gravPE. For example when I lift a book one usually takes the original position of the book to be the zero reference level for gravPE. Then as the original poster said, the work done in lifting the book will be the increase in gravPE in book.

But when the acceleration due to gravity is varying appreciably one must use
gravPE = -GMm/r.

As I lift a book the work done against the pull of gravity will increase its gravPE. Since r will be increasing, then GMm/r will decrease. But increasing r will INCREASE this PE since the gravPE is negative.

See this

If the direction pointing from the ground is positive(it usually is), then the direction of gravitational force is negative(they're opposite, ain't they?)
..(+)..........(-).....
.........................
...Λ............|.......
...|............|........
...|............|........
...|............|........
__|_______V_____

For this system
That is V = ∫Fdx = ∫m(-9.8)dx = -9.8mx [integrate over upper direction:x>o]----->we have negative potential.

It can also be this(if you want):

..(-)..........(+).....
.........................
...Λ............|.......
...|............|........
...|............|........
...|............|........
__|_______V_____

For this system
That is V = ∫Fdx = ∫m(9.8)dx = 9.8mx [integrate over upper direction:x<o]----->we also have negative potential.

Last edited: