# Homework Help: Newton's Law of Gravitation vs. Gravitational Potential Energy

1. Apr 15, 2007

### AznBoi

I'm confused about the two equations:
$$PE_{G}=G\frac{m1*m2}{r}$$

$$PE_{G}=mgh$$

When do you use either of these equations? Can these two equations be used interchangeably or not? Also, I think I remember reading somewhere that you need to use the 1st equation when the problem involves large masses and the 2nd equation when the problem is just talking about things on Earth.

Does anyone mind explaining when to use each of these equations and when not to? Why does it matter? Thanks.

2. Apr 15, 2007

### da_willem

The potential energy of two (spherical) masses a distance r apart actually contains a minus sign such that the closer the two masses are the less potential energy they have. You might know that you can freely add a constant to this expression as only differences in potential energy are observable.

$$-G \frac{m_1 m_2}{r}$$

When close to the earths surface (r=R+h, with R the radius of the eart and h the height above the earths surface) you can approximate

$$-1/r \approx -\frac{2}{R} + \frac{R+h}{R^2} = -\frac{1}{R} + \frac{h}{R^2}$$

Using this in the above expression for the potential energy (with m_2 = M is the earths mass) we get

$$G \frac{m M h}{R^2}$$

plus some constant that can be ignored (as remarked above). Now, calling the factor

$$g \equiv G \frac{M}{R^2} \approx 9,8 m/s^2$$

we get the second expression for the potential energy near the earth's surface:

$$mgh$$

Last edited: Apr 15, 2007
3. Apr 15, 2007

### da_willem

So concerning your question, the second expression you quote is an approximation of the first (more general) expression. It is a valid approximation only close to the surface of some mass where the approximation I made in the above derivation is applicable.