## SAT Physics Subject Test

**Chapter 4 ****Work, Energy, and Power**

**GRAVITATIONAL POTENTIAL ENERGY**

In the previous discussion, we took the surface of the earth to be our *U* = 0 reference level and assumed that the height, *h*, was small compared with the earth’s radius. In that case, the variation in *g* was negligible, so *g* was thought of as a constant.

**Work and Gravity**

By definition,*U*_{grav} = –*W*_{by grav}

But now we’ll take variations in *g* into account and develop a general equation for gravitational potential energy, one that isn’t restricted to small altitude changes.

Consider an object of mass *m* at a distance *r*_{1} from the center of the earth (or any spherical body) moving by some means to a position *r*_{2}.

How much work did the gravitational force perform during this displacement? The answer is given by the equation.

*W*_{by grav} = *GMm*

Therefore, since ∆*U*_{grav} = –*W*_{by grav}, we get

*U*_{2} –*U*_{1} = –*GMm*

Let’s choose our *U* = 0 reference at infinity. That is, we decide to allow *U*_{2} → 0 as *r*_{2} → ∞. Then this equation becomes

*U* = −

Notice that according to this equation (and our choice of *U* = 0 when *r* = ∞), the gravitational potential energy is always negative. This just means that energy has to be added to bring an object (mass *m*) bound to the gravitational field of *M* to a point very far from *M*, at which *U* = 0.

14. A satellite of mass *m* is in a circular orbit of radius *R* around the earth (radius *r*_{E}, mass *M*). What is its total mechanical energy (where *U*_{grav} is considered zero as *R* approaches infinity)?

Here’s How to Crack It

The mechanical energy, *E*, is the sum of the kinetic energy, *K*, and potential energy, *U*. You can calculate the kinetic energy, since you know that the centripetal force on the satellite is provided by the gravitational attraction of the earth.

Therefore