# Work done by the gravitational force

by eoghan
Tags: force, gravitational, work
 P: 181 Hi there! I'd like to calculate the work done by the gravitational force. I know the work is defined by the integration of a 1-form: $$L=\int_\gamma \omega$$ where $$\omega=F_xdx+F_ydy+F_zdz$$ This works fine in cartesian coordinates and I know how to integrate it, but what if I want to use spherical coordinates? Then I'd have: $$\omega=F_rdr+F_{\theta}d{\theta}+F_{\phi}{d\phi}=F_rdr$$ Suppose $$\gamma$$ is a curve defined in spherical coordinates (i.e. $$\vec\gamma=R(t)\hat r+\Theta(t)\hat\theta+\Phi(t)\hat\phi$$), how do I integrate the 1-form along $$\gamma$$?
Mentor
P: 11,038
 Quote by eoghan $$\omega=F_rdr+F_{\theta}d{\theta}+F_{\phi}{d\phi}=F_rdr$$
No, you need to use the line element in spherical coordinates:

$$d \vec l = dr \hat r + r d\theta \hat \theta + r \sin \theta d\phi \hat \phi$$

so that

$$\omega = F_r dr + F_\theta r d\theta + F_\phi r \sin \theta d\phi$$

Now, what are $F_r$, $F_\theta$, and $F_\phi$?
P: 181
 Quote by jtbell No, you need to use the line element in spherical coordinates: $$d \vec l = dr \hat r + r d\theta \hat \theta + r \sin \theta d\phi \hat \phi$$ so that $$\omega = F_r dr + F_\theta r d\theta + F_\phi r \sin \theta d\phi$$ Now, what are $F_r$, $F_\theta$, and $F_\phi$?
Then the integral is like this?

$$\int_\gamma \omega = \int_{t_0}^{t_1} \vec F \cdot\frac{d\vec l}{dt}dt=\int_{t_0}^{t_1} \left( F_r\frac{dr}{dt}+F_\theta r \frac{d\theta}{dt}+F_\phi rsin\theta\frac{d\phi}{dt}\right)dt$$

P: 112

## Work done by the gravitational force

Wouldn't the work done when moving between two points in a gravitational field just be the difference between the potential energies at those two points? You'd really only need to worry about the up direction...or r in spherical polar coordinates...if the coordinate origin is the earth's center.

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