Work done by the gravitational force

  • Thread starter eoghan
  • Start date
200
1
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:
[tex]L=\int_\gamma \omega[/tex]
where
[tex]\omega=F_xdx+F_ydy+F_zdz[/tex]

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:
[tex]\omega=F_rdr+F_{\theta}d{\theta}+F_{\phi}{d\phi}=F_rdr[/tex]
Suppose [tex]\gamma[/tex] is a curve defined in spherical coordinates (i.e. [tex]\vec\gamma=R(t)\hat r+\Theta(t)\hat\theta+\Phi(t)\hat\phi[/tex]),
how do I integrate the 1-form along [tex]\gamma[/tex]?
 

jtbell

Mentor
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[tex]\omega=F_rdr+F_{\theta}d{\theta}+F_{\phi}{d\phi}=F_rdr[/tex]
No, you need to use the line element in spherical coordinates:

[tex]d \vec l = dr \hat r + r d\theta \hat \theta + r \sin \theta d\phi \hat \phi[/tex]

so that

[tex]\omega = F_r dr + F_\theta r d\theta + F_\phi r \sin \theta d\phi[/tex]

Now, what are [itex]F_r[/itex], [itex]F_\theta[/itex], and [itex]F_\phi[/itex]?
 
Last edited:
200
1
No, you need to use the line element in spherical coordinates:

[tex]d \vec l = dr \hat r + r d\theta \hat \theta + r \sin \theta d\phi \hat \phi[/tex]

so that

[tex]\omega = F_r dr + F_\theta r d\theta + F_\phi r \sin \theta d\phi[/tex]

Now, what are [itex]F_r[/itex], [itex]F_\theta[/itex], and [itex]F_\phi[/itex]?
Then the integral is like this?

[tex]\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[/tex]
 
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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