Gravity as a vector Field

  • Thread starter schaefera
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  • #1
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In my book, it says that gravity can be thought of as a force in the form of this vector:

F= (-GMm)/(x2+y2+z2)*u

where u is a unit vector in the direction from the point to the origin. How would this be represented as a vector field (this is not a homework problem, just me wondering...)?

Is u, the unit vector, able to be split up into u= {(x)i + (y)j + (z)k}/(sqrt( x2+y2+z2), then you can sub in for that and get a vector field of the form

F=(-xGMm)/(((sqrt( x2+y2+z2)3) i + ... and so on?

Because then you can find the divergence of this vector field, but you can't find the divergence of that first equation I listed above because it's not explicitly a vector field...
 

Answers and Replies

  • #2
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You can take the divergence in the first case too. Just use the operator in spherical coordinates.
 
  • #3
HallsofIvy
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Yes,
[tex]F= -\frac{GmM}{x^2+ y^2+ z^2}\vec{u}[/tex]
and
[tex]u= \frac{1}{\sqrt{x^2+ y^2+ z^2}}(x\vec{i}+ y\vec{j}+ z\vec{k})[/itex]
so that
[tex]F= -\frac{GmM}{r^3}(x\vec{i}+ y\vec{j}+ z\vec{k})[/tex]
or
[tex]F= -\frac{GmM}{r^3}\vec{r}[/tex]
 

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