Representing Gravity as a Vector Field

AI Thread Summary
Gravity can be represented as a vector field using the formula F = -GMm/(x²+y²+z²) * u, where u is a unit vector pointing from the point to the origin. The unit vector u can be expressed as u = (x i + y j + z k) / sqrt(x²+y²+z²), allowing for the substitution into the vector field equation. This leads to the vector field representation F = -GmM/(r³) * (x i + y j + z k), where r is the distance from the origin. The divergence of this vector field can be calculated using spherical coordinates, providing insights into the field's behavior. Overall, this discussion clarifies how gravity can be modeled as a vector field and the mathematical implications of such a representation.
schaefera
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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...
 
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You can take the divergence in the first case too. Just use the operator in spherical coordinates.
 
Yes,
F= -\frac{GmM}{x^2+ y^2+ z^2}\vec{u}
and
u= \frac{1}{\sqrt{x^2+ y^2+ z^2}}(x\vec{i}+ y\vec{j}+ z\vec{k})[/itex]<br /> so that <br /> F= -\frac{GmM}{r^3}(x\vec{i}+ y\vec{j}+ z\vec{k})<br /> or<br /> F= -\frac{GmM}{r^3}\vec{r}
 
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