Representing Gravity as a Vector Field

Click For Summary
SUMMARY

The discussion centers on representing gravity as a vector field using the equation F = (-GMm)/(x²+y²+z²) * u, where u is a unit vector directed from a 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 formulation of the vector field F = -GmM/(r³) * (x i + y j + z k). The divergence of this vector field can be computed, contrasting with the original equation which is not explicitly a vector field. The operator in spherical coordinates can also be applied to find the divergence.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence
  • Familiarity with gravitational force equations
  • Knowledge of spherical coordinates in mathematics
  • Basic proficiency in physics, particularly classical mechanics
NEXT STEPS
  • Study vector fields in physics, focusing on gravitational fields
  • Learn about the divergence operator in spherical coordinates
  • Explore the implications of gravitational fields in astrophysics
  • Investigate the mathematical derivation of gravitational force equations
USEFUL FOR

Physicists, mathematicians, and students studying classical mechanics or vector calculus who are interested in the mathematical representation of gravitational forces.

schaefera
Messages
208
Reaction score
0
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...
 
Physics news on Phys.org
You can take the divergence in the first case too. Just use the operator in spherical coordinates.
 
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]<br /> so that <br /> [tex]F= -\frac{GmM}{r^3}(x\vec{i}+ y\vec{j}+ z\vec{k})[/tex]<br /> or<br /> [tex]F= -\frac{GmM}{r^3}\vec{r}[/tex][/tex]
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K