Newtonian gravitation as a vector

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  • #1
Hi guys! So I am a little congused about newtonian gravitation. I understand the equation to get a scalar for the strength of gravity, but when I plugged it into a vector field calculator (with random values of m and M), all the vectors only pointed in one way. I do understand that r-hat could be positive 1 or negative 1 depending where m and M are, but still, the arrows can only go in 2 directions. I'm not sure if this is a stupid question, but I am a little confused.
 

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  • #2
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What exactly is your question ?
 
  • #3
Nabeshin
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The vector equation for Newtonian gravitation reads:
[tex] \vec{F}_1 = \frac{G m_1 m_2}{|\vec{r}_{12}|^3} \vec{r}_{12} [/tex]

That is, the force on particle one (in a system of 1 and 2) is directed along the vector connecting it to the second particle ([itex] \vec{r}_{12} [/itex]). Given that, what exactly don't you understand?
 
  • #4
The vector equation for Newtonian gravitation reads:
[tex] \vec{F}_1 = \frac{G m_1 m_2}{|\vec{r}_{12}|^3} \vec{r}_{12} [/tex]

That is, the force on particle one (in a system of 1 and 2) is directed along the vector connecting it to the second particle ([itex] \vec{r}_{12} [/itex]). Given that, what exactly don't you understand?
So what would that be in the form of F(x,y,z)= F(x)i+F(y)j+F(z)k (or however you write it) using any random values of m and M. I am actually only 13 and I have only self tought myself some calculus (and I have watched videos), so I may be missing something in my math, but don't you need that kind of equation to get vector (x,y,z)?
 
  • #5
Nabeshin
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So what would that be in the form of F(x,y,z)= F(x)i+F(y)j+F(z)k (or however you write it) using any random values of m and M. I am actually only 13 and I have only self tought myself some calculus (and I have watched videos), so I may be missing something in my math, but don't you need that kind of equation to get vector (x,y,z)?

The vector in my equation [itex] \vec{r}_{12} = ( x_2 - x_1 , y_2 - y_1, z_2 - z_1 ) [/itex] written out in component form. Similarly the vector [itex] \vec{F} = (F_x,F_y, F_z ) [/itex]. Each component of the left hand side and right hand side are equal, which gives you your three equations. Does that make sense? Can you see how you might write that out in terms of your seemingly preferred i,j,k notation?
 
  • #6
The vector in my equation [itex] \vec{r}_{12} = ( x_2 - x_1 , y_2 - y_1, z_2 - z_1 ) [/itex] written out in component form. Similarly the vector [itex] \vec{F} = (F_x,F_y, F_z ) [/itex]. Each component of the left hand side and right hand side are equal, which gives you your three equations. Does that make sense? Can you see how you might write that out in terms of your seemingly preferred i,j,k notation?
I do understand that, however as I pug it into a vector field calculator the vector for the point which m is at does not necessarily point towards M (0,0,0).
 
  • #7
jbriggs444
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I do understand that, however as I pug it into a vector field calculator the vector for the point which m is at does not necessarily point towards M (0,0,0).
What exactly is a "vector field calculator" and what exactly are you pugging into it?
 
  • #8
What exactly is a "vector field calculator" and what exactly are you pugging into it?
I plugged in Newtonian gravitation into an online vector field graphing calculator.
 
  • #9
pbuk
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It is easiest to work in 2 dimensions and set ## m_2 ## at (0,0) so that the position vector of m_1 is ## \vec r = (x,y) ## and ## \vec r_{12} = - \vec r ##. You should then be able to get a 2D vector field plot that looks something like this. I'm not sure that helps you much though, or does it?
 
  • #10
It is easiest to work in 2 dimensions and set ## m_2 ## at (0,0) so that the position vector of m_1 is ## \vec r = (x,y) ## and ## \vec r_{12} = - \vec r ##. You should then be able to get a 2D vector field plot that looks something like this. I'm not sure that helps you much though, or does it?
That is what I did, however the equation you used was not Newtonian gravitation. When I plugged in Newtonian gravitation on that site (using random masses) that is not what it looked like.
 
  • #11
pbuk
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I think you will find that it was (I assumed ## Gm_1m_2 = 1 ## as all that affects is the scale of the arrows). What do you think I got wrong?
 
  • #12
I think you will find that it was (I assumed ## Gm_1m_2 = 1 ## as all that affects is the scale of the arrows). What do you think I got wrong?
Then it would be 1/(x^2+y^2), and it would not be to the 1.5th power.
 
  • #13
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Then it would be 1/(x^2+y^2), and it would not be to the 1.5th power.

the magnitude of the gravitational force is 1/r^2 = 1/(x^2 + y^2). but the x and y components are
- x/r^3 = -x (x^2 + y^2)^(3/2) and -y/r^3 = -y (x^2 + y^2)^(3/2)
 
  • #14
the magnitude of the gravitational force is 1/r^2 = 1/(x^2 + y^2). but the x and y components are
- x/r^3 = -x (x^2 + y^2)^(3/2) and -y/r^3 = -y (x^2 + y^2)^(3/2)
Could you show me the math behind that? If say GMm was 2 would it be -2x/r^3i-2y/r^3j? I think I see what the rule is for this. One final question: how would the formulas for each axis change in 3 dimensional space?
 
  • #15
pbuk
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Then it would be 1/(x^2+y^2), and it would not be to the 1.5th power.

No, note that the magnitude of the gravitational force is ## \frac{Gm_1m_2}{\lvert \vec r \rvert ^2} ## and the unit vector of its direction is ## \hat r = \frac{\vec r}{\lvert \vec r \rvert} ##, so we get
[tex] \vec{F}_1 = \frac{G m_1 m_2}{|\vec{r}|^2} \hat{r} = \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r} [/tex]
To get from this to ## Gm_1m_2(x^2 + y^2)^{-3/2} ## you need to know more about working with exponents in expressions like ## x^{-3} ## and ## \sqrt x = x^{(1/2)} ## than I did when I was 13, have you covered this in your studies yet?
 
  • #16
No, note that the magnitude of the gravitational force is ## \frac{Gm_1m_2}{\lvert \vec r \rvert ^2} ## and the unit vector of its direction is ## \hat r = \frac{\vec r}{\lvert \vec r \rvert} ##, so we get

To get from this to ## Gm_1m_2(x^2 + y^2)^{-3/2} ## you need to know more about working with exponents in expressions like ## x^{-3} ## and ## \sqrt x = x^{(1/2)} ## than I did when I was 13, have you covered this in your studies yet?
Yes, I am actually 3 grades ahead in math besides my personal studies, and I was confused because I didn't know about r^3.
 
  • #17
pbuk
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Hmmm, I'd like to post a small correction and make the derivation clearer:

$$ \vec{F} = -G \frac{m_1m_2}{|\vec{r}|^2} \hat r = -G \frac{m_1m_2}{|\vec{r}|^2} \frac{\vec r}{\lvert \vec r \rvert} = -G \frac{m_1m_2}{|\vec{r}|^3} \vec{r} $$
 

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