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- Thread starter Isaac0427
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What exactly is your question ?

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Nabeshin

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[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

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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)?

[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?

- #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

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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).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?

- #7

jbriggs444

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What exactly is a "vector field calculator" and what exactly are you pugging into it?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).

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I plugged in Newtonian gravitation into an online vector field graphing calculator.What exactly is a "vector field calculator" and what exactly are you pugging into it?

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pbuk

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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.

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pbuk

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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

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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?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)

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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

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?[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]

- #16

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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.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?

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pbuk

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

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