How to Integrate a Vector Equation in Physics?

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  • #1
Burnstryk

Homework Statement


This is not a homework question, just a general wonderment , how can I integrate the following wrt time?

Homework Equations


[tex]\dot{\textbf{r}}.\ddot{\textbf{r}} +G(m_1 + m_2)\frac{\dot{\textbf{r}}}{r^2} = 0 [/tex]

The Attempt at a Solution


The solution is given as [tex] \frac{1}{2}v^2 - \frac{G(m_1 + m_2)}{r} = C [/tex]

This is to find the vis viva eqn, r represents position vectors

I know that for the second part of the eqn, I integrate dr/dt and get -1/r when considering the r^2, it's the first part I'm having trouble with (scalar product)
 
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  • #2
What do you get if you differentiate ##\dot{\vec r}^2## with respect to time?
 
  • #3
Orodruin said:
What do you get if you differentiate ##\dot{\vec r}^2## with respect to time?

2 * v * a which makes sense, but I was wondering in general if there was a way to integrate a scalar product such as the one shown if the derivative of the solution wasn't so obvious.
 
  • #4
The scalar product is nothing but a sum of terms. You can integrate each term individually.
 
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  • #5
Burnstryk said:
2 * v * a which makes sense, but I was wondering in general if there was a way to integrate a scalar product such as the one shown if the derivative of the solution wasn't so obvious.

Maybe I know what your problem is. What you posted is already part of the solution - the "multiplcation by ##\dot{\textbf{r}}## trick" is a common integration technique for this type of problem. It is used, because it is known that it leads to an equation which can easily be integrated (because ##\dot{\textbf{r}} \cdot \ddot{\textbf{r}} ## is the derivative of ##\frac{1}{2}v^2##). Therefore, the reasoning is not "I have a scalar product, let's see how to integrate it" but "I know that for this kind of problem, a scalar product with ##\dot{\textbf{r}}## will help me solving it".

In your case, you have (btw, if it is supposed to be Newton's law of gravity, then there is a unit vector missing in your equation)

$$\ddot{\textbf{r}} = -\frac{GM}{r^2}\hat{\textbf{r}}$$,

and multiplication gives you

$$\dot{\textbf{r}} \cdot \ddot{\textbf{r}} = -\frac{GM}{r^2} \dot{\textbf{r}} \cdot \hat{\textbf{r}}$$

which can be easily integated with respect to time.
 
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  • #6
Orodruin said:
The scalar product is nothing but a sum of terms. You can integrate each term individually.

bluenoise said:
Maybe I know what your problem is. What you posted is already part of the solution - the "multiplcation by ##\dot{\textbf{r}}## trick" is a common integration technique for this type of problem. It is used, because it is known that it leads to an equation which can easily be integrated (because ##\dot{\textbf{r}} \cdot \ddot{\textbf{r}} ## is the derivative of ##\frac{1}{2}v^2##). Therefore, the reasoning is not "I have a scalar product, let's see how to integrate it" but "I know that for this kind of problem, a scalar product with ##\dot{\textbf{r}}## will help me solving it".

In your case, you have (btw, if it is supposed to be Newton's law of gravity, then there is a unit vector missing in your equation)

$$\ddot{\textbf{r}} = -\frac{GM}{r^2}\hat{\textbf{r}}$$,

and multiplication gives you

$$\dot{\textbf{r}} \cdot \ddot{\textbf{r}} = -\frac{GM}{r^2} \dot{\textbf{r}} \cdot \hat{\textbf{r}}$$

which can be easily integated with respect to time.

Thank you both, I understand now.
 

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