How to Integrate a Vector Equation in Physics?

[Burnstryk]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

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)

 

[Orodruin]

What do you get if you differentiate $\dot{\vec r}^2$ with respect to time?

 

[Burnstryk]

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.

 

[Orodruin]

The scalar product is nothing but a sum of terms. You can integrate each term individually.

 

[bluenoise]

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.

 

[Burnstryk]

The scalar product is nothing but a sum of terms. You can integrate each term individually.

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.