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Physics 3 - Return
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Physics 4 -
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Introduction

Physics 4 - Chapter 1
Motion
Orbital motion

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Second Edition of "Fizyka 3"
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Chapter 1. Basic concepts describing motion 1. Motion - general equations In general 't' will be used as a time variable, $\vec{r}$ as a position vector, and $\hat{r}$ as a unit vector. First, we define two values:
Speed $\vec{v} ≝ \frac{d \vec{r}}{dt}$	(1.1)
and
acceleration $\vec{a} ≝ \frac{d \vec{v}}{dt}$ .	(1.2)
Equation (1.2) can be also presented as
$d \vec{v} = \vec{a} \cdot d t$ - vector	(1.3)
And after multiplying respective sides of (1.1) and (1.3) we obtain:
$\vec{v} \cdot d \vec{v} = \vec{a} (\vec{r}) \cdot d \vec{r}$ - scalar	(1.4)
We can integrate formula (1.4), each side independently - right side along the trajectory from the initial position $\vec{r_{p}}$ to the final position $\vec{r_{k}}$ and the left side from the initial speed $\vec{v_{p}}$ to the final speed $\vec{r_{k}}$ , which results in:
$\frac{v_{k}^{2}}{2} - \frac{v_{p}^{2}}{2} = \int_{\vec{r_{p}}}^{\vec{r_{k}}} \vec{a} (\vec{r}) \cdot d \vec{r}$	(1.5)
We can multiply both sides of this equation by any value. If it just happens that this value is a mass m and if we denote $\vec{F} (\vec{r})$ as a product of $m \cdot \vec{a} (\vec{r})$ thus $\vec{F} (\vec{r}) = m \cdot \vec{a} (\vec{r})$ , we can see that we obtained a known relationship between kinetic energy and work:
$\frac{m \cdot v_{k}^{2}}{2} - \frac{m \cdot v_{p}^{2}}{2} = \int_{\vec{r_{p}}}^{\vec{r_{k}}} m \cdot \vec{a} (\vec{r}) \cdot d \vec{r} = \int_{\vec{r_{p}}}^{\vec{r_{k}}} \vec{F} (\vec{r}) \cdot d \vec{r}$ .	(1.6)
Please note that equation (1.5) represents general relationship and that means that any object (for example a mass m) at initial speed v_p , accelerated at $\vec{a} (\vec{r})$ at the distance of $\vec{r_{k}} - \vec{r_{p}}$ will reach final speed v_k . NOTE: For the purpose of this discussion, 2-nd Newtonian law is applied purely as a mathematical expression of the product: $m \cdot \vec{a} (\vec{r})$ by $\vec{F}$ . Similarly, integrating (1.3) in respect to $d \vec{v}$ and dt results in:
$\vec{v_{k}} - \vec{v_{p}} = \int_{t_{p}}^{t_{k}} \vec{a} (t) \cdot d t$ .	(1.7)
As already mentioned above, this relationship (1.7) applies to any object. Should this object be a mass m, then multiplying both sides by m we just obtained relationship between momentum and driving force.
$m \cdot \vec{v_{k}} - m \cdot \vec{v_{p}} = \int_{t_{p}}^{t_{k}} m \cdot \vec{a} (t) \cdot d t = \int_{t_{p}}^{t_{k}} \vec{F} (t) \cdot d t$	(1.8)
We derived relationships, both (1.6) and (1.8) from definitions of speed and acceleration (1.1) and (1.2), and from relationship between force, mass and acceleration. We have not introduced (defined) such terms as work, energy, and momentum, we just applied these names to expressions which appeared in. We will try to derive other relationships (motion laws) without using terms of work and energy. Why are we doing this? Well, most of the textbooks begin explanations regarding motion and mechanics by introducing terms of work and energy, and laws which rule them, while these terms and laws directly result from definitions (1.1) and (1.2). To derive energy conservation law, the equation (1.5) is written separately i times for each object (mass) m (indexed by i), and then individual equations are summed respectively by their sides.
$\sum_{i} (\frac{m_{i} \cdot v_{ki}^{2}}{2} - \frac{m_{i} \cdot v_{pi}^{2}}{2}) = \sum_{i} \int_{\vec{r_{pi}}}^{\vec{r_{ki}}} m_{i} \cdot \vec{a_{i}} (\vec{r_{i}}) \cdot d \vec{r_{i}} = \sum_{i} \int_{\vec{r_{pi}}}^{\vec{r_{ki}}} \vec{F_{i}} (\vec{r_{i}}) \cdot d \vec{r_{i}}$ .	(1.9)
In other words - apart from the concept of mass, m_i can be treated as coefficients of the linear combination of all equations that make up (1.9). If on the right side the interaction between all pairs is balanced, that is:
$\sum_{i} m_{i} \cdot \int_{\vec{r_{pi}}}^{\vec{r_{ki}}} \vec{a_{i}} (\vec{r_{i}}) \cdot d \vec{r_{i}} = \sum_{i} \int_{\vec{r_{pi}}}^{\vec{r_{ki}}} \vec{F_{i}} (\vec{r_{i}}) \cdot d \vec{r_{i}} = 0$ ,	(1.10)
then the equation (1.9) can be written as:
$\sum_{i} (\frac{m_{i} \cdot v_{ki}^{2}}{2} - \frac{m_{i} \cdot v_{pi}^{2}}{2}) = 0$ .	(1.11)
which is the same as following:
$\sum_{i} \frac{m_{i} \cdot v_{ki}^{2}}{2} = \sum_{i} \frac{m_{i} \cdot v_{pi}^{2}}{2}$	(1.12)
This means that the sum of kinetic energies of separated group of objects is constant. To obtain the law of momentum conservation, equation (1.8) should be written for each object separately and, similarly as before, summed up respectively by sides
$\sum_{i} (m_{i} \cdot \vec{v_{ki}} - m_{i} \cdot \vec{v_{pi}}) = \sum_{i} \int_{t_{p}}^{t_{k}} m_{i} \cdot \vec{a_{i}} (t) \cdot d t = \sum_{i} \int_{t_{p}}^{t_{k}} \vec{F_{i}} (t) \cdot d t$ .	(1.13)
If on the right side the sum of forces equals zero,
$\sum_{i} \int_{t_{p}}^{t_{k}} m_{i} \cdot \vec{a_{i}} (t) \cdot d t = \sum_{i} \int_{t_{p}}^{t_{k}} \vec{F_{i}} (t) \cdot d t = \int_{t_{p}}^{t_{k}} (\sum_{i} \vec{F_{i}} (t)) \cdot d t = 0$	(1.14)
then
$\sum_{i} (m_{i} \cdot \vec{v_{ki}} - m_{i} \cdot \vec{v_{pi}}) = 0$	(1.15)
and finally
$\sum_{i} m_{i} \cdot \vec{v_{ki}} = \sum_{i} m_{i} \cdot \vec{v_{pi}}$ ,	(1.16)
thus, momentum remains constant. It is worth noting an interesting relationship in equation (1.5) – assuming that acceleration would be a function of a position as per formula $\vec{a} (\vec{r}) = - \frac{b}{r^{2}} \cdot \hat{r}$ , where b = constant, and motion would occur from initial speed v_p = 0 and initial position r_p = ∞ to final speed v_k and final position r_k. In this case:
$\frac{v_{k}^{2}}{2} = - \int_{\infty}^{\vec{r_{k}}} \frac{b}{r^{2}} d \vec{r} = \frac{b}{r_{k}}$ .	(1.17)
This would be a speed of the object at the position r (as measured from the central point), which is approaching from infinity to central point (not taking into account any relativistic effects). After multiplying this equation by mass m, this equation also represents a kinetic energy required to leave a potential field such as in given expression.