<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/dc3aeca7-9dcc-4a62-baac-cceb7cc207b4/Conservative_forces.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/dc3aeca7-9dcc-4a62-baac-cceb7cc207b4/Conservative_forces.png" width="40px" /> Conservative forces: A force where the work done is path independent

• Properties:

1. Work is change in kinetic energy

   $$ \begin{aligned} W_{AB}&=\int^B_A\vec F\cdot\text d\vec r=m\int^B_A\frac{\text d\vec v}{\text dt} \cdot\text d \vec r \\ &=m\int^B_A\frac{\text d\vec v}{\text dt} \cdot\vec v\, \text dt=\frac 12 m\int^B_A\frac{\text d}{\text dt}(|\vec v|^2)\,\text dt\\ &=\frac 12 m(|\vec v_B|^2-|\vec v_A|^2) = \Delta E_k

   \end{aligned} $$

2. Relation between kinetic energy and potential since: $\text dE_p=-\vec F\cdot \text d\vec r$

   $$ \Delta E_k=\int^B_A\vec F\cdot \text d \vec r=-\int^B_A \text d E_p=E_p(\vec r_A)-E_p(\vec r_B)=-\Delta E_p $$

3. If a force $\vec F$ is conservative then:

   $$ \vec \nabla \times \vec F=0 $$

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Math notations

• take $\vec r=(x,y,z)=(r_1,r_2,r_3)$

$$ \begin{aligned} \vec \nabla&=\hat e_i \frac{\partial}{\partial r_i} \\ \vec \nabla\times\vec v&=\hat e_i\epsilon_{ijk}\frac{\partial v_i}{\partial r_j} &\text{ from: } \;A\times B=\hat e_i\epsilon_{ijk}A_jB_k\\ \vec \nabla\cdot\vec v&=\delta_{ij}\frac{\partial v_i}{\partial r_j}=\frac{\partial v_i}{\partial r_i} &\text{ from: }\; \; \, A\cdot B=A_iB_i\qquad\; \end{aligned} $$

Central forces

<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/ac3492c2-4f41-4695-adda-f092c899161a/Central_forces.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/ac3492c2-4f41-4695-adda-f092c899161a/Central_forces.png" width="40px" /> Central force: a force that only depends on $r=|\vec r|$ and is in the direction $\hat r$, that is:

$$ \vec F(\vec r)=f(r)\,\hat r=f(r)\,\frac{\vec r}r $$

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• Properties if a force is both conservative and a central force

from $\text dU=-\vec F\cdot \text d\vec r$ we can derive

$$ E_p(\vec r)-E_p(\vec r_0)=-\int^{\vec{r}}{\vec r_0}f(r)\,\hat r \cdot\text d\vec r=-\int^r{r_0}\text d r' f(r') $$

$r_0$ is arbitrary and is generally set to $\infin$ since $E_p(\infin)=0$

$-E_p(r)$ is the kinetic energy of a particle at $r$ which was initially at rest at $\infin$

Newtonian gravity

$$ \vec F=-G_N\frac{Mm}{r^2}\,\hat r $$

• From this we can calculate the following:

  • The potential

  $$ E_p(r)=G_NMm\int^r_\infin \frac{\text d r'}{(r')^2}=G_NMm\left[ -\frac{1}{r'} \right]^r_\infin=-G_N\frac{Mm}{r} $$

  • The force

  $$ \vec F=-\vec \nabla E_p=G_N Mm \vec \nabla\left( \frac 1r \right) $$

  • $g$ gravitational field strength, $\Phi$ gravitational potential

  $$ \begin{aligned} g(\vec r)&=\frac{1}{m}\vec F=-\frac{G_NM}{r^2}\,\hat r =-\vec \nabla \Phi \\ \Phi(\vec r)&=\frac{1}{m}E_p=-\frac{G_NM}{r} \end{aligned} $$

$E_p$ demonstration

1. At a distance $h$ above the surface of the earth the gravitational potential is:

$$ \Phi=-\frac{G_NM_E}{R_E+h}=-\frac{G_NM_E}{R_E}\left( 1+\frac{h}{R_E} \right)^{-1} $$

1. Using Taylors expansion we can approximate $(1+x)^{-1}\approx 1-x$ for small $x$

$$ \Phi(R_E+h)\approx-\frac{G_NM}{R_E}\left(1-\frac{h}{R_E}\right) $$

1. For $h\ll R_E$ We can define the change in gravitational potential to be