1.3 Let us start with something familiar, the Sun-Earth system
Fill out the following (feel free to Google):
Radius of Sun (\(R\)) | \(6.96\times10^{8}\) m |
Average distance from Sun to Earth (\(R_{e}\)) | \(1.50\times10^{11}\) m |
Mass of Sun (\(M_{s}\)) | \(1.99\times10^{30}\) kg |
Mass of Earth (\(M_{e}\)) | \(5.97\times10^{24}\) kg |
Gravitational constant (\(G\)) | \(6.67\times10^{-11}\) m3kg-1s-2 |
Linear momentum of Earth (\(p_{e}\)) | \(3\times10^{4}\) kg ms-1 |
(We can get \(p_{e}\) by knowing that the Earth circles around the Sun in about 365 days. How?)
Let \({\bf r}_{s}\) and \({\bf r}_{e}\) be the position vectors of the Sun and Earth respectively. Express \({\bf r}_{se}\) in terms of the two postition vectors. \[ {\bf r}_{se}={\bf r}_{e} - {\bf r}_{s} \]
Newton’s Law of Gravitation: | \(\displaystyle {\bf F}=-\frac{GMm}{r^{2}}\hat{{\bf r}}\) |
Force that the Sun exerts on the Earth: | \(\displaystyle {\bf F}_{se}=-\frac{GM_{s}M_{e}}{r^{2}_{se}}\hat{{\bf r}}_{se}\) |
Force that the Earth exerts on the Sun: | \(\displaystyle {\bf F}_{es}=\frac{GM_{s}M_{e}}{r^{2}_{se}}\hat{{\bf r}}_{se}\) |
Newton’s second law (express in terms of rate of change of momentum) | \(\displaystyle {\bf F}=\frac{d{\bf p}}{dt}\) |
Applying Euler’s scheme for solving differential equation numerically
\[ \begin{align*} d\textbf{p} &= \textbf{F}dt \\ \textbf{p}_{i + 1} &= \textbf{p}_{i} + d\textbf{p} \\ &= \textbf{p}_{i} + \textbf{F}_{i}dt \end{align*} \]
Relationship between momentum and position vector is given by:
\[{\bf p}=m{\displaystyle \frac{d{\bf r}}{dt}}\]
Applying Euler’s scheme for solving differential equation numerically.
\[ \begin{align*} d\textbf{r} &= \frac{\textbf{p}}{m} dt \\ \textbf{r}_{i + 1} &= \textbf{r}_{i} + d\textbf{r} \\ &= \textbf{r}_{i} + \frac{\textbf{p}}{m}dt \end{align*} \]
We will discuss the above briefly in class and do part 3 together as a group.
To be continued…