Please refer to the previous posts in this series if this post is to make any sense.
Believe it or not, I made a small error in the equations from the last post. I used kilometers instead of meters in part of the equation for the force due to gravity. The two equations for the thrust phase and the coasting phase are:\[ \begin{array}{l} {{F}{=}\hspace{0.33em}{2}{,}{800}{,}{000}\hspace{0.33em}{-}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400000}{{6400000}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}}\\ {{=}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{a}} \end{array} \]
and \[ \begin{array}{c} {{F}{=}{-}{9}{.}{8}{\left({\frac{6400000}{{6400000}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{a}} \end{array} \]
Now as mentioned before, these are rather difficult to solve. However, first class rocket scientists rarely solve equations by hand. They resort to numerical methods to solve them. Numerical methods means that they enter the equations into a program and then let the computer solve them.
I have done that and have plotted the results. Below is a plot of the distance the rocket is after t seconds:
Looks like we sent the rocket into space with quite a kick! At 270 seconds, the thrust stops and the rocket keeps going with no sign of slowing down. Its velocity when the thrust stops is greater than the escape velocity at that height. Escape velocity is the speed an object needs to break free of the earth’s gravity. The rocket never falls back to earth.
I also plotted the velocity of the rocket. See how the velocity remains relatively constant after thrust cutoff because gravity is too weak to slow it down:
So that’s all I will say about Newton’s laws, at least for this series of posts. Next week, what’s the deal with this clock?
