If the initial speed is not too great it is an arc of a sinusoid.Originally Posted by ThePerfectHacker
RonL
Assuming Newtonian gravity (which should suffice) for an object launched from the surface of the Earth we get the equationOriginally Posted by ThePerfectHacker
from energy conservation, where v0 is the initial (launch) speed, M is the mass of the Earth, R is the radius of the Earth, G is the Universal Gravitation constant, m is the mass of the object, and y is the height above the Earth. (Note: ) Or we could use Newton's 2nd:
(Positive downward.)
Neither equation is trivial to solve for y(t). I can't find my notes on how I did it, but I seem to recall solving the energy equation for y(t) at some point in the past. However, I don't recall getting any great inspiration from it either.
-Dan
The more I stare at the energy equation, the more I doubt I solved it. I may have solved a first or second approximation where y is small.
Thank you, though I can't claim credit...most of it came from one of my textbooks.Originally Posted by Quick
I was thinking that if you really wanted to do the project you could still model the equations even if you didn't have the derivations. The quadratic resistance case is a pretty ugly (oxymoron?) equation but easy enough to program on a fancy calculator, or in Excel if you wanted to make a graph.
-Dan
Note that this is the first integral of the Newton's 2nd equation in the previous post where this appears. (5 posts ago)
Solve this equation for v:
v is positive for an object moving upward and negative for moving downward, so we can split the solution into two pieces for upward and downward motion.
For simplicity, take (NOT the acceleration!) and . This equation is then
Take the first approximation to this equation where y is considered to be small. (ie. keep y to first order, y << R.) This equation becomes an approximate linear first order equation:
Letting and we see that
has the solution
(*)
Consider now the upward motion solution. (Use the "+" signs.) Inserting this into equation * we get that and , so
.
Now . Thus .
.
Back-subbing to get to the original variables:
+
+
To get the rest of the motion, solve y(T) for maximum height, then use the solution to the equation with the initial condition . (Or simply consider that the equation for y(t) for the whole motion will be symmetric about t = T.)
As I mentioned previously, it doesn't give me any great inspirations, though it's a nice exercise in applied differential equations.
-Dan
I was just thinking. I did this analysis a few years ago and it never occured to me until now that the solution needs to be forced to be continuous about the time t = T. That could add some interesting complications in solving for the downward motion equation. I'm not going to bother with it. Consider this as a problem "for the interested student."