A point-particle of mass is launched from the origin of the plane with speed at an angle with respect to the positive axis (with the portion of the velocity upward). Let the position of the particle at any particular time be .
There is another point particle of mass fixed at position . There is no ambient gravity.
What value of (in degrees) maximizes the value of when for the first time after launch?
Details and Assumptions:
1)
2)
3)
Universal gravitational constant
4)
5)
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Non-detailed solution. Will add more if requested.
Consider the particle of mass m at any general time t and its coordinates at that instants are r p = ( x ( t ) , y ( t ) ) = ( x , y ) . Now, the fixed mass is at the point r c = ( x M , y M ) .
The vector joining these two points is: r = r p − r c
The force of gravity experienced by the particle due to the fixed particle is:
F = ∣ r ∣ 3 G M m r
F = m x ¨ i ^ + m y ¨ j ^
Substitute the numbers and simplify to obtain the required expression
By Newton's law of gravitation, two coupled nonlinear differential equations are arrived at. They are:
x ¨ = ( ( 1 + x ) 2 + ( 2 + y ) 2 ) 3 / 2 − ( 1 + x ) y ¨ = ( ( 1 + x ) 2 + ( 2 + y ) 2 ) 3 / 2 − ( 2 + y )
x ( 0 ) = y ( 0 ) = 0 ; x ˙ ( 0 ) = v o cos θ ; y ˙ ( 0 ) = v o sin θ
These equations are solved numerically by varying the angle in each iteration. Now a faster way would have been to run a loop which spits out the optimal angle. However, due to laziness, I took the 'brute-force' approach. I varied the angle in steps of 10 degrees initially and then narrowed it down as I saw a pattern. The following plot illustrates the approach:
One can see that the x coordinate is maximum when θ = 3 0 o and lower when θ = 3 5 o . This allowed me to conclude that the maximising angle lies between 30 and 35 degrees. I continued this process until I narrowed down this interval and arrived at the answer of θ = 3 1 . 7 2 o . This took a while. There are more efficient ways to tackle this problem.