Assume the moon is a perfect sphere and a straight tunnel has been drilled through the center. How long would it take a 1kg ball dropped from one end of the tunnel to reach the center? Ignore all resistances.
If a second 1kg ball is dropped 10 seconds after the first one, when and where in the tunnel would they first meet?
Idealized Moon Stats:
 Diameter: 3480 kilometers
 Mass: 7.38x10^22 kilograms (uniform density)
If there are no errors, the first ball will reach the center at 1625.00 sec (27.08 min). And they will meet at 3255.00 sec (54.25 min).
PART 1
This relies on Newton's law of gravity that says that the force between two objects is F=GMm/r^2. The negative sign is to show that the force is an attractive one.
Also, one more thing is that when the ball is inside the moon, only the mass that is further in will exert a net force on it. By symmetry, all the mass on the "outside" will cancel each other out.
r=position of ball from center
R=radius of moon
M=mass of moon
M'=mass of moon exerting force on ball
m=mass of ball
First, let's find the mass of the moon exerting a force on the ball in terms of the giving variables.
M'=(density of moon)*(volume of mass exerting force)
=total mass of moon*(volume of mass exerting force)/(total
volume)
=M*r^3/R^3
Because of uniformity, the density is just the total mass divided by total volume. Also, it is assumed that the drilled hole is negligible.
F=GM'm/r^2
=GMmr/R^3
By Newton's Second Law,
F=ma=m*r'' (second derivative of r with respect to time)
=GMmr/R^3
r''+(GM/R^3)*r=0 (second order homogenous ODE)
r''+C^2*r=0 (letting C^2=GM/R^3)
r=C1*cos(C*t)+C2*sin(C*t), where C1, C2 are constants
r'=C*C1*sin(C*t)+C*C2*cos(C*t)
We need two initial initial conditions to solve this. This is given by the fact that the ball is dropped from rest and the initial position. So r'(0)=0 and r(0)=R.
r'(0)=C*C2 > C2=0
r=C1*cos(C*t)
r(0)=C1=R
So r=R*cos(C*t) where C=sqrt(GM/R^3)
G=6.67*10^(11) m^3/(kg*s)
M=7.38*10^22 kg
R=1740000 m
C=0.000967/s
The time for r=0 occurs when cos(C*t)=0. The first time that occurs is C*t=pi/2.
t=pi/(2C)
=1625.00 sec (27.08 min)
PART 2
For the second ball, all the differential equations are the same. It's just that the initial boundary conditions have changed. Instead of solving them again, we can just do a simple translation since everything occurs 10 seconds later.
r1=R*cos(C*t) (ball 1)
r2=R*cos(C*(t10))
r1=r2
cos(C*t)=cos(C*(t10))
cos(C*t)cos(C*(t10))=0
Since cos(A)cos(B)=2*sin((A+B)/2)*sin((AB)/2),
2*sin(C(t5))*sin(5)=0
sin(C*(t5))=0
C*(t5)=n*pi
t=5+n*pi/C
n=0 is meaningless because that occurs before t=10. The first answer occurs at n=1.
t=5+pi/C
=3255.00 sec (54.25 min)
This can also be estimated by geometry and symmetry. The first ball will have to reach the other end of the moon first at a total of 2*1625.00 seconds = 3250.00 seconds. The second ball will be 10 seconds behind it. Because the time range is so small, their speeds are relatively the same. Since they are 10 seconds apart and they travel at about the same speed towards each other, it takes them about 5 seconds to run into each other. That gives 3255.00 seconds, which is a hell of a good estimation.
BUT! It turns out that this is the exact answer. The answer for the first part is pi/2C. And the answer for the second part is 5+pi/C.
Edited on November 26, 2004, 7:33 pm
Edited on December 31, 2004, 1:20 am

Posted by np_rt
on 20041122 19:35:07 