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)
(In reply to
part 1: different answer by Charlie)
The following method is based on a surface acceleration to gravity of 1.6266 m/s^2 (from GM/r^2), and uses numerical integration of the velocity and distance traveled. Acceleration due to gravity is assumed to be proportional to the cube of the distance from the center of the moon. The time taken to reach the moon's center calculated by this method is 1917.6 seconds or 31.96 minutes. An integration interval of 1/16 second was used.
The program is
DEFDBL A-Z
acc0 = 1.6266 ' m/s
intvl = .0625
r = 1
dist = 0 ' meters fallen
vel = 0
DO
r = 1 - dist / (3480000 / 2)
acc = acc0 * r * r * r
newVel = vel + acc * intvl
newDist = dist + intvl * (vel + newVel) / 2
dist = newDist
vel = newVel
t = t + intvl
IF t = INT(t) AND t MOD 100 = 0 THEN
PRINT USING "##### ####.### ########.##"; t; vel; dist
END IF
LOOP UNTIL dist > 3480000 / 2
PRINT USING "#####.## ####.### ########.##"; t; vel; dist
PRINT t / 60
The resulting table of speeds and depths at intervals of 100 seconds and at the end is:
100 161.904 8114.07
200 319.341 32231.44
300 468.210 71695.94
400 605.085 125472.72
500 727.435 192227.72
600 833.726 270423.33
700 923.413 358418.49
800 996.832 454562.88
900 1055.037 557277.25
1000 1099.597 665115.17
1100 1132.399 776804.64
1200 1155.468 891270.63
1300 1170.815 1007641.01
1400 1180.327 1125239.39
1500 1185.685 1243568.29
1600 1188.312 1362285.73
1700 1189.343 1481177.83
1800 1189.609 1600129.13
1900 1189.634 1719091.94
1917.63 1189.634 1740059.24
31.96041666666667
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Posted by Charlie
on 2004-11-22 22:55:03 |
part 1: another method; another answer
