 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Moon Experiment (Posted on 2004-11-22) 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)

 See The Solution Submitted by Brian Smith Rating: 2.4000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) part 2 numerical solution | Comment 10 of 22 | The results of a numerical integration for the two bodies:

`interval = 1/4096 sec.t(sec)     vel 1 (m/s)  drop 1 (m)     vel 2 (m/s)  drop 2 (m)      10       16.266        81.33;          0.000         0.00  100      162.406      8126.63;        146.209      6583.55  200      323.295     32430.62;        307.317     29277.53  300      481.164     72684.94;        465.555     67951.31  400      634.539    128513.58;        619.445    122243.62  500      781.986    199395.05;        767.548    191647.32  600      922.129    284667.25;        908.481    275514.12  700     1053.658    383533.64;       1040.929    373060.63  800     1175.345    495070.73;       1163.653    483375.65  900     1286.054    618236.66;       1275.507    605428.75 1000     1384.749    751880.93;       1375.447    738079.84 1100     1470.510    894755.18;       1462.539    880089.82 1200     1542.534   1045524.82;       1535.970   1030132.18 1300     1600.150   1202781.54;       1595.053   1186805.40 1400     1642.819   1365056.38;       1639.237   1348645.98 1500     1670.142   1530833.57;       1668.108   1514142.18 1600     1681.864   1698564.57;       1681.398   1681748.12 1700     1677.877   1866682.61;       1678.983   1849898.19 1800     1658.216   2033617.33;       1660.883   2017021.70 1900     1623.066   2197809.37;       1627.270   2181557.57 2000     1572.755   2357725.04;       1578.456   2341968.86 2100     1507.753   2511870.58;       1514.898   2496757.20 2200     1428.667   2658806.10;       1437.190   2644476.70 2300     1336.237   2797159.09;       1346.056   2783747.52 2400     1231.324   2925637.21;       1242.350   2913268.75 2500     1114.909   3043040.34;       1127.038   3031830.52 2600      988.081   3148271.83;       1001.199   3138325.36 2700      852.022   3240348.72;        866.008   3231758.50 2800      708.005   3318410.91;        722.727   3311257.19 2900      557.375   3381729.23;        572.696   3376078.83 3000      401.538   3429712.23;        417.314   3425617.94 3100      241.950   3461911.70;        258.035   3459411.76 3200       80.102   3478026.87;         96.346   3477144.62 3249.264       -0.000   3480000.32;         16.265   3479919.00 3250       -1.197   3479999.88;         15.069   3479930.52 3251       -2.824   3479997.87;         13.442   3479944.78 3252       -4.450   3479994.23;         11.816   3479957.41 3253       -6.077   3479988.97;         10.189   3479968.41 3254       -7.703   3479982.08;          8.562   3479977.79 3254.264       -8.133   3479979.99;          8.133   3479979.99 54.23773600260417`

By 10 seconds, when the second ball is dropped, the first ball is 81.33 meters down and traveling at 16.266 m/s. At 3249.264 seconds, the first ball has reached the opposite side of the moon and reversed direction. The 0.32 meters of extra travel shown is due either to the finite nature of numerical integration or the inaccurate representation of the assumed surface gravity of the moon, due to a finite accuracy given based on the size and mass of the moon.

At the time that the first ball reverses direction, the balls are 81.32 meters apart, close to the expected 81.33 due to the symmetry with the beginning 81.33 meter head-start the first ball had.

The two balls meet 5 seconds after the first ball reversed direction, as expected.  That takes place 20.33 meters from where the first ball reversed direction, correctedly on the opposite surface of the moon.  This is one quarter the distance that had separated them at the time the first ball reversed direction, as at that time the first ball was traveling slowly, while the second was traveling relatively rapidly (16.265 m/s).

CLS
DEFDBL A-Z
acc0 = 1.62659334126# ' m/s
intvl = .03125 / 128
PRINT "interval = 1/"; LTRIM\$(STR\$(1 / intvl)); " sec."
r = 1
dist = 0    ' meters fallen
vel = 0
dist2 = 0
vel2 = 0

DO
pDiff = dist - dist2
pVel = vel
r = 1 - dist / (3480000 / 2)
acc = acc0 * r
newVel = vel + acc * intvl
newDist = dist + intvl * (vel + newVel) / 2
dist = newDist
vel = newVel
IF drop2 THEN
r2 = 1 - dist2 / (3480000 / 2)
acc2 = acc0 * r2
newVel2 = vel2 + acc2 * intvl
newDist2 = dist2 + intvl * (vel2 + newVel2) / 2
dist2 = newDist2
vel2 = newVel2
END IF
t = t + intvl
IF t = 10 THEN
drop2 = 1
END IF
IF t = INT(t) AND t MOD 100 = 0 OR t = 10 OR vel < 0 AND t = INT(t) THEN
PRINT USING "##### ########.### #########.##"; t; vel; dist;
PRINT USING ";   ########.### #########.##"; vel2; dist2
IF t = 10 THEN
drop2 = 1
END IF
END IF
IF vel * pVel < 0 THEN
PRINT USING "#####.### ########.### #########.##"; t; vel; dist;
PRINT USING ";   ########.### #########.##"; vel2; dist2
END IF
diff = dist - dist2

LOOP UNTIL diff * pDiff < 0
PRINT USING "#####.### ########.### #########.##"; t; vel; dist;
PRINT USING ";   ########.### #########.##"; vel2; dist2

PRINT t / 60

 Posted by Charlie on 2004-11-23 02:33:03 Please log in:
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