8 ms·

In truth, if both experiments are done independently, (so not to affect each other), and we measure how long it takes the two massive objects to hit, and both o

In truth, if both experiments are done independently, (so not to affect each other), and we measure how long it takes the two massive objects to hit, and both objects are allowed to move each other, then the pull from a feather moves the earth slightly less than the pull the bowling ball has on the earth, so technically they will be slightly different times. But that difference is not really measurable and probably barely calculable - though I'm sure someone on here may do that.

crdrost 5y ago
Physicisty type here! It's actually easy to calculate because it's so small. If you dropped them side-by-side the difference would require factoring in the curvature of the Earth and would really be way below anything you might measure... If you drop them in series then yes that does have another effect... the actual acceleration should be like g(1+m/M) instead of just g, so if a ball is 0.6 kg and Earth is 6e24 kg the fall time goes like √(2D/g) and so for a 10s (500 m) freefall in vacuum it changes the time by 0.5 yoctoseconds? Way smaller than any clock accuracy as far as I know... No the real inaccuracy is that the drag force in air will scale like the square of the velocity, m dv/dt = m g – ½ ρ A v² c Where c is a dimensionless geometric constant called the Drag Coefficient, you can handwave that for these rough sphere surfaces it's maybe 0.5 or so, ρ is the density of air which of course gets lower at high elevations, at sea level it’s 1.22 kg/m³, but decreases like 0.13 kg/m³/km or so. A is the cross-sectional area of the ball π r². If you pretend that these variables are fixed instead of changing then this is a soluble equation, defining U = √{2mg/(ρ A v² c)}, dv/dt = g – g (v/U)² v(t) = U tanh(g t/U) U is the terminal velocity. Note that for small times tanh( gt/U) ≈ gt/U and v(t) ≈ gt becomes independent of U and hence m... So that's my pet peeve: The two things on Earth should fall side-by-side in tandem at first, then slowly the tennis ball should lag behind! That's not what we see in this simulation, unfortunately.
- crdrost 5y ago
  Bonus points, I tried to do these calcs and the basketball likely has a terminal velocity ≈ 19.7 ± 0.3 m/s whereas the tennis ball likely has a terminal velocity ≈ 22.7 ± 0.4 m/s... As air resistance kicks in the tennis ball should pull ahead, not lag behind! GDoc: https://docs.google.com/spreadsheets/d/1AnUeQ0RR7a5GA6Nrtef7GCfqMwTv04ZLsH1_DWigVRo/edit?usp=drivesdk https://docs.google.com/spreadsheets/d/1AnUeQ0RR7a5GA6Nrtef7...
  - me_me_me 5y ago
    Extra credit if you consider object spinning or not to determine impact of Magnus effect

crdrost 5y ago

Physicisty type here! It's actually easy to calculate because it's so small. If you dropped them side-by-side the difference would require factoring in the curvature of the Earth and would really be way below anything you might measure... If you drop them in series then yes that does have another effect... the actual acceleration should be like g(1+m/M) instead of just g, so if a ball is 0.6 kg and Earth is 6e24 kg the fall time goes like √(2D/g) and so for a 10s (500 m) freefall in vacuum it changes the time by 0.5 yoctoseconds? Way smaller than any clock accuracy as far as I know... No the real inaccuracy is that the drag force in air will scale like the square of the velocity, m dv/dt = m g – ½ ρ A v² c Where c is a dimensionless geometric constant called the Drag Coefficient, you can handwave that for these rough sphere surfaces it's maybe 0.5 or so, ρ is the density of air which of course gets lower at high elevations, at sea level it’s 1.22 kg/m³, but decreases like 0.13 kg/m³/km or so. A is the cross-sectional area of the ball π r². If you pretend that these variables are fixed instead of changing then this is a soluble equation, defining U = √{2mg/(ρ A v² c)}, dv/dt = g – g (v/U)² v(t) = U tanh(g t/U) U is the terminal velocity. Note that for small times tanh( gt/U) ≈ gt/U and v(t) ≈ gt becomes independent of U and hence m... So that's my pet peeve: The two things on Earth should fall side-by-side in tandem at first, then slowly the tennis ball should lag behind! That's not what we see in this simulation, unfortunately.

crdrost 5y ago
Bonus points, I tried to do these calcs and the basketball likely has a terminal velocity ≈ 19.7 ± 0.3 m/s whereas the tennis ball likely has a terminal velocity ≈ 22.7 ± 0.4 m/s... As air resistance kicks in the tennis ball should pull ahead, not lag behind! GDoc: https://docs.google.com/spreadsheets/d/1AnUeQ0RR7a5GA6Nrtef7GCfqMwTv04ZLsH1_DWigVRo/edit?usp=drivesdk https://docs.google.com/spreadsheets/d/1AnUeQ0RR7a5GA6Nrtef7...
- me_me_me 5y ago
  Extra credit if you consider object spinning or not to determine impact of Magnus effect