Timeline for Why do objects not fall faster if you wait?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2021 at 14:49 | comment | added | CookieNinja | "Yes. That wouldn't be true in the spaceship frame and that is point of confusion, not an implication of an outside frame. Because if it were, OP wouldn't be asking why the same isn't true for the ball " - I didn't draw the same conclusion since he mentions that that the light would reach the bottom faster than 1 sec which also wouldn't be true for an inside observer. Everything points to a misunderstanding about reference point. | |
| Feb 14, 2021 at 14:46 | review | Low quality answers | |||
| Feb 15, 2021 at 9:30 | |||||
| Feb 14, 2021 at 14:43 | comment | added | CookieNinja | "you states that ΔV is the difference between the throwing velocity and the intial velocity of the ship." - I was trying to point out that if the ship is moving faster then the ball also be moving faster in the same direction meaning no difference in outcome. | |
| Feb 14, 2021 at 14:42 | comment | added | Vulgar Mechanick | Yes. That wouldn't be true in the spaceship frame and that is point of confusion, not an implication of an outside frame. Because if it were, OP wouldn't be asking why the same isn't true for the ball | |
| Feb 14, 2021 at 14:39 | comment | added | CookieNinja | "Because of constant acceleration, the velocity will increase by time, meaning if you send another light from your flashlight, it will arrive even sooner." This sentence in OPs question implies an outside observer. Because it wouldn't be true for inside observer in case of constant acceleration. | |
| Feb 14, 2021 at 14:39 | comment | added | Vulgar Mechanick | you states that ΔV is the difference between the throwing velocity and the intial velocity of the ship. The initial velocity of the ship has no place in the expression and inronically feeds into OP's confusion, instead of resolving it | |
| Feb 14, 2021 at 14:37 | comment | added | Vulgar Mechanick | The question is clearly from inside the spaceship. Secondly, the OP's misconception lies in the phyics of non inertial frames. Bringing special relativity into it is wrong and extremely misleading | |
| Feb 14, 2021 at 14:36 | comment | added | CookieNinja | ΔV could be 0 (relative to the ship) if you just release the ball but if you were to throw the ball towards the bottom it wouldn't be. | |
| Feb 14, 2021 at 14:33 | comment | added | CookieNinja | From inside the spaceship you won't be able to tell the difference between the first light pulse and the second one assuming constant acceleration. "If a ship is moving with any constant velocity in any direction, the time for the photon to reach the bottom will ALWAYS be 1 second" - from inside the ship yes. From the outside the light moving in the opposite direction of the ship will have to be constant and meet the bottom faster than 1 sec because the bottom is coming to meet it. | |
| Feb 14, 2021 at 14:29 | comment | added | Vulgar Mechanick | The expression for ΔV is also incorrect. the ball already has the same velocity as the floor at the point of release. And velocities are relative. What is your "throwing velocity" or "initial velocity of spaceship" relative to? | |
| Feb 14, 2021 at 14:25 | review | First posts | |||
| Feb 14, 2021 at 14:42 | |||||
| Feb 14, 2021 at 14:24 | comment | added | Vulgar Mechanick | this question is extremely misleading. OP's question has nothing to do with the invariance of the speed of light. "the time it takes to reach the bottom is dependent on how fast the bottom is moving to meet it." This is incorrect. If a ship is moving with any constant velocity in any direction, the time for the photon to reach the bottom will ALWAYS be 1 second | |
| Feb 14, 2021 at 14:17 | history | answered | CookieNinja | CC BY-SA 4.0 |