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I am currently reading the book "universe in a nutshell". I reached the point where Einsteins Equivalence principle is explained. It states, that you can't distinguish if your in a space shuttle that moves at constant acceleration or if you are in a parking space shuttle in a gravitational force field.

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It is then concluded, if the space shuttle has the height of 299.792.458 meters, then a flashlight at the top of the space shuttle would reach the bottom earlier then a second (although light-speed is 299.792.458 meters per second), because the space shuttle is in motion and the bottom moves towards the light.

Because of constant acceleration, the velocity will increase by time, meaning if you send another light from your flashlight, it will arrive even sooner.

Since the same argument works for a ball that I could throw from the top, it would imply to me that just by letting time pass, a ball would fall quicker to the ground. In other words, the gravity on earth increases by each second. I know this can't be right, so what I am getting wrong here?

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I think You are confused about why the ball falls in the first place. The ball does NOT fall "because the space ship is in motion". The ball falls down because it is an acceleration motion as opposed to motion (in a straight line) with uniform velocity.

Consider a spaceship moving "upwards" (in the direction of the roof of your room-like ship) with a constant velocity v relative to the earth . Everything in your ship has the same velocity v so there is no relative motion. Now, when you release that ball from your hand (without giving it an impulse) there is no change in the velocity of the ball and thus the relative velocity between the floor and the ball remains 0. The ball simply floats.

Now, consider a spaceship accelerating at 9.81 ms^-2 upwards due a thrust force acting on it. You hold a ball stationary above a height. Lets see why everything appears stationary from an inertial frame. The ship accelerates at g due to thrust. You accelerate at g due to the normal force exerted on you by the ship. The ball accelerates at g due to frictional force from your hands. Everything has the same acceleration (and velocity) and there is no relative motion.

Now, at some time T, when everything has a velocity V relative to the earth, you release the ball, which causes the friction accelerating it to disappear. As a result, the ball maintains its velocity v whereas everything else continues to experience an increase in velocity ΔV=a Δt relative to the earth and to the ball, causing it to "fall" to the ground.

Coming back to your question, whether or not the ball falls depends on if the ship is accelerating, and the time taken for the ball to fall depends on increase in velocity of the floor relative to the earth, NOT its instantaneous velocity at the point of release

No matter how much time you let pass, the increase in velocity over the course of the fall always remains the same (assuming acceleration doesnt change)

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The speed of light is constant from any reference point, the same can't be said about the ball.

The light pulse will only appear to reach the bottom faster from the point of view of an outside observer. From inside the ship you won't be able to tell the difference in either case. When you send a light pulse from the top it will travel towards the bottom of the space ship at c from the outside perspective. This means that the time it takes to reach the bottom is dependent on how fast the bottom is moving to meet it. The ball however will move towards the bottom of the spaceship at ΔV (the difference between your throwing velocity and the initial velocity of the space ship). Thus the gains you make by waiting will be counteracted by a higher initial velocity of the accelerating spaceship.

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    $\begingroup$ 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 $\endgroup$ Commented Feb 14, 2021 at 14:24
  • $\begingroup$ 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? $\endgroup$ Commented Feb 14, 2021 at 14:29
  • $\begingroup$ 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. $\endgroup$ Commented Feb 14, 2021 at 14:33
  • $\begingroup$ 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. $\endgroup$ Commented Feb 14, 2021 at 14:36
  • $\begingroup$ 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 $\endgroup$ Commented Feb 14, 2021 at 14:37

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