MOTIONS OF FREELY MOVING OBJECTS IN A ROTATING SPACE-STATION

MOTIONS OF FREELY MOVING OBJECTS IN A ROTATING SPACE-STATION Physics & Comology Space Technology
	Fig. 1 shows a ball B, hanging in a string, vertically 'above' an observer O. The circumferential speed of the rotating hull is V1 and so B has the speed V2, lower than V1. When the ball is released from the string in the shown position B, in that very moment, no more forces are working on it any longer and it will move along a straight path B-B' (1st Law of Newton). When the ball touches down in B', the arc B-B", that the end of the string describes in the same time and at the same speed V2, must then have the same length as the distance B-B' and for this, the hull has rotated over the angle a₀. So has the position of the observer O', still standing vertically under the string. The length of the arc O-O' is thus longer than that of O-B' and the ball touches down at a distance O'-B' away from the observer. [ a₀ = tana_B => a₀ > a_B ] Relative the observer, the ball fell down along the curved path B"-B', not vertically down, but touching vertically down, as follows from the vectors V2', V2" and V3 (a impact = a bounce). If the bounce was fully elastic, basically the situation of Fig.2 applies, with the difference that the ball does not get the component V1, but V2" instead (all seen relative the observer). In Fig.2 an observer throws a ball straight up with the speed V2. In the very moment the ball loses contact with the observer's hand, no more forces are working on it and it will have the resulting speed V in the shown direction. From geometry follows that its straight path B-B' is shorter than the arc distance O-B' and as it travels it with a higher speed than V1, it will touch down in B', before the observer gets there. Relative the observer the ball describes the shown curve from O' to B' - a juggler from Earth would have to learn his profession again. Depending on what direction the observer is facing, if we say the ball touches down ahead of him in Fig.2, then it does so behind him in Fig 1 (and vice versa). If we combine these two motions, to represent a diver, standing on the edge of a platform over a swimming pool, where he jumps straight up, his body will in principle move as shown in Fig 3. If he's lucky, he may just hit the water, if not, smash on the edge of the pool and get hurt (or worse). If we invert the situation, the diver and the platform facing the opposite direction, he will miss the pool completely and fall backwards on the platform instead. If he would not jump just straight up, but forward as well, he may fall down in the pool along a rather vertical path, or whatever path, depending on how he jumps.
Just imagine Newton would have grown up in a huge rotating space-station, say the size of the Moon, or even the Earth and be unaware of the world outside, his laws of motion would have become totally different. If the same would apply on A. Einstein, gravity and inertia would not have been equivalent for him...... We could thus say that a rotating world is sort of an other universe - Newton was right, rotation IS absolute! (with this statement I have "disqualified" myself for being taken seriously, if it helps?) Backgrounds of Relativity

Fig. 1 shows a ball B, hanging in a string, vertically 'above' an observer O. The circumferential speed of the rotating hull is V1 and so B has the speed V2, lower than V1.

When the ball is released from the string in the shown position B, in that very moment, no more forces are working on it any longer and it will move along a straight path B-B' (1st Law of Newton).

When the ball touches down in B', the arc B-B", that the end of the string describes in the same time and at the same speed V2, must then have the same length as the distance B-B' and for this, the hull has rotated over the angle a₀.

So has the position of the observer O', still standing vertically under the string. The length of the arc O-O' is thus longer than that of O-B' and the ball touches down at a distance O'-B' away from the observer.
[ a₀ = tana_B => a₀ > a_B ]

Relative the observer, the ball fell down along the curved path B"-B', not vertically down, but touching vertically down, as follows from the vectors V2', V2" and V3 (a impact = a bounce). If the bounce was fully elastic, basically the situation of Fig.2 applies, with the difference that the ball does not get the component V1, but V2" instead (all seen relative the observer).

In Fig.2 an observer throws a ball straight up with the speed V2. In the very moment the ball loses contact with the observer's hand, no more forces are working on it and it will have the resulting speed V in the shown direction. From geometry follows that its straight path B-B' is shorter than the arc distance O-B' and as it travels it with a higher speed than V1, it will touch down in B', before the observer gets there. Relative the observer the ball describes the shown curve from O' to B' - a juggler from Earth would have to learn his profession again.

Depending on what direction the observer is facing, if we say the ball touches down ahead of him in Fig.2, then it does so behind him in Fig 1 (and vice versa). If we combine these two motions, to represent a diver, standing on the edge of a platform over a swimming pool, where he jumps straight up, his body will in principle move as shown in Fig 3. If he's lucky, he may just hit the water, if not, smash on the edge of the pool and get hurt (or worse).

If we invert the situation, the diver and the platform facing the opposite direction, he will miss the pool completely and fall backwards on the platform instead. If he would not jump just straight up, but forward as well, he may fall down in the pool along a rather vertical path, or whatever path, depending on how he jumps.