ChatterBank3 mins ago
gravitation
if a hole is made in the earth from top to bottom and a small stone is thrown, what is the status of that stone
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There would need to be a damping force for it to stop oscillating.
Removing the air would do this in theory.so it would oscillate for ever in the same way that the moon orbits us forever.
However forever is a long time and there are other significant assumptions as well as the practicalities. The Earth is not a perfectly spherical uniformly distributed mass
for example.
Removing the air would do this in theory.so it would oscillate for ever in the same way that the moon orbits us forever.
However forever is a long time and there are other significant assumptions as well as the practicalities. The Earth is not a perfectly spherical uniformly distributed mass
for example.
Ummm, can I change my answer to "It wouldn't oscillate forever"?
If the pipe contained a perfect vacuum, and if the pipe coincided with the axis of rotation of the Earth, and if the axis of rotation of the Earth were in a constant direction, and if ...etc. then the stone would accelerate as it approached the centre of the Earth, decellerate as it approached the surface on the other side, and oscillate for ever.
In practice, Coriollis' force will always jam the stone against the side of the tube and it will slowly slide to the centre of the Earth, held back by friction.
As you probably know, Coriollis' force is the apparent force which arises because an object will have a certain angular momentum cause by the rotation of the Earth (except at the poles). If you drop the object so that it is nearer the centre of the Earth, the object tries to keep the same angular momentum, but, because it is now nearer the centre of the Earth, its radius is smaller, so the only way that the product of radius and angular velocity can stay the same is for its angular velocity to increase i.e. although the piece of pipe near the stone rotates once in 24 hours, the stone will rotate in 23 hours (say). Consequently it hits the side of the pipe.
If the pipe contained a perfect vacuum, and if the pipe coincided with the axis of rotation of the Earth, and if the axis of rotation of the Earth were in a constant direction, and if ...etc. then the stone would accelerate as it approached the centre of the Earth, decellerate as it approached the surface on the other side, and oscillate for ever.
In practice, Coriollis' force will always jam the stone against the side of the tube and it will slowly slide to the centre of the Earth, held back by friction.
As you probably know, Coriollis' force is the apparent force which arises because an object will have a certain angular momentum cause by the rotation of the Earth (except at the poles). If you drop the object so that it is nearer the centre of the Earth, the object tries to keep the same angular momentum, but, because it is now nearer the centre of the Earth, its radius is smaller, so the only way that the product of radius and angular velocity can stay the same is for its angular velocity to increase i.e. although the piece of pipe near the stone rotates once in 24 hours, the stone will rotate in 23 hours (say). Consequently it hits the side of the pipe.
In order to stop oscillating the system needs to lose energy. It can do that if there is air in the tube, the energy goes into making the air warmer, or it can scrape against the side of the tube, again it loses energy by making the tube warmer. There are other ways that the stone can lose energy e.g. if there is a magnetic field.
If there is no way for the stone to lose energy what happens is that the sum of its potential energy (due to its position) and its kinetic energy (due to its speed) remains constant. So if you drop it from the surface of the Earth its initial kinetic energy is zero and its initial potential energy is 1 (in some units). As it falls the kinetic energy (speed) increases and the potential energy (height) decreases. At the centre of the Earth its potential energy will be zero and its kinetic energy will be 1. So it has just enough speed to take it up to the surface again, where the potential energy will again be 1 and the kinetic energy (speed) willl be zero. The cycle then repeats exactly the same.
If there is no way for the stone to lose energy what happens is that the sum of its potential energy (due to its position) and its kinetic energy (due to its speed) remains constant. So if you drop it from the surface of the Earth its initial kinetic energy is zero and its initial potential energy is 1 (in some units). As it falls the kinetic energy (speed) increases and the potential energy (height) decreases. At the centre of the Earth its potential energy will be zero and its kinetic energy will be 1. So it has just enough speed to take it up to the surface again, where the potential energy will again be 1 and the kinetic energy (speed) willl be zero. The cycle then repeats exactly the same.
