Let's set a reference point very far away from the Earth which holds unmoved.
If I am at the equator and the Earth starts to accelerate to the escaping speed on the surface, my guess is I will be shaken off the Earth. My relative speed to the materials that make the Earth is zero near me and big to the materials on the other side of the Earth thus a net relative speed.
If I am at the north pole and the earth rotates faster then I should stay feeling the same gravitational force. My relative speed to the materials of Earth is zero cause they cancel each other at different locations.
If I speed up and shoot off the earth, my relative speed to the earth is big as Newton has told us.
Now what happens if I jump up into the air and the Earth starts rotating faster and faster? My guess is I will fall back to the ground and hit hard. Again my relative speed to the materials that made Earth is zero in this case because they cancel each other again.
My question is, does the effective force I feel from the materials that make Earth has any correlation with their relative speed to me or not? More particularly, relative angular momentum and gravity.
All these are standard exercises in Newtonian mechanics and are addressed most simply by writing down kinetic and potential energy, in some convenient reference frame, deducing the action and then performing the desired change of coordinates.
Jixin Chen
The gravitational force is centered on the Earth's center and is independent of the Earth rotation.
JES
The motion of an object in a central potential (of which the Newtonian potential is an example) is another standard exercise, which leads to the expression of how the angular momentum of the object affects its motion...Cf. for instance https://people.umass.edu/~bvs/Two_body.pdf
These aren't research issues anymore... They're the prerequisite for research.
Stellan Gustafsson and Stam Nicolis ,
Thank you both for following the discussion and referring to the references.
My question was when one of the objects is rotating during gravitational attraction.
I have no problem integrating the mass of the body and reaching the conclusion that it is equivalent to all materials located at the center of the mass when staying still. My guess is this is also true for Einstein's general relativity.
However, if the object has a size and carries angular momentum, then each part of the object will carry very different effective gravitational attraction to the other object due to the centrifugal repulsion force. If we integrate all parts now the effective force between the two bodies no longer follows the equation we have derived from staying still. This is because the centrifugal force is independent of the direction of the rotation and thus cannot cancel each other from different parts.
Is this a conventional/conceptional/logic error or this is something real? If it is a true effect, we may update the calculation of the galaxy with a rotation speed, eliminating dark matter and explaining why some galaxies have and some others have no dark matter. Or develop a device to provide anti-gravity lift. Such a device still obeys energy conservation but no need to push with explosives.
It is relatively easy to test the speculation in astronomy. If a galaxy is thin, it is likely the core spins faster thus more dark matter in the galaxy although my speculation is less effective gravity.
A gyroscope may work as an anti-gravity device.
The centrifugal force is mv^2/r and at 11200 m/s on Earth's surface, it equals gravitational pulling force. So the antigravity force is v^2/11200^2 of gravitational force at rest.
Thus if we spin up a gyroscope of 1 m radius at 2000 cycle/s, the edge will reach 12560 m/s and be ready to take off the ground. If we place it horizontally, then all parts are escaping gravity and we may have a UFO. But 2000 cycle/s is very difficult to reach with our current technology. A very fast gyroscope say 20000 cycles per minute and 5 cm diameter will reach an edge speed of only 100 m/s providing 100^2/11200^2 = 0.00007 force.
If we build a space ship with faster ones, we may use solar gravity to realize interstellar traveling.
Dear Jixin Chen ,
A high velocity gyroscope is not an antigravity device, although it might have some interesting effects.
What Dark Matter is concerned, it doesn't exist IMO. I believe that Newton's constant G is not constant but depends on the geometry of the source.
So, a very simple experiment can demonstrate this. You can use a simple torsion balance and calculate the constant G with different geometries, especially a flat ellipse, to simulate a galaxy.
JES
Stellan Gustafsson
If the weight of the gyroscope is the same before and after spin, let’s analyze by part. At the edge of the disk the speed reach 11 km/s about 100x bullet speed, if a small piece fall, it will shoot straight and orbital the Earth, ie during its fly it will feel zero force at any time including the time it just fall off the edge. So my guess is it is also free of force before falling off the edge.
Thus all the edge pieces are zero gravity.
getting into the center the speed gets smaller but there is no reason it feels more gravity. Let’s assume we designed one with no center just a ring to be easy analysis.
So the total spinning gyroscope is antigravity and if one let it go, it becomes zero gravity on the Earth surface. Unless some unknown force emerges and keep the same weight before spin.
Dear Jixin Chen
Sorry, but you are wrong, this is not antigravity! A piece escaping from your wheel will fly horizontally so if the velocity is sufficient, it can fly away from the Earth escaping from the gravitational pull but, this is normal kinematic.
In fact, this way of thinking is the same as for the Isaac Newton's development of gravitation (although his was thinking on a canon ball).
I keep thinking and coming up with a reason that a gyroscope may not be anti-gravity. The part shooting off the surface is following a curved trajectory following the circumstance of the earth rather than a straight line in space. Thus when it force to turn, it is converting from one curved trajectory to another, which require force to push it down.
Another one is the spinning parts can be replaced with a symmetry argument that it is replaced by its neighbor such that nothing is changed. So Although spins it can be treated as if still.
- Badges
- Science topic
- Similar topics
- Correction