Correctly calculating gravity classically.

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In summary, the conversation discusses the formula for calculating gravity and how it is a good approximation for point masses but falls short when applied to objects with non-zero volume. The speaker proposes a solution by considering the sum of the surface of all spheres at equal distance from a central point, taking into account the mass and center of mass of each surface. This formula is then compared to Newtonian physics and found to give the same results. The speaker also wonders why this formula is not used in classical gravitation and suggests that it may be due to the complexity of developing a formula to describe the distribution of mass in a system. They invite comments and feedback from someone with more knowledge on the subject.
  • #1
Stupid Theory
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I'm someone who's always been a person who has casually interested in math. Recently, after a long break, I've regained a bit of this interest and looked into how gravity is calculated. I'm sure most know the G*m1*m2/r^2, and while a good approximation of gravity, it is not completely correct. We see this when the real world has slight differences from what is expected by this formula and what actually happens. I gave a great deal of thought to why this formula finally falls short and came to an answer. We make an assumption by assuming that calculating by the center of mass has no effect on how gravity interacts. I tested this theory by comparing the results of multiple gravitating forces on each other, and how it differs from if just one object had been gravitating. There is no part in this formula which accounts for the size and mass distribution of objects.

So we come to part two, I thought about how gravity would need to be calculated in order to account for this difference. The answer is a rather interesting and simple idea. At any equal distance from a central point, all mass will have an equal effect of gravity, simply in different vectors. Why not instead of thinking of gravity as a force between two points, think of it as the sum of the surface of all spheres 0 to infinity from your central point. Now, we just need to figure out, what is the gravitational that each sphere's surface will have? Surpringly simple, if the Center of mass for this surface is R(x) distance from the center, and surface is at a radius of x, and the mass of this surface is M(x), the formula is:

F= G*M1*M(x)*R(x)/x^3 applied at vector of M1->R(x) So the total gravitational force on any object for any number of other objects at any distances, would be the sum 0 to infinity of said object. Substitute in points, and this gives the same answer as Newtonian physics as R(x)/x^3 will cancel to 1/x^2 if R(x) and x are equal.

And finally, part 3. Albeit exceedingly more complex, since you need to develop a formula to describe the distribution of mass in the system, why isn't a formula like this used for classical gravitation? It predicts small changes from standard classical models with a quick overview. The biggest one is that objects that get very close have their gravitational pull in a way 'dispersed' into less coherent directions. This would especially be noticeable in objects with highly eccentric orbits. Gravitation in these situations would be slightly less than expected in Newtonian, and approach but never quite reach Newtonian as objects get far. This explains mercury's precession, and explains the pioneer anomaly in classical mechanics.

I'd love to hear comments from someone more knowledgeable, as I have an entirely self taught knowledge on this subject. Please let me know if I need to give examples. Thanks for reading.
 
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  • #2
Stupid Theory said:
I'm someone who's always been a person who has casually interested in math. Recently, after a long break, I've regained a bit of this interest and looked into how gravity is calculated. I'm sure most know the G*m1*m2/r^2, and while a good approximation of gravity, it is not completely correct.
Yes, it is- if it applied correctly. That is the formula for the gravitational force between two point masses. It becomes approximate when you approximate two masses with non-zero volume as by point masses.

We see this when the real world has slight differences from what is expected by this formula and what actually happens. I gave a great deal of thought to why this formula finally falls short and came to an answer. We make an assumption by assuming that calculating by the center of mass has no effect on how gravity interacts.[\quote]
What? On the contrary, the "error" in that approximation is in assuming that the center of mass is the ponly thing that counts!

I tested this theory by comparing the results of multiple gravitating forces on each other, and how it differs from if just one object had been gravitating. There is no part in this formula which accounts for the size and mass distribution of objects.

So we come to part two, I thought about how gravity would need to be calculated in order to account for this difference. The answer is a rather interesting and simple idea. At any equal distance from a central point, all mass will have an equal effect of gravity, simply in different vectors. Why not instead of thinking of gravity as a force between two points, think of it as the sum of the surface of all spheres 0 to infinity from your central point. Now, we just need to figure out, what is the gravitational that each sphere's surface will have? Surpringly simple, if the Center of mass for this surface is R(x) distance from the center, and surface is at a radius of x, and the mass of this surface is M(x), the formula is:

F= G*M1*M(x)*R(x)/x^3 applied at vector of M1->R(x) So the total gravitational force on any object for any number of other objects at any distances, would be the sum 0 to infinity of said object. Substitute in points, and this gives the same answer as Newtonian physics as R(x)/x^3 will cancel to 1/x^2 if R(x) and x are equal.

And finally, part 3. Albeit exceedingly more complex, since you need to develop a formula to describe the distribution of mass in the system, why isn't a formula like this used for classical gravitation?
Why do you think it isn't? It is a standard problem in any Calculus text to use integration to find the gravitational field of a non-point mass.

It predicts small changes from standard classical models with a quick overview. The biggest one is that objects that get very close have their gravitational pull in a way 'dispersed' into less coherent directions. This would especially be noticeable in objects with highly eccentric orbits. Gravitation in these situations would be slightly less than expected in Newtonian, and approach but never quite reach Newtonian as objects get far. This explains mercury's precession, and explains the pioneer anomaly in classical mechanics.

I'd love to hear comments from someone more knowledgeable, as I have an entirely self taught knowledge on this subject. Please let me know if I need to give examples. Thanks for reading.
 
  • #3
ST - Frankly, I do not know what you are talking about. Newtonian theory simply accommodates the behavior of one accumulated mass in the presence of another accumulated mass. It is accurate enough to explain all visible planetary motion except that of Mercury. Mercury is close enough to the sun that relativistic calculations are required.
 
  • #4
What? On the contrary, the "error" in that approximation is in assuming that the center of mass is the ponly thing that counts!
I don't understand. You say on the contrary and seem to be agreeing. I have dyslexia, did I say something that makes no sense. Of course its not the only thing that counts, the formula has many parts, that to me seems to be the only one that was a mistake.


Why do you think it isn't? It is a standard problem in any Calculus text to use integration to find the gravitational field of a non-point mass.
Interesting. I am reading up an shell theorem. Given that apparently my theory already has a name, why are anomalous precession taken as a error in classical physics, instead of information on the distribution of mass within objects?
 
  • #5


I find your exploration of gravity and its calculation to be very interesting. It is clear that you have put a lot of thought and effort into understanding this topic, and your proposed formula does seem to offer some new insights into the classical understanding of gravity.

However, I would caution against dismissing the classical formula for gravity (G*m1*m2/r^2) as just an approximation. While it may not fully capture all aspects of gravity, it has been proven to be accurate in many real-world situations and has been used successfully in countless scientific studies and experiments.

Additionally, your proposed formula may work well in certain scenarios, but it is important to consider its limitations. For example, calculating the gravitational force of objects with complex and irregular mass distributions may prove to be quite challenging using your formula. In contrast, the classical formula can be applied to a wide range of scenarios with relative ease.

Furthermore, as a scientist, it is important to always seek evidence and support for our theories and ideas. While your formula may offer some explanations for observed phenomena, it is necessary to conduct experiments and gather data to validate its accuracy and applicability.

In conclusion, your exploration of gravity and its calculation is commendable and offers some interesting insights. However, it is important to consider the strengths and limitations of both the classical formula and your proposed formula, and to continue conducting research and experiments to further our understanding of this fundamental force.
 

FAQ: Correctly calculating gravity classically.

1. What is the formula for calculating gravity classically?

The formula for calculating gravity classically is F = G(m1m2)/r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the objects.

2. How is classical gravity different from other types of gravity?

Classical gravity, also known as Newtonian gravity, is based on the laws of motion and the inverse-square law, while other types of gravity, such as Einstein's general theory of relativity, take into account the curvature of spacetime.

3. Can classical gravity accurately predict the motion of all objects in the universe?

No, classical gravity is limited in its ability to predict the motion of objects in the universe. It works well for objects with relatively small masses and speeds, but it cannot account for the effects of extreme gravity or high speeds.

4. How is gravity related to mass and distance?

Gravity is directly proportional to the mass of the objects and inversely proportional to the square of the distance between them. This means that the greater the mass of the objects and the closer they are, the stronger the force of gravity between them.

5. What is the significance of the gravitational constant in calculating gravity classically?

The gravitational constant, denoted by G, is a fundamental constant that relates the mass, distance, and force in the formula for classical gravity. It allows us to quantify the strength of the gravitational force between two objects and is essential in accurately calculating this force.

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