PSM5

I started wondering why we don't use an intrinsic gravity number in which we can say this planet has a mass of 'x' kg therefore it has an intrinsic gravitational pull of 'x' newtons, similarly to voltage in a battery.

Are you familiar with the standard gravitational parameter? It doesn't have units of force, but it is a direct measure of a body's "gravitational charge".

As juan noted, it also adjusts the value of the strength of the gravitational field

Yes, I was agreeing with him on that point.

If a body A exerts a force on body B then its the active gravitational mass which is acting by A and its the passive gravitational mass that is being acted on B. Therefore we let

 

M = active gravitational mass of body A

m = passive gravitational mass of bdy B

 

The force on B due to A is then F = GMm/r²

 

The gravitational constant G then determines the strength of the force. M and m are found to be equal to the inertial mass of body A and B respectively.

I agree with your post. However, I would just like to add that the use of the term passive gravitational mass in the universal law of gravitation is just a convention. It is not necessary. The split of gravitational mass into it's current two forms, active and passive, did not even exist before 1957. For example, when you apply Newton's second law to the universal law of gravitation to get the acceleration of m_p, you are canceling out the passive gravitational mass m_p in the universal law of gravitation with the inertial mass m_i in Newton's second law. If they were not the same then you should not be able to do this. Having three different types of mass can be quite confusing.

I simply am having trouble understanding why the main variables involved in these processes simply aren't enough to satisfy their respective equations, why is a proportionality constant so necessary?

It's not really necessary. It just makes the units come out right. If units are not important for what you're trying to show then you can just set it's value to 1 and leave it out. An example is this wikipedia page showing the proportional equivalence of the three types of mass:

http://en.wikipedia.org/wiki/Equivalence_principle#Active.2C_passive.2C_and_inertial_masses

As juanrga pointed out, there is another way to look at this. However, I'm not sure if this would be an academically correct way of looking at it. And that way is to consider the value of G to be a constant of proportionallity between active gravitational mass and inertial mass. For example, if two universes (A and B) were identical in every way except one had a higher or lower measured value of G than the other, then the gravitational force between two bodies in A and the gravitational force between two bodies in B would be different, even though the inertial masses of all four bodies were the same. In other words, G determines how strong the gravitaional field is for any particular unit of inertial mass.

I would think that questions about gravity on a science forum should probably stick with the technically correct answers, even if that means providing a disclaimer that it's a miniscule difference.

Both of Janus's scenarios are technically correct.

In common language usage, does anyone make the distinction?

I will try.

The first is called the universality of free fall. It is the acceleration of either body (the earth or the falling object) relative to their common center of mass.

[math]A_{m1}=\frac{Gm2}{r^2}[/math]

[math]A_{m2}=-\frac{Gm1}{r^2}[/math]

The second is called the relative acceleration. It is the acceleration of either body relative to the other.

[math]A_{rel}=\frac{G(m1+m2)}{r^2}[/math]

The difference between the two scenarios is only in the frame of reference. The time to impact will be the same either way. I think the reason Janus declaired the first scenario to be a more technical usage is because the frame of reference for the first scenario is inertial, the second is not.

Okay, I think I'm understanding your thought experiment now. I'm just not sure how it relates to the equivalence principle proper. Your OP claim that there is a difference between the acceleration of an object in free fall and an accelerating rocket is true for macroscopic objects. From the same Wikipedia page that you linked:

So the original equivalence principle, as described by Einstein, concluded that free-fall and inertial motion were physically equivalent. This form of the equivalence principle can be stated as follows. An observer in a windowless room cannot distinguish between being on the surface of the Earth, and being in a spaceship in deep space accelerating at 1g. This is not strictly true, because massive bodies give rise to tidal effects (caused by variations in the strength and direction of the gravitational field) which are absent from an accelerating spaceship in deep space.

I have read some text that claim the equivalence principle applies only to point masses. But that was not part of Einstein's original.

Let's suppose the rocket is rotating.

How could it manage to get an acceleration corresponding to a force from the outside to the inside (centripetal)?

Take in mind that the answer must correspond to the Earth's hypothetical movement' date=' but also to any material body, since all material bodies are the source of some gravitational field.[/quote']

I wasn't trying to answer a question because I do not completely understand what your point is. But I did find your post interesting and that is why I commented. A rotating accelerating rocket will have a force that is opposite from that of gravity. But they do share something in common. The direction of their forces both converge at a center.Just thought it was an interesting observation.

 

So, if I take a massive body, and I blow it in an accelerated way, I will recreate a kind of gravity.

That's the part I don't understand.

The equivalence principle applies only to point masses.

What if the accelerating rocket were rotating? What would be it's direction? The milky way galaxy rotates, so how can you know that the rocket is not rotating?