Celestia Forums

Hi

This post is about a question: "Why masses enter gravitation formula by multiplication?"
There is a theoretical reason for this to happen, or it is just the outcome of empirical calculations?

I find this way of representing physical interactions formally similar with the one used in kemical kinetics, where the reagent concentrations are multiply together. In the chemical case a reason for multiplication is quite obviously related to statistical considerations about the density of molecules of each reagent. But in Gravity I cant figure out something similar, or if this could be the case, Newton's formula will mean a calculation of the probability of the event where two masses meet up to happen. Am I wrong?

Thanks for any bit of explanation

taulero

NB: I apologize for my English (ciao form Italy)

You are referring to the calculation of the gravitational force between two bodies of mass M and m, with separation r:

F = GMm/r?

Yes?

Remember that gravity has the property of being a force per mass, measured in N/kg or in the equivalent acceleration, m/s?. All bodies experience the same acceleration in a gravitational field, regardless of their mass (as Galileo, according to legend, observed at the leaning tower in Pisa ).
So GM/r? is the acceleration that the first body induces in the second. We must multiply that by the mass of the second body to calculate the force responsible for that acceleration.

When we calculate orbital periods for the pair of bodies, we are balancing one force per mass (gravity) against another (the centrifugal pseudoforce), so the multiplication disappears and we are left with the more expected sum of the two masses.

Grant

Tanks
for your so clear explanation!
I was a bit afraid posting a sort of naive kind of question, but fotunately it have found someone who understood it, and took the time to answer it.

be good

ciao
from tau (ITA)

That "simple" question is interesting, since it is pretty basic and can be answered at various different levels. . Let me stay at a similar one as Grant used and just add a few further comments...

Firstly, the bodies with position vectors [tex]\vec{x_i}[/tex] and masses [tex]m_i[/tex] are supposed to satisfy Newtons equations of motion. The first thing to do next, is to change variables: along with the relative positions [tex]\vec{x_i} -\vec{x_j},\ i\ne j[/tex], let us introduce the center-of-mass (=barycenter) position vector,

[tex]\vec{R} = \sum_i m_i\vec{x_i}/M[/tex],

where the total mass

[tex]M=\sum_i m_i[/tex]

of the system appears to be concentrated! In fact, one can readily derive from the original equations of the system that the center-of-mass vector satisfies the following Newton equation of motion:

[tex]\frac{d^2\, \vec{R}}{dt^2} =\frac{\vec{F}_{ext}}{M}[/tex]

an expression of Newton's second law, which states that the center-of-mass at [tex]\vec{R}(t)[/tex] moves as though it possessed the total mass of the system and were acted upon by the total external force [tex]\vec{F}_{ext}[/tex]. Consequently, if [tex]\vec{F}_{ext}=0[/tex] then the center-off mass moves with constant velocity, as we all know.

Concerning the acceleration [tex]\frac{d^2\, \vec{R}}{dt^2}[/tex] of the center-of-mass, it's indeed the sum of the masses, NOT their product that matters.

For a simple two-body gravitating system, for example, one encounters besides the total mass [tex]M=m_1+m_2[/tex] (that comes with the center-off-mass motion), the so-called reduced mass,

[tex]\frac{1}{m_{red}} = \frac{1}{m_1} +\frac{1}{m_2},[/tex]

or equivalently

[tex]m_{red} = \frac{m_1\,m_2}{M},[/tex]

such that the product of masses as appearing in the force between the two bodies, factors once more, into the product of the reduced mass and the total mass,

[tex]F =\frac{G m_{red}\, M}{r^2}[/tex]

Fridger

Tanks a lot
I would have never thought all that by myself!
I'm happy I posted here the rather naif question
Peraphs it is looking at planets and stars which makes peoples so gentle?
I guess the answer is "Yessss"

take care

from a sunny day in Italy