# Cannot this result be deduced?

In the physics book I am reading, Mecánica elemental by Juan Roederer, the concept of gravitational mass is introduced by a series of ideal experiments:

Body $$O$$ is fixed at the origin and body $$1$$ is put at different distances $$r, r', r''$$ from it. The gravitational force acting upon body $$1$$ is measured at each position, let $$f_1 (r), f_1 (r'), f_1 (r'')$$ be the magnitudes of these forces. Then body $$1$$ is replaced with body $$2$$, and then $$3$$, one at a time, and the same measurements are made.

From these experiments we find out that the quotient $$\frac {f_2 (r)}{f_1 (r)} = \frac {f_2 (r')}{f_1(r')} = \dots$$ is a constant which will be represented as $$\mu_{21}$$ and defined as gravitational mass of body $$2$$ w.r.t. body $$1$$. And similarly we get $$\frac {f_3 (r)}{f_1 (r)} = \frac {f_3 (r')}{f_1(r')} = \dots = \mu_{31}$$.

In a similar fashion, when bodies $$2$$ and $$3$$ are compared, we get $$\frac {f_3 (r)}{f_2 (r)} = \frac {f_3 (r')}{f_2(r')} = \dots = \mu_{32}$$ and finally, the text says that we experimentally verify $$\mu_{32} = \frac{\mu_{31}}{\mu_{21}}$$ (which seems to me a pretty obvious result) specially mentioning that this result can not (the word "not" is actually emphasized) be deduced from the previous facts. This confuses me very much, because I might get this result from the previous ones with very simple math, so I believe that I am missing something about the physics here.

• Can you type the relevant passage from the textbook into your question? Include the Spanish original and the English translation if you feel confident translating. That would help us to determine what is confusing about the textbook. – Mark H Jun 2 '20 at 5:39
• I've read this in Spanish and I think @Javi has done a good job in translating it. The text doesn't say anything more. – Urb Jun 2 '20 at 17:52

Gravitational mass of body 2 w.r.t. body 1 should be determined from gravitational force of body 2 acting on body 1. The way you wrote it seems that gravitational mass of body 2 w.r.t. body 1 is ratio of forces of two bodies in the same gravitational field. In this case, you are correct that the result $$\mu_{32}=\mu_{31}/\mu_{21}$$ is just consequence of the definition.
In the first case however there is one mass that determines ratio of forces which are produced on different bodies in same gravitational field and another mass that is ratio of forces produced on the same body by gravitational fields of different bodies. The gravitational force acting on body 1 in gravitational field of body 2 could look like this: $$\vec{F}_{12}=\mu_1\nu_2\vec{f}(\vec{r}_1,\vec{r}_2)$$ where the $$\vec{f}$$ is pure function of coordinates of objects and it does not depend on what kind of object we have, $$\mu$$ is gravitational mass of the body in given gravitational field and $$\nu$$ is gravitational charge of the body which produces the gravitational field. In general case this law would not abide to third law of motion, nor would a momentum be conserved. It is actually straightforward exercise to show that there is no way to retain third law of motion if ratio $$\mu/\nu$$ is object dependent (which we need to assume is the case, otherwise we could just redefine one of the mass and be left with just the second). But the third law of motion is empiric fact and you cannot refuse the proposed force just because it does not follow it.
• It looks like the textbook defines the gravitational mass of body $A$ w.r.t. body $B$ as $\mu_{AB} = f_A/f_B$ but, as you mention, both $f_A$ and $f_B$ are forces measured in the gravitational field of a different body which is fixed at the origin. – Javi Jun 3 '20 at 2:03