How Do I Solve for Mass Using the Proportional Relationship in Astronomy?

  • Thread starter Thread starter Calpalned
  • Start date Start date
  • Tags Tags
    Proportionality
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Calpalned
Messages
297
Reaction score
6

Homework Statement


In an astronomy problem, I am given ##P^2 \propto \frac {a^3}{M_s}## and I must solve for M. How do I use an equation like this? Is this not the same as ##P^2 = \frac {a^3}{M_s}##

Homework Equations


N/A

The Attempt at a Solution


My attempt is to treat the given equation as ##P^2 = \frac {a^3}{M_s}## and solve for ##M_s = \frac{a^3}{P^2}##
 
on Phys.org
With ##
P^2 \propto \frac {a^3}{M_s}## you have ##
M_s \propto \frac {a^3}{P^2}## so if you know M at some P and a you can calculate (solve for) M when given M for some other P and/or a, but that's all you can do. So you need a kind of 'reference point '
 
  • Like
Likes   Reactions: Calpalned
BvU said:
With ##
P^2 \propto \frac {a^3}{M_s}## you have ##
M_s \propto \frac {a^3}{P^2}## so if you know M at some P and a you can calculate (solve for) M when given M for some other P and/or a, but that's all you can do. So you need a kind of 'reference point '
Understand, so the ##\propto## symbol requires some kind of reference point.
 
Instaed of taking this as P^2 = (A^3)/M [I'm dropping the subscript for easy of writing] as you suggest, I would take it as P^2 = K*(A^3)/M where k is a "constant of proportionality" and can only be solved for when, as BvU says, you have a reference point. Anyway, it gives you M = K * (A^3)/(P^2) which, again, is only a full solution once you have a reference point that allows you to calculate K.
 
Calpalned said:
Understand, so the ##\propto## symbol requires some kind of reference point.
Look at Newton's Law of Gravity, for example. First he figured out that the forces were proportional to the product of the masses divided by the square of the distance between them, THEN he (well, someone later actually) figured out the value for M which is the constant of proportionality for that equation.

And note that M is determined empirically, by getting lots of reference points measured very accurately and I'm not sure who got the modern value but it wasn't Newton.
 
Calpalned said:
Understand, so the ##\propto## symbol requires some kind of reference point.

No, a "reference point" is not necessarily required. You could, instead, be given the "constant of proportionality" or some other type of extra information that would enable you to know the equality form. Certainly, having a reference point is likely the most common form of such extra information, but it is not the only form.