4
$\begingroup$

More specifically: if you know your movement relative to a body, could you use the inverse square law to calculate distance? Since if you are distance $D$ away and you move $d$, $\Delta L \propto (D-d)^2-D^2=2dD-d^2$, I'm assuming you should be able to calculate D linearly.

$\endgroup$
5
$\begingroup$

It is an interesting question, but it won't work because the luminosity of astrophysical sources is not constant enough, and because measurements cannot be taken with the required precision on long timescales.

Let's assume that you have an astrophysical sources that has a constant luminosity (note, I think you have confused luminosity with apparent brightness) and which has a perfectly measured line of sight speed $v$.

At some particular moment $$ F(t) = \frac{L}{4\pi D(t)^2}\ ,\tag*{(1)}$$ where $L$ is the luminosity, $D(t)$ is the (changing) distance and $F(t)$ is the observed flux.

We could observe how the brightness changed with time $$ \frac{dF}{dt} = \dot{F} = -\frac{2L}{4\pi D^3}v\ . \tag*{(2)}$$

If the luminosity is unknown, we can eliminate it by combining equations (1) and (2) and rearranging to give $D$ in term of the observables, $F$, $\dot{F}$ and $v$. $$ D = -\frac{2F}{\dot{F}}v\ , \tag*{(3)}$$ where a positive $v$ indicates movement away from us.

So it could work in principle, but let's put in some numbers. Let's say we have a source at $D=10$ pc and let's give it a big stellar velocity, say $v = 100$ km/s. That would mean (from equation 3), that $\dot{F}/F = 6\times 10^{-13}$ s$^{-1}$ or $$\frac{\dot{F}}{F} \sim 2\times 10^{-5} \left(\frac{D}{10\ {\rm pc}}\right)^{-1} \left(\frac{v}{100\ {\rm km/s}}\right)\ {\rm year}^{-1}$$.

So you would have to have instrumentation capable of measuring a change in brightness of 2 parts in 100,000 every year, assuming that the luminosity didn't change at all. And the required precision gets more demanding for larger distances or for slower velocities.

The Sun's (a normal star) brightness varies by 1-2 orders of magnitude more than this every year and there just isn't any astronomical instrumentation that has the capability or stability to make such measurements. Therefore it is unfeasible for at least two reasons.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ That makes a lot of sense! Thanks so much. $\endgroup$ – Erez Israeli Miller Apr 13 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.