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Is it possible to calculate the radius of the sun with minimal amount of physics?

I couldn’t find resources on the internet that gives me an answer using a theoretical formula.

You are given values such as mass of the earth as well as the sun. The earth sun distance and diameter of the earth. You can assume the moon doesn’t exist so that it does not affect the equations. The value of g for the earth is also given.

Is this possible or I’m I asking a question while giving too little information (number of variables)?

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  • $\begingroup$ That's too little information. The absolute scale of the solar system (including the size of the sun) was only possible to be determined after either the solar occultations by the Moon or Venus transits were taken into account. If this is homework, then maybe there's another piece of info there? The ratio of $g_{\rm Earth}/g_{\rm Sun}$ maybe? Or some other observational quantity, like the angular diameter of the sun? $\endgroup$ – AtmosphericPrisonEscape Mar 7 at 16:41
  • $\begingroup$ Seems there is one piece of data missing. $\endgroup$ – Anders Sandberg Mar 7 at 16:48
  • $\begingroup$ @AtmosphericPrisonEscape not a homework question. Just thought about this. $\endgroup$ – physics2000 Mar 7 at 16:56
  • $\begingroup$ @AndersSandberg which one? $\endgroup$ – physics2000 Mar 7 at 16:58
  • $\begingroup$ @AtmosphericPrisonEscape actually my astrophysics teacher gave us this question. He told us to consider the position of the moon during a total solar eclipse. I did it by using the property of similar triangles and found the answer. But I wondered whether it’s possible to derive a purely theoretical equation that gives out the answer. $\endgroup$ – physics2000 Mar 7 at 17:03
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From purely orbital data - no .

It's an important feature of classical gravity that you can treat masses as point sources, so the Earth's orbit would be unchanged if the sun collapsed to a blackhole (with the same mass)

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  • $\begingroup$ Yes, but it’s possible to calculate the gravitational potential for different solids, hollow objects. Can this help in any way? $\endgroup$ – physics2000 Mar 7 at 16:58
  • $\begingroup$ Gravitational potential and gravitational field outside a spherical object depend only on the mass of the object and distance from the center, not its size. $\endgroup$ – Bill N Mar 7 at 19:04
  • $\begingroup$ @physics2000 Take a look at the shell theorem. $\endgroup$ – JMac Mar 7 at 19:15
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Since you know the distance to the sun, you can find its diameter because you can see it - meaning, from its apparent size. (This of course assumes that you are allowed to measure that apparent size.)

Measure the width of your house $d_{house}$, then walk 1km away and measure the apparent width $d_{house,apparent}$, for example by stretching out your arm with a ruled in your hand. Divide the two numbers and you now have a scale - a factor for the apparent width increase per kilometer:

$$\frac {d_{house}}{d_{house,apparent}} = f_{house}$$

Multiply this per-kilometer factor $f_{house}$ with the distance to the sun $r$, and you have a factor for how many times larger the sun is than its apparent size:

$$f_{sun}=rf_{house}$$

Then measure the sun's apparent size $d_{sun,apparent}$ in the same way. And multiply the factor to get the sun diameter $d_{sun}$:

$$d_{sun}=f_{sun} d_{sun,apparent}$$

Theoretically, this does not give the exact diameter since you don't see the sun across its entire diameter due to its curvature (the curvature "blocks" a bit of the diameter from our view) but the sun is so far away that your measurements alone will have a larger uncertainty.

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  • $\begingroup$ The only issue I see with this answer is that observing the apparent size of the sun wasn't listed as one of the methods we could use to determine size, as given in the question at least. $\endgroup$ – JMac Mar 7 at 19:00
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    $\begingroup$ @JMac True. I answered according to the "minimal amount of physics" requirement and assumed that I am able to do simple physics here on the planet... $\endgroup$ – Steeven Mar 7 at 19:04
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    $\begingroup$ It really depends on if he was trying to figure out something conceptual given his conditions or not. I had assumed the question was more related to seeing if the orbital information about our planet could let us determine the diameter of the sun. See also his note on the bottom of the question "Is this possible or I’m I asking a question while giving too little information (number of variables)?" The way the question was worded to me, it seems like he wanted to constrain it to the variables he listed, or figure out why they would be insufficient. $\endgroup$ – JMac Mar 7 at 19:10
  • $\begingroup$ @Steeven your solution for sure helps. But does not suffice. My question is about why doesn’t there exist a simple Newtonian solution to the radius of a star?(see my earlier comments for clarification). Your answer takes into account the practical perspective of the answer which does not really add weight to the solution that I want. $\endgroup$ – physics2000 Mar 7 at 19:51
  • $\begingroup$ @JMac I don’t wish to constrain it. If more variables are needed then fine. It’s just that I don’t wish to give too many variables like the g value of the sun such that the problem is no longer a problem anymore. Hence my side note. And yes, you are right that I wanted to know how our orbital data could relate to the radius of the sun. $\endgroup$ – physics2000 Mar 7 at 19:56
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You are given values such as mass of the earth as well as the sun. The earth sun distance and diameter of the earth.

Someone had to do an experiment to give you those values, so it should be allowable for you to make an observation of the angular diameter of the Sun. This is easy using a pinhole or reflecting from a plane mirror on to a screen. The diameter of the image divided by the distance from the pinhole/mirror is equal to the diameter of the sun divided by the earth-sun distance: $$D_{\mathrm{sun}}=D_{\mathrm{image}}\frac{\ell_{\mathrm{earth-sun}}}{\ell_{\mathrm{image-mirror}}}$$, where the $\ell$s are the distances. Simple geometrical optics.

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