How to determine a planet's size given gravity

1. Sep 14, 2016

ckirmser

If I have a given gravity and the density of a planet is that of Earth, how do I determine its size (mass and radius) in Earth units?

I know that g = m/r2 and that rho (density) = m/v; that v = 4/3*pi*r3, so m = rho*v, giving;
g = rho*4/3*pi*r.

But, this falls apart. I'm sure it's because I'm putting it into Earth terms. I mean, if I were to put in for Earth, I'd have;

1 = 1*4/3*pi*1
And, that ain't it.

So, if I presume a density equal to that of Earth and I'm given a gravity, how do I find the radius? What nice, neat little formula will give that? Is there one?

Thanx!

2. Sep 14, 2016

haruspex

Right, but to avoid confusion I'll add subscripts to that (g is usually reserved for Earth's surface gravity):
$g_p=\rho\frac 43\pi r_p$
I do not understand your reasoning for putting 1 for each of those variables.
You have two equations, the one above for an arbitrary planet and the corresponding equation for the specific case of Earth. Write out the second one using an appropriate subscript instead of p.

3. Sep 15, 2016

ckirmser

I put the 1s in to represent that I want the values relative to Earth. So, g -> 1 for Earth's gravity, rho -> 1 for Earth's density, r -> 1 for Earth's radius. I've got a feeling this particular assumption is what's screwing me over.

So, I assume Planet X has Earth's density. If Planet X has a g of, say, 1.25, how can this formula - if it's the right one to use - give me Planet X's radius?

OK, here's my thinking. I start with;

(Cool! Copying equations from Word worked!)
Then, since I'm trying to find Planet X's radius, I have to turn mass into radius. So, for the general equation for Earth, I stick in the equation for density. The radius cubed for density cancels out the radius squared for gravity, leaving me with rho and a radius;

Now, turning it into ratios to put Planet X's values in terms of Earth's (I thought of this after reading your post);

So, now to find for Planet x's radius;

Since I'm assuming Planet X's density to be the same as Earth's, the Rho ratio cancels out. Then, I'm given Planet X's gravity in terms of Earth's which leaves;

As Earth's radius is 1 - using ratios - that term goes away. Calculating this out, I get Planet X's radius as 0.2984 x Earth's. Okie dokie. All well and good. Sorta. I mean, Planet X has a higher gravity than Earth, so its radius should be greater. So, I stick in data for Mars.

Mars has a gravity of 0.38 that of Earth. I know the density isn't the same, but it's close enough for what I'm trying to accomplish. So, I toss in 0.38 and get a radius of 0.0907 Earth's for Mars when it is actually 0.532 Earth's. That's waaay off. But, the Wiki example for Surface Gravity uses this formula;

They just toss in Mar's mass and radius in terms of Earth's - 0.107 and 0.532 respectively - and they get 0.38g for Martian gravity. All I'm doing is tossing in density to try to get rid of mass and I don't think Mar's density is all that much off from Earth's, not to cause the difference in radius values that I'm getting.

So, I've done something wrong, but I don't know what.

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4. Sep 15, 2016

Bandersnatch

1. In step 3, there's a mistake. All constants must cancel out.
2. This won't change your end result, but why are you not including the gravitational constant in your gravitational acceleration equation?
You seem to have already figured it out that this is wrong thinking. You can never substitute 1 for those values without specifically including your units. For example, it's always 1 Earth's density, which has a concrete value in g/cm^3 that is definitely not 1. You can forgret about it if, like in step 3, you express both planets' density in terms of density of one of them. E.g. one is 1 Earth density, the other is 1.5 Earth density - the units [Earth density] cancel out, and you're left with a simple unitless ratio.
Same with other variables.

In the end, you should arrive at a 1:1 relationship between the radius and surface gravity.

5. Sep 15, 2016

ckirmser

In my last second to last equation, I thought I had done that; taken the densities out because I'm assuming that Planet X's density is equal to Earth's.

How would you correct my equation? What else did I do wrong?

6. Sep 15, 2016

Bandersnatch

$4/3 \pi$ needs to cancel out. You did not write eq 3 correctly because that expression is missing in the denominator.

And how did you arrive at that 1.25 factor?

7. Sep 15, 2016

haruspex

I understand that you want to work in terms of ratios, but that is not the right way. The safe way is to write out both equations and divide the one by the other.
This is almost the same mistake. As Bandersnatch notes, if you do it properly the constant disappears.

8. Sep 16, 2016

ckirmser

D'oh!

OK, I'll see what that does.

That was just snagged out of the air for the gravity of Planet X.

What I'm doing is trying to mix rules for two different RPGs. In one of them, planetary size is given as a simply number without giving any sort of indication what that number means. The other has planets with hard planet sizes in km. So, I'm trying to take the gravity for a planet given in Game #1 and figure out it's dimensions for Game #2. I just invented 1.25 as the gravity for a sample Planet X.

Last edited: Sep 16, 2016
9. Sep 16, 2016

ckirmser

OK, let’s try this again.

This time, I’m getting rid of ratios and working straight from Newton. Thus, my general equation is;

(1)

So, setting this up for Planet X, I’d have;

(2)

I toss in the equation for Density to get an r expression and I have;

(3)

Doing the same for Earth, I have;

(4)

Now, although I’ve defined the gravity of Planet X to be 1.25 Earth’s, I’m going to substitute that with the gravity of Mars (0.378g). That way, there’s some real data to use to verify the results. So, putting in Mars’ gravity, I have;

(5)

Putting in the equations for gravity for each object, gives me;

(6)

The constants cancel, leaving;

(7)

Since I’m assuming that Planet X has the same density as Earth (rX = r⊕) the density factors – I presume – would be considered the same as constants and would cancel;

(8)

OK, so, if I’ve done this right, rX should be the radius of Mars. Earth’s mean radius, r⊕, is 6,371km. So, Mars’ should be 0.378•6,371, or 2,408.24km.

Mars true mean radius is 3,389.5km. So, I’ve done something wrong – unless Mars’ density is notably different from Earth’s. So, OK, Mars’ density is 3.934g/cm2. Earth’s is 5.514g/cm2. When I move going from Eq. 7 to Eq. 8, I’d get;

(9)

Alrighty then, I’ll toss in this new factor, 0.530. 0.530•6,371 = 3,377.61.

Well, cool! I guess it does work out. With the exponents, I didn't think it'd be a linear relationship.

Thanx for the assists, guys! You forced me to work it out to the bare bones, rather than take short-cuts (I believe it was Merry who said, “short cuts make long delays”). I need that tattooed on the inside of my eyelids.

Thanx again!

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