How to determine a planet's size given gravity

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In summary: Earth given in Game #2...to try to come up with a planetary size in km for Earth.I'm trying to sorta justify having Earth's size as 1, since I'm trying to equate the gravity and density of Earth to those of the planet in the first RPG. I know it's a klugey way of going about it, but it's the only way I can think of right now.1. In step 3, there's a mistake. All constants must cancel out.In summary, the conversation discusses the process of determining the size (mass and radius) of a planet in Earth units, given its gravity and density. The equations g = m/r2 and rho = m
  • #1
ckirmser
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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!
 
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  • #2
ckirmser said:
g = rho*4/3*pi*r.
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##
ckirmser said:
1 = 1*4/3*pi*1
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
I put the 1s into 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;

upload_2016-9-15_10-4-21.png

(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;

upload_2016-9-15_10-11-35.png


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

upload_2016-9-15_10-15-44.png


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

upload_2016-9-15_10-24-13.png


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;

upload_2016-9-15_10-29-22.png


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;

upload_2016-9-15_10-35-52.png


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
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?
ckirmser said:
I put the 1s into 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.
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
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
##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?
 
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  • #7
ckirmser said:
I put the 1s into represent that I want the values relative to Earth
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.
ckirmser said:
Now, turning it into ratios to put Planet X's values in terms of Earth's (I thought of this after reading your post);

upload_2016-9-15_10-15-44-png.105983.png
This is almost the same mistake. As Bandersnatch notes, if you do it properly the constant disappears.
 
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  • #8
Bandersnatch said:
##4/3 \pi## needs to cancel out. You did not write eq 3 correctly because that expression is missing in the denominator.

D'oh!

OK, I'll see what that does.

And how did you arrive at that 1.25 factor?

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.
 
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  • #9
OK, let’s try this again.

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

upload_2016-9-16_14-6-52.png
(1)

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

upload_2016-9-16_14-6-52.png
(2)

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

upload_2016-9-16_14-6-52.png
(3)

Doing the same for Earth, I have;

upload_2016-9-16_14-6-52.png
(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;

upload_2016-9-16_14-6-52.png
(5)

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

upload_2016-9-16_14-6-52.png
(6)

The constants cancel, leaving;

upload_2016-9-16_14-6-52.png
(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;

upload_2016-9-16_14-6-52.png
(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
upload_2016-9-16_14-6-52.png
going from Eq. 7 to Eq. 8, I’d get;

upload_2016-9-16_14-6-52.png
(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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Related to How to determine a planet's size given gravity

1. How does gravity affect the size of a planet?

Gravity is directly related to the mass of an object. The greater the mass of a planet, the stronger its gravitational pull. Therefore, a planet's size can be determined by measuring its gravitational force.

2. What is the formula for determining a planet's size based on its gravity?

The formula for determining a planet's size based on its gravity is: Size = (Gravity * Distance^2) / (4 * π * Gravitational Constant), where Distance is the distance between the planet and the object being affected by its gravity, and the Gravitational Constant is a universal constant that relates to the strength of gravity.

3. Can a planet's size be accurately determined using only its gravity?

No, other factors such as the density and composition of the planet also play a role in determining its size. However, gravity is a key factor in determining a planet's size and can provide a good estimate.

4. How do scientists measure a planet's gravitational force?

Scientists use tools such as satellites, telescopes, and spacecraft to measure a planet's gravitational force. They can also measure the gravitational pull on objects in space and use that information to calculate the planet's size.

5. What other methods can be used to determine a planet's size besides gravity?

In addition to gravity, scientists can use techniques such as radar imaging, spectroscopy, and transit photometry to determine a planet's size. These methods involve analyzing the planet's surface features, atmospheric composition, and transit in front of its host star, respectively.

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