# Simple formula to relate planet mass with size?

sonhouse
Science 20 Nov '12 07:23
1. sonhouse
Fast and Curious
20 Nov '12 07:23
I just read about a newly discovered planet around a star Kappa Andromedae, about 170 light years out where the planet is about 60 AU away from the star and masses 7 Jupiters. How can you go from that to a radius? Assuming the planets have the same density of course, just the simplest case for a rough comparison. Given equal density I would assume that would factor out of the equation so you could compare say, Mercury with another planet with IT's density but say 10 times more massive.
2. joe shmo
Strange Egg
20 Nov '12 11:51
Originally posted by sonhouse
I just read about a newly discovered planet around a star Kappa Andromedae, about 170 light years out where the planet is about 60 AU away from the star and masses 7 Jupiters. How can you go from that to a radius? Assuming the planets have the same density of course, just the simplest case for a rough comparison. Given equal density I would assume that woul ...[text shortened]... could compare say, Mercury with another planet with IT's density but say 10 times more massive.
Well,

Assume a sphere

m= mass_ Jupiter
p = density_ Jupiter
r = radius of planet
n = number of times more massive than Jupiter

n*m = 4/3*p*PI*r^3

solve for "r"

r = ((3*n*m)/(4*PI*p))^(1/3)

r= (n^(1/3))*(3*m/(4*PI*p))^(1/3)

r = (n^(1/3))*(r_ Jupiter)

so if it is 7 times more massive than Jupiter than the radius is

r = (7^(1/3))*(r_Jupiter)

r= 1.913*(r_ Jupiter)
3. sonhouse
Fast and Curious
20 Nov '12 17:58
Originally posted by joe shmo
Well,

Assume a sphere

m= mass_ Jupiter
p = density_ Jupiter
r = radius of planet
n = number of times more massive than Jupiter

n*m = 4/3*p*PI*r^3

solve for "r"

r = ((3*n*m)/(4*PI*p))^(1/3)

r= (n^(1/3))*(3*m/(4*PI*p))^(1/3)

r = (n^(1/3))*(r_ Jupiter)

so if it is 7 times more massive than Jupiter than the radius is

r = (7^(1/3))*(r_Jupiter)

r= 1.913*(r_ Jupiter)
Ah, so it's the cube root of the difference in size. Thanks! So it would be roughly twice the size of Jupiter assuming the same density. That's getting up there since Jove is about 80,000 (about 130,000 km) miles across so this new planet would be something like 150,000 (240,000 Km)miles across. That's getting within reach of the size of the sun which clocks in at about 880,000 miles (1,400,000 Km) across!
4. joe shmo
Strange Egg
20 Nov '12 20:37
Originally posted by sonhouse
Ah, so it's the cube root of the difference in size. Thanks! So it would be roughly twice the size of Jupiter assuming the same density. That's getting up there since Jove is about 80,000 (about 130,000 km) miles across so this new planet would be something like 150,000 (240,000 Km)miles across. That's getting within reach of the size of the sun which clocks in at about 880,000 miles (1,400,000 Km) across!
5. sonhouse
Fast and Curious
21 Nov '12 13:081 edit
Originally posted by joe shmo
It occurred to me by just doing the cube root of the difference and comparing the real sizes, for instance if you did that for the sun Vs Jupiter, you could get a rough idea of the density difference. So the sun, 1.4 E6 Km diameter Vs 1.28E5 Km works out to be about 11 times bigger than Jupiter but if you take the cube root of that, the sun would be only about 2.2 times bigger than Jupiter so the density of the sun must be a LOT less than the density of Jupiter.

So it looks like the density of Jupiter is about 5 times that of the sun. That is solely based on the numbers. I have no idea what the real numbers are, but I bet it's not far off what I said, about 5 times more dense than the sun.
6. joe shmo
Strange Egg
21 Nov '12 23:324 edits
Originally posted by sonhouse
It occurred to me by just doing the cube root of the difference and comparing the real sizes, for instance if you did that for the sun Vs Jupiter, you could get a rough idea of the density difference. So the sun, 1.4 E6 Km diameter Vs 1.28E5 Km works out to be about 11 times bigger than Jupiter but if you take the cube root of that, the sun would be only ab ...[text shortened]... eal numbers are, but I bet it's not far off what I said, about 5 times more dense than the sun.
I don't think that works, the formula doesn't compare densities, because it was derived assuming the densities of the two compared planets were equal...If you want to compare densities based on size its going to be a whole new formula. I'll work it out for you.

let

p_A = density of planet A
p_B = density of planet B

the ratio, density of planet B: density of planet A = n'

n' = p_B/p_A (n' will be the number of times more dense or less dense "planet B" is than "planet A" )

now each of the respective densities of the planet are as follows

p_A = m_A/V_A ( that is the mass of planet A divided by the volume of planet A, and the same goes for planet B)
p_B = m_B/V_B

now we are still trying to find the ratio n', so we back substitute as follows.

n' = (m_B/V_B)/(m_A/V_A)

once again assuming a sphere

V_A = 4/3*PI*r_A^3

V_B = 4/3*PI*r_B^3

so we back substitute these equations into the most recent equation for n' and after simplification

n' = (m_B/m_A)*(r_A/r_B)^3

now we assume that planet B is n times more massive than planet A, which means

m_B = n*m_B

back substitute and simplify and we arrive at the final equation

n' = n*(r_A/r_B)^3

!AND A FURTHER NOTE!:
all the variables on the right you should know...that is to say, don't solve for r_A from the first equation we derived earlier when we assumed the planets A and B had equal densites. For If you do, n' will just be a bogus result.
7. 22 Nov '12 01:30
Originally posted by joe shmo
I don't think that works, the formula doesn't compare densities, because it was derived assuming the densities of the two compared planets were equal...If you want to compare densities based on size its going to be a whole new formula. I'll work it out for you.

let

p_A = density of planet A
p_B = density of planet B

the ratio, density of planet B: ...[text shortened]... nets A and B had equal densites. For If you do, n' will just be a bogus result.
Small typo...

now we assume that planet B is n times more massive than planet A, which means

m_B = n*m_B

I think you mean m_B = n*m_A

Otherwise good work.
8. joe shmo
Strange Egg
22 Nov '12 02:22
Originally posted by googlefudge
Small typo...

now we assume that planet B is n times more massive than planet A, which means

m_B = n*m_B

I think you mean m_B = n*m_A

Otherwise good work.
Yep, thanks for the read through!
9. joe shmo
Strange Egg
22 Nov '12 02:561 edit
!AND A FURTHER NOTE!:
all the variables on the right you should know...that is to say, don't solve for r_A from the first equation we derived earlier when we assumed the planets A and B had equal densites. For If you do, n' will just be a bogus result.

To further explain this note, I would like to add that the first equation derived in this thread is in fact just a particular solution to the latest equation (n' = 1). So if you were to substitute the the first into the last, regardless of what any of the variables (n, r_A,r_B) you will always arrive at the solution n' =1 (indicating that the densities of planets A & B are equal)
10. sonhouse
Fast and Curious
22 Nov '12 07:00
Originally posted by joe shmo
!AND A FURTHER NOTE!:
all the variables on the right you should know...that is to say, don't solve for r_A from the first equation we derived earlier when we assumed the planets A and B had equal densites. For If you do, n' will just be a bogus result.

To further explain this note, I would like to add that the first equation derived in t ...[text shortened]... ways arrive at the solution n' =1 (indicating that the densities of planets A & B are equal)
Thanks again for the work! What do you do for a living? Are you a math professor or something?
11. joe shmo
Strange Egg
22 Nov '12 16:06
Originally posted by sonhouse
Thanks again for the work! What do you do for a living? Are you a math professor or something?
No, I'm not a Math professor. Currently I'm unemployed, hopefully not for much longer. I just have a Bachelors in Mechanical Engineering, and a desire to understand things well enough so that if the opportunity should arise, I can help explain them to others.

I'm lucky I found this site a few years back when I had (what would probably have been) a fleeting interest intellectual activities. Before I did, I virtually new nothing of math, or physics. I would say in no small part, RHP's problem solvers facilitated the math/physics bug ( that I think I originally obtained from some plug and chug equations I had to use at my job ), which inevitably lead me to college, where I could really dig in and change my brain. Thanks people. (In fact you were probably one them, but most of the others don't seem to be around anymore). So I guess I'm just doing what I can to keep the good problem solving karma going!
12. sonhouse
Fast and Curious
22 Nov '12 17:32
Originally posted by joe shmo
No, I'm not a Math professor. Currently I'm unemployed, hopefully not for much longer. I just have a Bachelors in Mechanical Engineering, and a desire to understand things well enough so that if the opportunity should arise, I can help explain them to others.

I'm lucky I found this site a few years back when I had (what would probably have been) a fl ...[text shortened]... more). So I guess I'm just doing what I can to keep the good problem solving karma going!
Works for meðŸ™‚