Gravitational acceleration based on density

  • #1
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Im trying to create or find some way to calculate gravitational acceleration on planet surface based solely on density and mass of planet.

This should be based on average density and it can be also some approximation, without calculating minor effects like rotation, planet bulge, relativity and so on.

So it should look like this:


g= G*M*d*X

where

g – gravitational acceleration on the surface of planet

G – gravitational constant

M – mass of planet (in kg)

d - average density of planet (in kg/m3)

X – missing link which I wasnt able to derive

Important thing is, that Im trying to get rid of radius of planet and use only density, not sure if it is possible.

Similar topics, which I looked in, but wasnt able to find there what I need:

https://www.physicsforums.com/threads/gravity-as-function-of-density.521163/

https://en.wikipedia.org/wiki/Gauss's_law_for_gravity#Differential_form
 

Answers and Replies

  • #2
nasu
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Assuming a spherical and homogeneous planet, the radius can be calculated when you know mass and density.
So if your formula contains a radius, you can just replace it by the expression of radius as a function of mass and density.
 
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  • #3
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Assuming a spherical and homogeneous planet, the radius can be calculated when you know mass and density.
So if your formula contains a radius, you can just replace it by the expression of radius as a function of mass and density.
density = M/[(4/3)pi*r^3]
d/M= (4/3)pi*r^3
d/(M*(4/3)pi=r^3
(d/(M*(4/3)pi)^-3=r

g=M*G/r^2
g=M*G/((d/(M*(4/3)pi)^-3)^2
g=M*G/((d/(M*(4/3)pi)^-3)^2

Something like this?

Probably it can be further simplified, but my math knowledge is quite poor :)
 
  • #4
Ibix
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The first line is fine. The left hand side of the second line is upside down, and you need to take the cube root in the fourth line (raise to the power 1/3, not -3). If you correct those and carry them through your approach seems fine.
 
  • #5
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The first line is fine. The left hand side of the second line is upside down, and you need to take the cube root in the fourth line (raise to the power 1/3, not -3). If you correct those and carry them through your approach seems fine.
Thanks, I see the mistake, will try to correct

Corrected version:
density = M/[(4/3)pi*r^3]
d/M= 1/((4/3)pi*r^3)
(d*(4/3)pi)/M=1/r^3
((d*(4/3)pi)/M)*r^3=1
r^3=1/((d*(4/3)pi)/M)
r =(1/((d*(4/3)pi)/M))^-3

g=M*G/r^2
g=M*G/((1/((d*(4/3)pi)/M))^-3)^2
 
  • #6
Ibix
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You've still taken the cube root wrong. ##r^3=3M/4\pi d## means that ##r=(3M/4\pi d)^{1/3}##. You've written ##r=(3M/4\pi d)^{-3}##, which isn't right. If you correct that, you're fine.

You can simplify the expression a lot by combining fractions. For example the first expression can be re-written as ##d=3M/4\pi r^3##, which is easy to parse even without brackets. You can also use LaTeX to write maths on this forum, which is a lot clearer. Click the reply button to see my maths quoted, or check out the link on the left below the reply box.
 
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  • #7
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You've still taken the cube root wrong. ##r^3=3M/4\pi d## means that ##r=(3M/4\pi d)^{1/3}##. You've written ##r=(3M/4\pi d)^{-3}##, which isn't right. If you correct that, you're fine.

You can simplify the expression a lot by combining fractions. For example the first expression can be re-written as ##d=3M/4\pi r^3##, which is easy to parse even without brackets. You can also use LaTeX to write maths on this forum, which is a lot clearer. Click the reply button to see my maths quoted, or check out the link on the left below the reply box.
OK, my first try of Latex using what have you written

##d=3M/4\pi r^3##
##d/3M=1/4\pi r^3##
##4\pi d/3M=1/\ r^3##
##r^3=3M/4\pi d##
##r=(3M/4\pi d)^{1/3}##

##g=MG/r^2##
##g=MG/((3M/4\pi d)^{1/3})^2##
 
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  • #8
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additional simplification, but not sure if I did it right

##d=3M/4\pi r^3##
##d/3M=1/4\pi r^3##
##4\pi d/3M=1/\ r^3##
##r^3=3M/4\pi d##
##r=(3M/4\pi d)^{1/3}##

##g=MG/r^2##
##g=MG/((3M/4\pi d)^{1/3})^2##
##g=MG/(3M/4\pi d)^{2/3}##
##g=(MG/1)/((3M)^{2/3}/(4\pi d)^{2/3})##
##g=(G(4\pi d)^{2/3})/(3^{2/3}M^{-1/3})##
##g=G(M^{1/3})(\pi d)^{2/3}(3^{2/3})##
##g=G(M^{1/3})(\pi d)^{2/3}(1/3)##
 
  • #9
jbriggs444
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##g=(G(4\pi d)^{2/3})/(3^{2/3}M^{-1/3})##
##g=G(M^{1/3})(\pi d)^{2/3}(3^{2/3})##
You seem to have lost the 4 here.
##g=G(M^{1/3})(\pi d)^{2/3}(3^{2/3})##
##g=G(M^{1/3})(\pi d)^{2/3}(1/3)##
The square of the cube root of three is supposedly equal to one third?
 
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  • #10
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You seem to have lost the 4 here.

The square of the cube root of three is supposedly equal to one third?
yes, you are right with the lost 4

also the cube root of three is wrong, thanks for spotting it

Corrected version

##d=3M/4\pi r^3##
##d/3M=1/4\pi r^3##
##4\pi d/3M=1/\ r^3##
##r^3=3M/4\pi d##
##r=(3M/4\pi d)^{1/3}##

##g=MG/r^2##
##g=MG/((3M/4\pi d)^{1/3})^2##
##g=MG/(3M/4\pi d)^{2/3}##
##g=(MG/1)/((3M)^{2/3}/(4\pi d)^{2/3})##
##g=(G(4\pi d)^{2/3})/(3^{2/3}M^{-1/3})##
##g=G(M^{1/3})(4\pi d)^{2/3}(3^{2/3})##
 
  • #11
jbriggs444
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##g=MG/(3M/4\pi d)^{2/3}##
##g=(MG/1)/((3M)^{2/3}/(4\pi d)^{2/3})##
##g=(G(4\pi d)^{2/3})/(3^{2/3}M^{-1/3})##
##g=G(M^{1/3})(4\pi d)^{2/3}(3^{2/3})##
Personally, I would have reformatted this. You have nested parenthesized expressions all on the same line. Hard to parse by eye. Instead of stringing things all on one line, I'd have used the \frac item to format fractions with a numerator up top, a denominator on the bottom. Let's see if that makes things more legible...

$$g=MG/(3M/4\pi d)^{2/3}$$

[Changing fraction format]

$$g=\frac {MG} {(3M/4\pi d)^{2/3}}$$

[Switching up the approach in addition to the notation...]

$$g=MG \frac{(4\pi d)^{2/3}}{(3M)^{2/3}}$$

[Combining the M term. Power of 1 in numerator and 2/3 in denominator = 1/3 in numerator alone]

$$g=M^{1/3}G \frac{(4\pi d)^{2/3}}{3^{2/3}}$$

[Negating the exponent on ##3^{2/3}## and putting that term in the numerator

$$g=M^{1/3}G\ (4\pi d)^{2/3}\ 3^{-2/3}$$

[Re-ordering the terms to match as closely as possible what you had concluded]

$$g=GM^{1/3}\ (4\pi d)^{2/3}\ 3^{-2/3}$$

Looks like you had failed to negate the sign on the exponent of ##3^{2/3}## when you moved it from denominator to numerator.

Edit: Of course a good final step would be to validate this formula, for instance by plugging in numbers for the mass and density of the earth.

Additional edit: Qualitatively, the formula is right. Gravitational acceleration should be directly proportional to mass. You have mass ##M^{1/3}## and density ##d^{2/3}## in the formula. Total exponent = 1, so the formula agrees that there is a direct proportion. Gravitation acceleration should be inversely proportional to the square of radius. Density scales as the inverse cube of radius and you have density in the formula with a 2/3 exponent. So the formula agrees with this as well. You have G in there as a direct proportion. That is consistent too. The rest is just a rolled-up constant of proportionality. If the validation test comes out good, the formula just has to be right.
 
Last edited:
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  • #12
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Personally, I would have reformatted this. You have nested parenthesized expressions all on the same line. Hard to parse by eye. Instead of stringing things all on one line, I'd have used the \frac item to format fractions with a numerator up top, a denominator on the bottom. Let's see if that makes things more legible...

$$g=MG/(3M/4\pi d)^{2/3}$$

[Changing fraction format]

$$g=\frac {MG} {(3M/4\pi d)^{2/3}}$$

[Switching up the approach in addition to the notation...]

$$g=MG \frac{(4\pi d)^{2/3}}{(3M)^{2/3}}$$

[Combining the M term. Power of 1 in numerator and 2/3 in denominator = 1/3 in numerator alone]

$$g=M^{1/3}G \frac{(4\pi d)^{2/3}}{3^{2/3}}$$

[Negating the exponent on ##3^{2/3}## and putting that term in the numerator

$$g=M^{1/3}G\ (4\pi d)^{2/3}\ 3^{-2/3}$$

[Re-ordering the terms to match as closely as possible what you had concluded]

$$g=GM^{1/3}\ (4\pi d)^{2/3}\ 3^{-2/3}$$

Looks like you had failed to negate the sign on the exponent of ##3^{2/3}## when you moved it from denominator to numerator.

Edit: Of course a good final step would be to validate this formula, for instance by plugging in numbers for the mass and density of the earth.

Additional edit: Qualitatively, the formula is right. Gravitational acceleration should be directly proportional to mass. You have mass ##M^{1/3}## and density ##d^{2/3}## in the formula. Total exponent = 1, so the formula agrees that there is a direct proportion. Gravitation acceleration should be inversely proportional to the square of radius. Density scales as the inverse cube of radius and you have density in the formula with a 2/3 exponent. So the formula agrees with this as well. You have G in there as a direct proportion. That is consistent too. The rest is just a rolled-up constant of proportionality. If the validation test comes out good, the formula just has to be right.
Thanks a lot, have checked it and came to g=9,81 which is pretty accurate
 

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