# Gravitational acceleration based on density

• Prometeus
In summary: K M^{1/3}\ d^{2/3}$$In summary, we are trying to find a formula for gravitational acceleration on the surface of a planet based solely on the planet's density and mass. This formula should be based on the average density and should not take into account other effects like rotation, planet bulge, or relativity. After some calculations, we have derived the following formula: g = GM^(1/3) * (4πd)^(2/3) * 3^(-2/3). This formula has been validated qualitatively and appears to be correct. Prometeus 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 I am 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 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. Dale and Prometeus nasu said: 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 :) 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. Ibix said: 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 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. Prometeus Ibix said: 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## Ibix 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)## Prometeus said: ##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? Prometeus jbriggs444 said: 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})## Prometeus said: ##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}$$Looks like you had failed to negate the sign on the exponent of ##3^{2/3}## when you moved it from denominator to numerator. Edit: Of course a good final step would be to validate this formula, for instance by plugging in numbers for the mass and density of the earth. Additional edit: Qualitatively, the formula is right. Gravitational acceleration should be directly proportional to mass. You have mass ##M^{1/3}## and density ##d^{2/3}## in the formula. Total exponent = 1, so the formula agrees that there is a direct proportion. Gravitation acceleration should be inversely proportional to the square of radius. Density scales as the inverse cube of radius and you have density in the formula with a 2/3 exponent. So the formula agrees with this as well. You have G in there as a direct proportion. That is consistent too. The rest is just a rolled-up constant of proportionality. If the validation test comes out good, the formula just has to be right. Last edited: Prometeus jbriggs444 said: Personally, I would have reformatted this. You have nested parenthesized expressions all on the same line. Hard to parse by eye. Instead of stringing things all on one line, I'd have used the \frac item to format fractions with a numerator up top, a denominator on the bottom. Let's see if that makes things more legible...$$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

## 1. What is gravitational acceleration based on density?

Gravitational acceleration based on density is a concept that describes the acceleration of an object due to the influence of gravity and its density. It is a measure of how much an object is affected by gravity based on its mass and volume.

## 2. How is gravitational acceleration based on density calculated?

Gravitational acceleration based on density is calculated by dividing the force of gravity acting on an object by its mass. This results in a value that represents the acceleration of the object due to gravity.

## 3. How does density affect gravitational acceleration?

Density plays a significant role in determining the strength of gravitational acceleration. Objects with higher densities experience a greater gravitational pull compared to objects with lower densities. This is because the mass of an object is directly proportional to its density.

## 4. Does the location of an object affect gravitational acceleration based on density?

Yes, the location of an object can affect its gravitational acceleration based on density. This is because the strength of gravity varies depending on the distance from the center of mass of the object. Therefore, the same object may experience different gravitational accelerations at different locations.

## 5. What real-life applications does gravitational acceleration based on density have?

Gravitational acceleration based on density has various real-life applications, such as in the field of geophysics to study the Earth's interior, in the field of astrophysics to study the properties of celestial bodies, and in engineering to design and test structures that can withstand gravitational forces.

• Classical Physics
Replies
54
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
268
• Cosmology
Replies
19
Views
947
Replies
15
Views
2K
• Classical Physics
Replies
6
Views
665
• Introductory Physics Homework Help
Replies
4
Views
739
• Classical Physics
Replies
7
Views
1K
• Mechanics
Replies
32
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
5K
• Mechanics
Replies
12
Views
1K