Finding mass of a planet when density varies with radial distance

AI Thread Summary
To find the mass of a planet with varying density, the density function ρ(r) must be integrated over the volume of the planet, accounting for spherical shells. The correct approach involves using the volume element dV = 4πr²dr and integrating from the center (r = 0) to the surface (r = Ro). The total mass is calculated as M = ∫₀^{Ro} ρ(r) 4πr² dr. The integration should yield a total mass of approximately 4.1 x 10²⁴ kg, indicating that previous calculations may have errors in setup or execution. Properly applying these principles will lead to the correct mass and weight calculations for a one-kilogram mass on the planet's surface.
Smartguy94
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Homework Statement



The density of a certain planet varies with radial distance as:
ρ (r)= ρo(1- αr/Ro)
where Ro= 3.98 x 106 m is the radius of the planet, ρo= 4980 kg/m3 is its central density, and α = 0.17. What is the total mass of this planet ?

Calculate the weight of a one kilogram mass located on the surface of this planet.

Homework Equations



ρ = m/v

The Attempt at a Solution



since ρ= m/v

then

m = ρv
v of a planet = 4/3∏ R^3
so i substitute the given info on the equations and got

m = ρo(1- αr/Ro) * 4/3∏ R^3
calculate all of those and I realize that i don't know what r is.

so I assume that r = Ro but I got the wrong answer

any help?
 
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Since the density changes with radius (r), you will need to set up an integration over the radius r and sum the mass shells from the center (r = 0) to the surface (r = Ro).
 
gneill said:
Since the density changes with radius (r), you will need to set up an integration over the radius r and sum the mass shells from the center (r = 0) to the surface (r = Ro).

i integrate the ρ as a function of r

3.98x10^6
∫ ρo(1- αr/Ro)dr
0

and got 3641700 as my answer

since ρv = m
and v = 4/3 ∏ Ro^3
m = 6.07x10^13

but its wrong, what did i do wrong?
 
Smartguy94 said:
i integrate the ρ as a function of r

3.98x10^6
∫ ρo(1- αr/Ro)dr
0

and got 3641700 as my answer

since ρv = m
and v = 4/3 ∏ Ro^3
m = 6.07x10^13

but its wrong, what did i do wrong?

The ρ(r) function only gives you the density at a given radius. It does not address the geometry of the object. You need to determine the mass of each shell of matter of differential thickness dr, and sum them all up (integrate). In this case the shells will be spherical, and of thickness dr. So first find an expression for the mass of a spherical shell of thickness dr.
 
gneill said:
The ρ(r) function only gives you the density at a given radius. It does not address the geometry of the object. You need to determine the mass of each shell of matter of differential thickness dr, and sum them all up (integrate). In this case the shells will be spherical, and of thickness dr. So first find an expression for the mass of a spherical shell of thickness dr.

m = ρv
∫m = ∫ρv ?
which results to same exact answer as before?
 
Since the density only varies as the radius, and not as the latitude and/or longitude, the volume element is
dV=4πr2dr

After all, the total volume of a sphere of radius R0 can be obtained from \displaystyle V=\int_0^{R_0}4\pi r^2\,dr\,.

So the total mass of the planet is \displaystyle M=\int_0^{R_0}(ρ(r))4\pi r^2\,dr\,.
 
SammyS said:
Since the density only varies as the radius, and not as the latitude and/or longitude, the volume element is
dV=4πr2dr

After all, the total volume of a sphere of radius R0 can be obtained from \displaystyle V=\int_0^{R_0}4\pi r^2\,dr\,.

So the total mass of the planet is \displaystyle M=\int_0^{R_0}(ρ(r))4\pi r^2\,dr\,.


i did that integration and got 1.147x10^24 and its still wrong...
 
Can you show us how you did the integration? I got \approx 4.1 \times 10^{24}
 
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