Finding mass of a planet when density varies with radial distance

1. Dec 15, 2011

Smartguy94

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

ρ = m/v

3. 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?

2. Dec 15, 2011

Staff: Mentor

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).

3. Dec 15, 2011

Smartguy94

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?

4. Dec 15, 2011

Staff: Mentor

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.

5. Dec 15, 2011

Smartguy94

m = ρv
∫m = ∫ρv ???
which results to same exact answer as before?

6. Dec 15, 2011

SammyS

Staff Emeritus
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 $\displaystyle M=\int_0^{R_0}(ρ(r))4\pi r^2\,dr\,.$

7. Dec 15, 2011

Smartguy94

i did that integration and got 1.147x10^24 and its still wrong...

8. Dec 16, 2011

JHamm

Can you show us how you did the integration? I got $\approx 4.1 \times 10^{24}$