Finding mass of a planet when density varies with radial distance

Click For Summary

Homework Help Overview

The discussion revolves around calculating the mass of a planet with a density that varies with radial distance, described by the equation ρ(r) = ρo(1 - αr/Ro). Participants are tasked with determining the total mass of the planet and the weight of a one-kilogram mass at its surface.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to integrate the density function over the radius to account for varying density. There are attempts to set up the integration and calculate the mass of spherical shells, but confusion arises regarding the correct application of the integration process.

Discussion Status

Several participants have attempted integration to find the total mass, but discrepancies in results indicate a lack of consensus on the correct approach. Some have provided guidance on considering the geometry of the planet and the mass of differential shells, while others express uncertainty about their calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is an ongoing exploration of assumptions related to the integration process and the interpretation of the density function.

Smartguy94
Messages
40
Reaction score
0

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?
 
Physics news on Phys.org
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}
 

Similar threads

Replies
4
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 62 ·
3
Replies
62
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
1K