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How to calculate the mass of earth using triple integrals?
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How to calculate the mass of earth using triple integrals?
The mass density of earth is $5.51 ,g/m^3$
The radius : $6353$ to $6384 ,km$ (average : $6371 ,km$)
However, I don't even know how to start, what's the density function I suppose to use and what's the limits of integration?
If anyone can help?
multivariable-calculus physics
edited Jul 29 at 11:10
user529760
519216
asked Jul 29 at 9:37
Aleph0
6
add a comment |Â
up vote
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favorite
How to calculate the mass of earth using triple integrals?
The mass density of earth is $5.51 ,g/m^3$
The radius : $6353$ to $6384 ,km$ (average : $6371 ,km$)
However, I don't even know how to start, what's the density function I suppose to use and what's the limits of integration?
If anyone can help?
multivariable-calculus physics
edited Jul 29 at 11:10
user529760
519216
asked Jul 29 at 9:37
Aleph0
6
Why would you need triple integrals? You have an average radius and an average density, you can calculate the mass approximately as $$m approx frac 4 3 pi rho r^3. $$ Anything more would be overkill.
– giobrach
Jul 29 at 9:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to calculate the mass of earth using triple integrals?
The mass density of earth is $5.51 ,g/m^3$
The radius : $6353$ to $6384 ,km$ (average : $6371 ,km$)
However, I don't even know how to start, what's the density function I suppose to use and what's the limits of integration?
If anyone can help?
multivariable-calculus physics
edited Jul 29 at 11:10
user529760
519216
asked Jul 29 at 9:37
Aleph0
6
How to calculate the mass of earth using triple integrals?
The mass density of earth is $5.51 ,g/m^3$
The radius : $6353$ to $6384 ,km$ (average : $6371 ,km$)
However, I don't even know how to start, what's the density function I suppose to use and what's the limits of integration?
If anyone can help?
multivariable-calculus physics
edited Jul 29 at 11:10
user529760
519216
asked Jul 29 at 9:37
Aleph0
6
edited Jul 29 at 11:10
user529760
519216
edited Jul 29 at 11:10
user529760
519216
edited Jul 29 at 11:10
user529760
519216
519216
asked Jul 29 at 9:37
Aleph0
6
asked Jul 29 at 9:37
Aleph0
6
asked Jul 29 at 9:37
Aleph0
6
6
Why would you need triple integrals? You have an average radius and an average density, you can calculate the mass approximately as $$m approx frac 4 3 pi rho r^3. $$ Anything more would be overkill.
– giobrach
Jul 29 at 9:45
add a comment |Â
Why would you need triple integrals? You have an average radius and an average density, you can calculate the mass approximately as $$m approx frac 4 3 pi rho r^3. $$ Anything more would be overkill.
– giobrach
Jul 29 at 9:45
Why would you need triple integrals? You have an average radius and an average density, you can calculate the mass approximately as $$m approx frac 4 3 pi rho r^3. $$ Anything more would be overkill.
– giobrach
Jul 29 at 9:45
Why would you need triple integrals? You have an average radius and an average density, you can calculate the mass approximately as $$m approx frac 4 3 pi rho r^3. $$ Anything more would be overkill.
– giobrach
Jul 29 at 9:45
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Mass = volume $times$ mass density
Volume of a sphere is $$ V=4/3 pi R^3 $$Where R is the radius.
You do not need triple integral to find the approximate mass of the earth.
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
1
down vote
You situation does not need triple integrals: see my comment. However, had you had information for a variable mass density (i.e. $rho$ as a function of position, generally non-constant) you would have had to use integration.
Suppose that Earth is a sphere with radius $R$. If it has mass density $rho : B_0(R) to mathbb R$ (where $B_0(R)$ is the ball of radius $R$ centered in the origin), then the total mass may be calculated as
$$ m = int_B_0(R) rho(x,y,z) dV .$$
Then you may change variables to spherical
$$begincases
x= rcosthetacosvarphi \
y = rsinthetacosvarphi \
z = rsinvarphi
endcases $$
so that the determinant of the transformation is
$$|J| = r^2cosvarphi. $$
If you assume that the mass density is spherically symmetric, i.e. it depends on the radial coordinate only, your integral becomes
$$m = int_-pi/2^+pi/2 int_0^2pi int_0^R rho(r) r^2cosvarphi dr dtheta dvarphi$$
which is
$$m = 2pi cdot int_-pi/2^+pi/2 cosvarphi dvarphi cdot int_0^R r^2 rho(r) dr = 4pi int r^3 rho(r) dr.$$
You can see that if $rho$ is constant you revert to the original formula, but now you may plug in whichever kind of radially-dependent mass density function and get the total mass.
answered Jul 29 at 10:09
giobrach
2,504418
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Mass = volume $times$ mass density
Volume of a sphere is $$ V=4/3 pi R^3 $$Where R is the radius.
You do not need triple integral to find the approximate mass of the earth.
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
1
down vote
Mass = volume $times$ mass density
Volume of a sphere is $$ V=4/3 pi R^3 $$Where R is the radius.
You do not need triple integral to find the approximate mass of the earth.
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Mass = volume $times$ mass density
Volume of a sphere is $$ V=4/3 pi R^3 $$Where R is the radius.
You do not need triple integral to find the approximate mass of the earth.
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
Mass = volume $times$ mass density
Volume of a sphere is $$ V=4/3 pi R^3 $$Where R is the radius.
You do not need triple integral to find the approximate mass of the earth.
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 9:45
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
up vote
1
down vote
You situation does not need triple integrals: see my comment. However, had you had information for a variable mass density (i.e. $rho$ as a function of position, generally non-constant) you would have had to use integration.
Suppose that Earth is a sphere with radius $R$. If it has mass density $rho : B_0(R) to mathbb R$ (where $B_0(R)$ is the ball of radius $R$ centered in the origin), then the total mass may be calculated as
$$ m = int_B_0(R) rho(x,y,z) dV .$$
Then you may change variables to spherical
$$begincases
x= rcosthetacosvarphi \
y = rsinthetacosvarphi \
z = rsinvarphi
endcases $$
so that the determinant of the transformation is
$$|J| = r^2cosvarphi. $$
If you assume that the mass density is spherically symmetric, i.e. it depends on the radial coordinate only, your integral becomes
$$m = int_-pi/2^+pi/2 int_0^2pi int_0^R rho(r) r^2cosvarphi dr dtheta dvarphi$$
which is
$$m = 2pi cdot int_-pi/2^+pi/2 cosvarphi dvarphi cdot int_0^R r^2 rho(r) dr = 4pi int r^3 rho(r) dr.$$
You can see that if $rho$ is constant you revert to the original formula, but now you may plug in whichever kind of radially-dependent mass density function and get the total mass.
answered Jul 29 at 10:09
giobrach
2,504418
add a comment |Â
up vote
1
down vote
You situation does not need triple integrals: see my comment. However, had you had information for a variable mass density (i.e. $rho$ as a function of position, generally non-constant) you would have had to use integration.
Suppose that Earth is a sphere with radius $R$. If it has mass density $rho : B_0(R) to mathbb R$ (where $B_0(R)$ is the ball of radius $R$ centered in the origin), then the total mass may be calculated as
$$ m = int_B_0(R) rho(x,y,z) dV .$$
Then you may change variables to spherical
$$begincases
x= rcosthetacosvarphi \
y = rsinthetacosvarphi \
z = rsinvarphi
endcases $$
so that the determinant of the transformation is
$$|J| = r^2cosvarphi. $$
If you assume that the mass density is spherically symmetric, i.e. it depends on the radial coordinate only, your integral becomes
$$m = int_-pi/2^+pi/2 int_0^2pi int_0^R rho(r) r^2cosvarphi dr dtheta dvarphi$$
which is
$$m = 2pi cdot int_-pi/2^+pi/2 cosvarphi dvarphi cdot int_0^R r^2 rho(r) dr = 4pi int r^3 rho(r) dr.$$
You can see that if $rho$ is constant you revert to the original formula, but now you may plug in whichever kind of radially-dependent mass density function and get the total mass.
answered Jul 29 at 10:09
giobrach
2,504418
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You situation does not need triple integrals: see my comment. However, had you had information for a variable mass density (i.e. $rho$ as a function of position, generally non-constant) you would have had to use integration.
Suppose that Earth is a sphere with radius $R$. If it has mass density $rho : B_0(R) to mathbb R$ (where $B_0(R)$ is the ball of radius $R$ centered in the origin), then the total mass may be calculated as
$$ m = int_B_0(R) rho(x,y,z) dV .$$
Then you may change variables to spherical
$$begincases
x= rcosthetacosvarphi \
y = rsinthetacosvarphi \
z = rsinvarphi
endcases $$
so that the determinant of the transformation is
$$|J| = r^2cosvarphi. $$
If you assume that the mass density is spherically symmetric, i.e. it depends on the radial coordinate only, your integral becomes
$$m = int_-pi/2^+pi/2 int_0^2pi int_0^R rho(r) r^2cosvarphi dr dtheta dvarphi$$
which is
$$m = 2pi cdot int_-pi/2^+pi/2 cosvarphi dvarphi cdot int_0^R r^2 rho(r) dr = 4pi int r^3 rho(r) dr.$$
You can see that if $rho$ is constant you revert to the original formula, but now you may plug in whichever kind of radially-dependent mass density function and get the total mass.
answered Jul 29 at 10:09
giobrach
2,504418
You situation does not need triple integrals: see my comment. However, had you had information for a variable mass density (i.e. $rho$ as a function of position, generally non-constant) you would have had to use integration.
Suppose that Earth is a sphere with radius $R$. If it has mass density $rho : B_0(R) to mathbb R$ (where $B_0(R)$ is the ball of radius $R$ centered in the origin), then the total mass may be calculated as
$$ m = int_B_0(R) rho(x,y,z) dV .$$
Then you may change variables to spherical
$$begincases
x= rcosthetacosvarphi \
y = rsinthetacosvarphi \
z = rsinvarphi
endcases $$
so that the determinant of the transformation is
$$|J| = r^2cosvarphi. $$
If you assume that the mass density is spherically symmetric, i.e. it depends on the radial coordinate only, your integral becomes
$$m = int_-pi/2^+pi/2 int_0^2pi int_0^R rho(r) r^2cosvarphi dr dtheta dvarphi$$
which is
$$m = 2pi cdot int_-pi/2^+pi/2 cosvarphi dvarphi cdot int_0^R r^2 rho(r) dr = 4pi int r^3 rho(r) dr.$$
You can see that if $rho$ is constant you revert to the original formula, but now you may plug in whichever kind of radially-dependent mass density function and get the total mass.
answered Jul 29 at 10:09
giobrach
2,504418
answered Jul 29 at 10:09
giobrach
2,504418
answered Jul 29 at 10:09
giobrach
2,504418
answered Jul 29 at 10:09
giobrach
2,504418
2,504418
add a comment |Â
add a comment |Â
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Why would you need triple integrals? You have an average radius and an average density, you can calculate the mass approximately as $$m approx frac 4 3 pi rho r^3. $$ Anything more would be overkill.
– giobrach
Jul 29 at 9:45