Triple integral to find the gravity inside a solid sphere

[pleasehelp12]

gravity inside a solid sphere

Homework Statement

I'm having a hard time setting up a triple integral to find the force of gravity inside a solid sphere. I've done a similar proof in physics before with gravity inside a spherical shell, but it only required a single integral. In this problem the answer must be derived using a triple integral.

Homework Equations

Newton's law of gravity:

|[tex]\vec{F_{grav}}[/tex]|= [tex]\frac{GmpdV}{r^{2}}[/tex]

The Attempt at a Solution

I believe i found a way to set up the triple integral using spherical coordinates on another physics forum thread, but I don't understand how to get the integrand. Can someone please explain the way the intergrand was derived in the following integral?

And, if you want to prove it mathematically using Newton's general law of gravitation and calculus techniques, here is most of it:
1. In spherical polar coordinates, let the position of a mass particle inside the ball be given by [itex](\hat{r},\theta,\phi)[/itex] (measured from the C.M)
where [itex]\phi[/itex] is the angle between the vertical and the particle's position vector.
Let the density be constant for all sphere particles and the radius of the sphere R.


2. Consider a test particle P having mass m and position vector [itex]r\vec{k}[/itex], i.e, a distance r along the "vertical"

3. We need to sum up all forces acting on P from sphere particles, i.e, compute the integral:
[tex]\vec{F}=-G\rho{m}\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\hat{r}^{2}\sin\phi((r-\hat{r}\cos\phi)\vec{k}-\hat{r}(\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j}))}{(\hat{r}^{2}+r^{2}-2r\hat{r}\cos\phi)^{\frac{3}{2}}}d\theta{d\phi}d\hat{r}[/tex]
where G is the universal gravitation constant and [itex]\rho[/itex] is the density of sphere particles.

4. It is easy to see that the horizontal plane components vanishes; the [itex]\phi[/itex]-integration is then best handled by integration by parts.
In the [itex]\hat{r}[/itex] integration, take care of whether you have [itex]r<\hat{r}[/itex] or [itex]r>\hat{r}[/itex]

 

[JOE SUPPLE BRUNS]

gravity inside a solid sphere

Homework Statement

I'm having a hard time setting up a triple integral to find the force of gravity inside a solid sphere. I've done a similar proof in physics before with gravity inside a spherical shell, but it only required a single integral. In this problem the answer must be derived using a triple integral.

Homework Equations

Newton's law of gravity:

|[tex]\vec{F_{grav}}[/tex]|= [tex]\frac{GmpdV}{r^{2}}[/tex]

The Attempt at a Solution

I believe i found a way to set up the triple integral using spherical coordinates on another physics forum thread, but I don't understand how to get the integrand. Can someone please explain the way the intergrand was derived in the following integral?

gravity inside a solid sphere

Homework Statement

I'm having a hard time setting up a triple integral to find the force of gravity inside a solid sphere. I've done a similar proof in physics before with gravity inside a spherical shell, but it only required a single integral. In this problem the answer must be derived using a triple integral.

Homework Equations

Newton's law of gravity:

|[tex]\vec{F_{grav}}[/tex]|= [tex]\frac{GmpdV}{r^{2}}[/tex]

The Attempt at a Solution

I believe i found a way to set up the triple integral using spherical coordinates on another physics forum thread, but I don't understand how to get the integrand. Can someone please explain the way the intergrand was derived in the following integral?

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html#wtls

 

[Farang]

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html#wtls

Only took 8 years...

 

[JOE SUPPLE BRUNS]

THANK FOR NOTICING . ALSO SEE GRAVITATION CHAPTER IN HALLIDAY RESNICK INTRO PHYSICS USED TEXTBOOKS

 

[JOE SUPPLE BRUNS]

THIS PROOF IS FOR INVERSE SQUARE MATH EQUATIONS.
DIVIDE BY DISTANCE r BETWEEN
CENTER OF MASS 1 AND
CENTER OF MASS 2
SQUARED.
YOUR WEIGHT F ON PLANET EARTH SURFACE IS
m1 = YOUR BODY MASS
m2 = PLANET EARTH MASS

r = 6378 KM = 4000 MILES = NOT 0 KM = NOT 0 MILES =
PLANET EARTH RADIUS + HEIGHT OF m1

NOW CALC F = WEIGHT

ALSO APPLIES TO ELECTRIC FORCE
INVERSE SQUARE MATH EQUATION

EXCEPT
GRAVITY IS ONLY ATTRACTION FORCE BUT
ELECTROSTATIC CAN BE
ATTRACTION FORCE OR
REPULSION FORCE

 

[Farang]

Has your shift button stuck or are you normally a loud person?

 

[TeethWhitener]

Only took 8 years...

Here's a link to the page on how to use the open practice problem forums:
https://www.physicsforums.com/threads/read-me-how-to-use-this-forum.855656/
It's a repository of old unanswered threads that are open for anyone to respond to, without the typical limitations that are generally placed on responses in the homework forums. Many of us in the forum like to solve these problems in our spare time. To keep our knives sharp, so to speak.

 

[berkeman]

Thread is closed.