Go Back   Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Nov 4th 2013, 06:16 AM   #1
Junior Member
 
Join Date: Nov 2013
Posts: 4
calculus of gravitational force

Good morning,

in cases where it is not possible to neglect the extension of an irregularly shaped body as we can calculate the gravitational force exerted on the earth of it? Of course the body will be large enough that it can not consider the local Earth's field as a field of parallel forces.
I know that we should integrate over the whole mass of the body, but this is in general rather complex. There is a more practical approach? for example by successive approximations?

thank you
jjorg is offline   Reply With Quote
Old Nov 4th 2013, 08:38 AM   #2
MBW
Senior Member
 
MBW's Avatar
 
Join Date: Apr 2008
Location: Bedford, England
Posts: 668
Integration IS a succession of approximations, taken to the limit as the approximation approaches zero.
Thus the answer is Yes,
breaking the shape of the body down into a number of smaller, simpler, shapes, is indeed possible,
And can definitely be more practical in many circumstances.
MBW is offline   Reply With Quote
Old Nov 5th 2013, 03:29 AM   #3
Junior Member
 
Join Date: Nov 2013
Posts: 4
Thank you.
A question:
consider a hypothetical very large space station (like those in Star Wars's movies...). A station of a (necessarily) irregular shape.
Now, how do you calculate the earth gravitational force?
jjorg is offline   Reply With Quote
Old Nov 6th 2013, 07:50 AM   #4
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,287
To get the exact answer yuo would need to integrate over the volume of the object:


For a spherical object (like the Death Star or the earth) with homogeous mass distribution you can simply use the familiar formula of GMm/r^2. And if the distance between the objects is great enough for all practical purposes you can use the familiar formula with little error. It's really only if you have an irregular shaped satellite whose orbital radius is similar order of magnituide as the size of the satellite that you would need to worry about such details.
ChipB is offline   Reply With Quote
Old Dec 22nd 2013, 09:56 AM   #5
Pmb
Physics Team
 
Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,533
Originally Posted by ChipB View Post
For a spherical object (like the Death Star or the earth) with homogeous mass distribution you can simply use the familiar formula of GMm/r^2.
If you wish to find the force on a spherical body due to the field of a spherical planet then you still have to integrate over the spherical body. With homogeous mass distribution the mass density merely is pulled through the integral sign. Just because the body is spherical it doesn't mean that you can treat it as if it was a point object in a gravitational field. I believe that you're thinking of the shell theorem. If so then I think that you're misapplying it.

Last edited by Pmb; Dec 22nd 2013 at 10:01 AM.
Pmb is offline   Reply With Quote
Old Dec 22nd 2013, 02:01 PM   #6
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,287
Originally Posted by Pmb View Post
I believe that you're thinking of the shell theorem. If so then I think that you're misapplying it.
Yes I am thinking of the shell theorom, which applies to a point outside a homogenous sphere. See http://en.wikipedia.org/wiki/Shell_theorem

It would seem to me to apply equally to the sum of all points of both spherical objects in orbit about each other. If you disagree I'd like to see your analysis.
ChipB is offline   Reply With Quote
Old Dec 22nd 2013, 03:07 PM   #7
Pmb
Physics Team
 
Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,533
Originally Posted by ChipB View Post
Yes I am thinking of the shell theorom, which applies to a point outside a homogenous sphere. See http://en.wikipedia.org/wiki/Shell_theorem

Yes. That's the theorem I was talking about when I made that comment.

This isnít a simple problem. There are a few intricacies to consider and Iím not 100% sure of what the result is. To make sure I wasnít making a correction without being certain I contacted a friend of mine who has a great deal more expertise in gravitational physics than I do just to make doubly sure. He seems to agree with me. However he had the same idea that I had and thatís to consider what Newtonís Law would say with respect to this. So right now Iím still not certain. So let me spell out where Iím at;

When I read this thread I first assumed you were right. It seemed quite obvious to me given the shell theorem. However the more I thought about it the more I doubted my original assumption. I recalled exactly what the theorem states Ė The gravitational force exerted on a point object outside of a spherically symmetric distribution of matter is the same as the gravitational force exerted by a point object having the same mass. Or stated another way, the gravitational field outside of a spherically symmetric distribution of matter is the same as if the source was a point object located at the geometric center of the distribution.

Does this mean the opposite is true as well? Newtonís third law would seem to imply it does. However I donít fully trust Newtonís third law. Itís not a true law since it doesnít always hold true. There are examples where it fails, most notably when one tries to apply it to the force between two charges in motion. In this case weíre trying to apply it to field theory and Iím not 100% convinced that it holds true.

Consider this Ė Let our gravitational source of mass M be at the origin of our coordinate system at (0,0,0). Let the center of the sphere, which has a radius R, be located at a distance z[sub]c[/sub] from the center of the gravitational source on the z-axis. The center of the sphere would try to accelerate at the rate of g = GM/z[sub]c[/sub]^2. One side of the sphere would be a distance z[sub]c[/sub] Ė R from the origin so that this side of the sphere would try to accelerate at the rate of g = GM/(z[sub]c[/sub] Ė R)^2. The other side of the sphere would be at a location z[sub]c[/sub] + R and would to accelerate at the rate of g = GM/(z[sub]c[/sub] + R)^2. This means that tidal forces are set up inside the sphere. The total force acting on the sphere is then found by integrating these forces.

This is not a simple thing to integrate. I havenít tried it yet and Iím not looking forward to it but I canít imagine that itíd be an easy thing to do. However it would be instructive.

Before I responded to this thread I contacted a friend of my who has a great deal more experience in gravitational physics than I do. at first he said the same thing that you did. Then I asked him the following question
I thought that only applied to how a sphere acts like a point source of gravity, not how the sphere responds to a point source. E.g. part of the sphere is at its center so that part would accelerate in a g-field of a point source as accelerating at a rate of g = GM/r^2. However other portions of the sphere are closer to the source and want to accelerate faster while other portions of the sphere are located further away and will want to accelerate slower. So why does it want to accelerate as if it was all condensed at the center?


When he read that he wrote back and said
Ah, I misunderstood your question. As to your question concerning the exact response of a spherical body to the gravitational field of a point mass, I think we may reason in the following way.
Newton's theory is a linear theory. Hence the gravitational acceleration at the center of the sphere is equal to the sum of the gravitational field due to its own mass and that of the point particle. Assuming that the sphere is rigid so that it is not deformed due to tidal forces, the acceleration of gravity at the center of the sphere due to its own mass vanishes. Hence the acceleration of the sphere is equal to the integral of the force of gravity on all mass elements of the sphere in the field of the point mass divided by the mass of the sphere. Only the component along the line connecting the center of the sphere with the point mass matters. It is an interesting mathematical exercise to calculate this integral exactly. I have not seen it, and we have a lot of Christmas program in the family so I have no possibility of trying to perform it during the coming week. I you manage to do it I would be happy to see the calculation. It would be strange if no one has performed it.

However he also sent the following comment
Newton's 3. Law gives the expected answer to the integration.


So I'm going to give it a try and see what I get.
Pmb is offline   Reply With Quote
Old Dec 22nd 2013, 07:19 PM   #8
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,287
To me it seems quite straight forward: as long as the "other body" is spherical and homogenous it may as well be a point mass.

Imagine two spherical homogenous bodies revolving in orbit about each other. The shell theorem states that body A is subjected to gravitational force from body B as if B was a point mass, and also that body B is subjected to force from body A as if A was a point mass. Therefore each acts as if the other was a point mass. So it doesn't matter how big one is relative to the other, or how far apart they are - both feel a net force from the other that's equivalent to both being point masses.

I do agree that this is a bit simplistic in that it only addresses net forces on each body, and not the differential forces from near side to far side (i.e. tidal forces).
ChipB is offline   Reply With Quote
Old Dec 23rd 2013, 06:35 PM   #9
Pmb
Physics Team
 
Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,533
Originally Posted by ChipB
To me it seems quite straight forward: as long as the "other body" is spherical and homogenous it may as well be a point mass.
It may seem straight forward but it wasn't easy for Newton to prove this. We simply have the benefit of Newton's hindsight.

Let me state this for prosperity - In a region containing only a gravitational field and no mass, a sphere falls with the same acceleration as it would have if it were a pointlike particle. Proving this was one of the difficulties that Newton had to solve before publishing his treatment of planetary motion.

This is known as Newton's Theorem

A friend of mine is a text book author. He wrote one of those freshman physics textbooks as well as special relativity and modern physics texts and an advanced general relativity text. He pointed me in the right direction. He tells me that it's simple to prove by general theorems of potential theory. I'm going to try to solve this problem when I'm feeling better.

Last edited by Pmb; Dec 30th 2013 at 07:39 AM.
Pmb is offline   Reply With Quote
Old Dec 25th 2013, 03:09 PM   #10
Pmb
Physics Team
 
Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,533
This is odd. I asked two different friends of mine each of whom teaches graduate mechanics and I got two different responses. The one who agrees with my original assumption responded with an example similar to the one I gave. He wrote
A spherically symmetric mass distribution does produce, according to Gauss' law, the same field as a point charge of the same mass located at its center. However the rate of fall of a spherically symmetric mass towards a point mass is not he same as the rate of fall of a point mass at the center of the sphere. A simple thought experiment can show this to be the case. Imagine a pair of point masses m/2separated by a rigid massless rod, of length d. Let the axis that joins the 2 masses be aligned along a radius r from a fixed point mass M. Suppose the pair were released from rest at infinity. Then the velocity at r would be v = SQRT(2GMr/(r-d/2)^2). Were the two at their midpoint, that is, d=0, the velocity would be v = SQRT(2GM/r). Clearly the separated pair fall at a faster rate than an equivalent point mass at their mid point.
I don't understand his calculation though. Here's what I get for the same example;

The total energy of the dumbbell when aligned along the z-axis is given by the sum of the kinetic energy of each dumbbell end and the sum of the potential energies of each end. If the dumbbell starts from rest at infinity we have the total energy equal to zero. Therefore E = K + V = 0. Or


(1/2)mv^2 + (1/2)mv^2 - GMm/(r - d/2)^2 - GM/(r + d/2)^2 = 0

or upon solving for v

v = sqrt(GM[1/(r - d/2)^2 + 1/(r + d/2)^2])

This is different than my friend got. Can someone tell me where I went wrong? Thank you.

Last edited by Pmb; Dec 30th 2013 at 07:23 AM.
Pmb is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Tags
calculus, force, gravitational



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Resistive Force Calculus Help Jokerpoker Kinematics and Dynamics 20 Dec 21st 2013 06:46 PM
Gravitational Force? mushroomranger Kinematics and Dynamics 4 Mar 21st 2012 04:23 PM
gravitational force wikianswers Kinematics and Dynamics 3 Jan 13th 2011 08:32 PM
gravitational force Jwd41190 Kinematics and Dynamics 1 Oct 9th 2009 12:13 AM
Gravitational force hana1 Kinematics and Dynamics 3 Jun 27th 2009 03:17 AM


Facebook Twitter Google+ RSS Feed