Markus
Dec24-04, 06:47 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Thank you for your answer. Yes, I think you are right. The center of\ngravity is probably the correct notion here. However, I still have\nconfused ideas on the subject. What I\'m trying to understand is which\npoint is it, in an extended object in a (classical) gravity field,\nwhere we have zero tidal perturbation?\n\nMore precisely, consider an extended irregular body (not a sphere or\nsome object with symmetries, say for instance a cloud of particles) in\na gravity field (not a spherical either, say the gravitational field\nof an irregular galaxy). This field exerts a tidal perturbation on\nthis "cloud" of particles. Now, we know that tidal forces are the\nmanifestation of the gradient of the gravitational field. This is\nhowever a "differential" definition. I\'m trying to look at it from an\n"integral" point of view as: F_{t}(R) = F(R)-<F>, where F_{t}(R)\nstands for the tidal force acting upon a particle of the body\nundergoing that perturbation in some position R, F(R) the\ngravitational force acting in R, and <F> the mean gravitational force\nacting upon the whole body. I want to reduce <F> to a quantity\nequivalent to a force in some precise point internal to the body\nwithout loss of generality, i.e. to <F> = F(point in space). That\n"point in space" should be that where tidal forces are zero.\n\nThen, from the above discussion should we infere that the mean\ngravitational force acting on the tidally perturbed system is per\ndefinition equivalent to the force acting in its center of gravity\n(the average location of the weight, which in case of uniform gravity\nfields, or first order approximations coincides with the center of\nmass), and that this is also the place where we have zero tidal force?\n\nI\'m not at all sure.... can you or someone else furnish some further\nhints?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Thank you for your answer. Yes, I think you are right. The center of
gravity is probably the correct notion here. However, I still have
confused ideas on the subject. What I'm trying to understand is which
point is it, in an extended object in a (classical) gravity field,
where we have zero tidal perturbation?
More precisely, consider an extended irregular body (not a sphere or
some object with symmetries, say for instance a cloud of particles) in
a gravity field (not a spherical either, say the gravitational field
of an irregular galaxy). This field exerts a tidal perturbation on
this "cloud" of particles. Now, we know that tidal forces are the
manifestation of the gradient of the gravitational field. This is
however a "differential" definition. I'm trying to look at it from an
"integral" point of view as: F_{t}(R) = F(R)-<F>, where F_{t}(R)
stands for the tidal force acting upon a particle of the body
undergoing that perturbation in some position R, F(R) the
gravitational force acting in R, and <F> the mean gravitational force
acting upon the whole body. I want to reduce <F> to a quantity
equivalent to a force in some precise point internal to the body
without loss of generality, i.e. to <F> = F(point in space). That
"point in space" should be that where tidal forces are zero.
Then, from the above discussion should we infere that the mean
gravitational force acting on the tidally perturbed system is per
definition equivalent to the force acting in its center of gravity
(the average location of the weight, which in case of uniform gravity
fields, or first order approximations coincides with the center of
mass), and that this is also the place where we have zero tidal force?
I'm not at all sure.... can you or someone else furnish some further
hints?
gravity is probably the correct notion here. However, I still have
confused ideas on the subject. What I'm trying to understand is which
point is it, in an extended object in a (classical) gravity field,
where we have zero tidal perturbation?
More precisely, consider an extended irregular body (not a sphere or
some object with symmetries, say for instance a cloud of particles) in
a gravity field (not a spherical either, say the gravitational field
of an irregular galaxy). This field exerts a tidal perturbation on
this "cloud" of particles. Now, we know that tidal forces are the
manifestation of the gradient of the gravitational field. This is
however a "differential" definition. I'm trying to look at it from an
"integral" point of view as: F_{t}(R) = F(R)-<F>, where F_{t}(R)
stands for the tidal force acting upon a particle of the body
undergoing that perturbation in some position R, F(R) the
gravitational force acting in R, and <F> the mean gravitational force
acting upon the whole body. I want to reduce <F> to a quantity
equivalent to a force in some precise point internal to the body
without loss of generality, i.e. to <F> = F(point in space). That
"point in space" should be that where tidal forces are zero.
Then, from the above discussion should we infere that the mean
gravitational force acting on the tidally perturbed system is per
definition equivalent to the force acting in its center of gravity
(the average location of the weight, which in case of uniform gravity
fields, or first order approximations coincides with the center of
mass), and that this is also the place where we have zero tidal force?
I'm not at all sure.... can you or someone else furnish some further
hints?