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d bakker
science forum beginner
Joined: 05 Jun 2005
Posts: 1
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 Posted: Sun Jun 05, 2005 6:36 pm Post subject:
burning question about gravitation
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who can help me out of this inconsistency?
if you look at the gravitational force on a mass on the surface of the earth
you use:
Fz= G.m1.m2/r^2
If you redefine the problem and say: lets divide the earth in two parts and
sum the contribution of the parts then you should get the same gravitational
force, but my equality gives a different force.
Fz= G.m1.m2/(2.(r-d)^2) + G.m1.m2/(2.(r+d)^2) where d is the distance
between the center of gravity of the halfsphere and the midpoint of the
earth.
What am i doing wrong?
Door |
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N:dlzc D:aol T:com (dlzc)
science forum Guru
Joined: 25 Mar 2005
Posts: 2835
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| Posted: Sun Jun 05, 2005 7:54 pm Post subject:
Re: burning question about gravitation
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Dear d bakker:
"d bakker" <bakker.d@hccnet.nl> wrote in message
news:42a36298$0$154$3a628fcd@reader1.nntp.hccnet.nl...
| Quote: | who can help me out of this inconsistency?
if you look at the gravitational force on a mass
on the surface of the earth you use:
Fz= G.m1.m2/r^2
If you redefine the problem and say: lets divide
the earth in two parts and sum the contribution
of the parts then you should get the same
gravitational force, but my equality gives a
different force.
Fz= G.m1.m2/(2.(r-d)^2) + G.m1.m2/(2.(r+d)^2)
where d is the distance between the center of
gravity of the halfsphere and the midpoint of the
earth.
What am i doing wrong?
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Maybe how you arrived at "center of gravity".
David A. Smith |
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tadchem
science forum Guru
Joined: 03 May 2005
Posts: 1348
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| Posted: Sun Jun 05, 2005 10:41 pm Post subject:
Re: burning question about gravitation
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"d bakker" <bakker.d@hccnet.nl> wrote in message
news:42a36298$0$154$3a628fcd@reader1.nntp.hccnet.nl...
| Quote: | who can help me out of this inconsistency?
if you look at the gravitational force on a mass on the surface of the
earth
you use:
Fz= G.m1.m2/r^2
If you redefine the problem and say: lets divide the earth in two parts
and
sum the contribution of the parts then you should get the same
gravitational
force, but my equality gives a different force.
Fz= G.m1.m2/(2.(r-d)^2) + G.m1.m2/(2.(r+d)^2) where d is the distance
between the center of gravity of the halfsphere and the midpoint of the
earth.
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Your equations assume that each 'halfsphere' is spherically symmetric.
| Quote: | What am i doing wrong?
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Operating a human brain without the proper license...
Tom Davidson
Richmond, VA |
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PD
science forum Guru
Joined: 03 May 2005
Posts: 4363
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| Posted: Sun Jun 05, 2005 11:48 pm Post subject:
Re: burning question about gravitation
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d bakker wrote:
| Quote: | who can help me out of this inconsistency?
if you look at the gravitational force on a mass on the surface of the earth
you use:
Fz= G.m1.m2/r^2
If you redefine the problem and say: lets divide the earth in two parts and
sum the contribution of the parts then you should get the same gravitational
force, but my equality gives a different force.
Fz= G.m1.m2/(2.(r-d)^2) + G.m1.m2/(2.(r+d)^2) where d is the distance
between the center of gravity of the halfsphere and the midpoint of the
earth.
What am i doing wrong?
Door
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F = G*m1*m2/r^2 assumes spherically symmetric m1 and m2. You don't have
that for your m1's anymore, and so the force law must be modified. Not
to worry. You assume that each half-sphere is assembled from a bunch of
infinitesimals, each a different distance from m2, and for which the
law you're using works, and then you add up the forces from all the
infinitesimals. To do the sum requires an integral. Are you up for
that?
(This is actually interesting, because the form of the function will be
different depending on whether m2 is hovering over the convex part of
the hemisphere or over the flat part of the hemisphere, and yet when
you add up the contributions from both halves, m2 hovering over the
convex part of one and over the flat part of the other, you need to
recover the simple law above. That is, the asymmetric term must cancel
out. Moreover, that cancellation must occur regardless of the azimuth
of m2's location with respect to the plane between the two hemispheres!
It's almost worth doing the integrals to see that magic happen!)
PD |
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