the "speed" of pedestrians on the brooklyn bridge
Uno
policeman who was adding in the radial velocity of the earth to his
calculations. I didn't think about much until Tony Bologna and crew
decided to arrest 700 Americans on the Brooklyn bridge.
I've joked on facebook that nasa will cover this event before network
news will, and depending on respondent, that could become true literally.
$ gfortran -Wall -Wextra brook1.f90 -o out
$ ./out
pi is 0.78539819
rotation of earth at radius in mph is 1020.0054
w.r.t. center of planet, speed of any point in brooklyn is
179.85957
$ gfortran -Wall -Wextra brook1.f90 -o out
$ ./out
pi is 3.1415927
rotation of earth at radius in mph is 1020.0054
w.r.t. center of planet, speed of any point in brooklyn is
774.17438
$ cat brook1.f90
implicit none
real :: laditude, speed1, speed2, pi
pi = 4.0 * atan(1.0)
print *, "pi is ", pi
speed1 = 456.1 !m/s
speed1 = speed1 * 3600 !m/hr
speed1 = speed1 * 3.28 / 5280 !mi/hr
print *, "rotation of earth at radius in mph is ", speed1
laditude = 40.624722 ! north latitude for brooklyn in degrees
laditude = laditude * pi /180.0 !radians
speed2 = speed1 * cos(laditude)
print *, "w.r.t. center of planet, speed of any point in brooklyn is ",
speed2
end program
! gfortran -Wall -Wextra brook1.f90 -o out
$
What I realized is that it makes a huge difference in this type of
hyperbole which way your "street" is pointing. A north-south street is
unaffected, while an east-west one feels its full brunt. Are intenrnet
map capabilities sufficiently advanced that we can know what angle the
brooklyn bridge goes at. By my eye, it could be 20% w.r.t. east west.
q1) Can you determine the inclination of the brooklyn bridge?
q2) does fortran have a += operator like perl or c?
Where were you for the revolution? The winners are going to ask the
loser a lot of questions: they'll be a captive audience.
--
Uno
glen herrmannsfeldt
> I went to comedy gig on thursday, where the comic talked about a
> policeman who was adding in the radial velocity of the earth to his
> calculations. I didn't think about much until Tony Bologna and crew
> decided to arrest 700 Americans on the Brooklyn bridge.
Reminds me about reading not so long ago, in the early days
of Newtonian gravity there was question about the inverse
square law as regards the moon. It seesm that the orbit of the
moon was measurably wrong, and at first the solution was to
blame the inverse square law. In the end, it turned out to be
due to the slight oblate spheroidal (fatter at the equator) shape
of the earth that wasn't considered in the calculation.
Anyway, after you include the tangential velocity for the earth's
rotation you next add the velocity from its orbit around the sun,
and then the sun around the CM of the milky way.
-- glen
Uno
> Uno<U...@example.invalid> wrote:
>> I went to comedy gig on thursday, where the comic talked about a
>> policeman who was adding in the radial velocity of the earth to his
>> calculations. I didn't think about much until Tony Bologna and crew
>> decided to arrest 700 Americans on the Brooklyn bridge.
>
> That is tangential velocity. Radial velocity is up or down.
this region was the radial velocity of building's 7 demise. This is better:
$ gfortran -Wall -Wextra brook1.f90 -o out
$ ./out
pi is 3.1415927
w.r.t. CM planet, speed of any point in brooklyn is 774.17438
implicit none
real :: laditude, speed1, speed2, pi
pi = 4.0 * atan(1.0)
print *, "pi is ", pi
speed1 = 456.1 !m/s
speed1 = speed1 * 3600 !m/hr
speed1 = speed1 * 3.28 / 5280 !mi/hr
laditude = laditude * pi /180.0 !radians
speed2 = speed1 * cos(laditude)
! gfortran -Wall -Wextra brook1.f90 -o out
$
>
> of Newtonian gravity there was question about the inverse
> square law as regards the moon. It seesm that the orbit of the
> moon was measurably wrong, and at first the solution was to
> blame the inverse square law. In the end, it turned out to be
> due to the slight oblate spheroidal (fatter at the equator) shape
> of the earth that wasn't considered in the calculation.
ask you this, glen, what "causes" this bulge? I used to think it had to
do with forces that held "us" in orbit as we spin around on our axis,
and I thought of these forces as between us and the sun.
The fact that this phenomenon induced the "error" you mention makes me
wonder if I have it all backwards. (I can hear Richard sharpening up
the knife right there.)
How oblate would the earth be if the moon were just booted away by some
celestial jokester?
> Anyway, after you include the tangential velocity for the earth's
> rotation you next add the velocity from its orbit around the sun,
> and then the sun around the CM of the milky way.
galactic boondocks moreso than at its center. Cheers from NM.
--
Uno
"And it all started with a big bang." I got on a billboard talking
about this today. San Mateo and Montgomery.
glen herrmannsfeldt
(snip, I wrote)
>> of Newtonian gravity there was question about the inverse
>> square law as regards the moon. It seesm that the orbit of the
>> moon was measurably wrong, and at first the solution was to
>> blame the inverse square law. In the end, it turned out to be
>> due to the slight oblate spheroidal (fatter at the equator) shape
>> of the earth that wasn't considered in the calculation.
> I saw this value 0.0033528 associated with earth's flattening. Let me
> ask you this, glen, what "causes" this bulge? I used to think it had to
> do with forces that held "us" in orbit as we spin around on our axis,
> and I thought of these forces as between us and the sun.
> The fact that this phenomenon induced the "error" you mention makes me
> wonder if I have it all backwards. (I can hear Richard sharpening up
> the knife right there.)
> How oblate would the earth be if the moon were just booted away by some
> celestial jokester?
The ellipse axis slowly rotates, and that rate is affected
by the earth shape.
>> Anyway, after you include the tangential velocity for the earth's
>> rotation you next add the velocity from its orbit around the sun,
>> and then the sun around the CM of the milky way.
> and then the frilliness of the dark matter that affects us here in the
> galactic boondocks moreso than at its center. Cheers from NM.
Daniel Carrera
> I saw this value 0.0033528 associated with earth's flattening. Let me
> ask you this, glen, what "causes" this bulge? I used to think it had to
> do with forces that held "us" in orbit as we spin around on our axis,
> and I thought of these forces as between us and the sun.
Like the other planets, Earth has enough mass to be in hydrostatic
equilibrium. That means that Earth's shape is almost perfectly
determined by the net forces acting on it, and the primary ones are
Earth's self-gravity and its rotation.
Due to Earth's rotation, the "effective" gravitational force that you
feel at Earth's equator is about 0.5% lower than at the poles, and
Earth's shape reflects this.
All planets have some amount of flattening around their rotational axis,
and it has nothing to do with the Sun really. The tidal forces of the
Sun on Earth are small; the moon would have a much stronger effect, and
it is still secondary to Earth's rotation.
> How oblate would the earth be if the moon were just booted away by some
> celestial jokester?
Earth's shape is very tiny; much smaller than the effect of Earth's
rotation.
Cheers,
Daniel.
Daniel Carrera
> The mantle is fluid, so centrifugal force from the rotation.
equilibrium. There comes a point where an object has enough mass that
the only shape it can have is the one determined by self gravity and
rotation.
Daniel.
Uno
> In brief, because of Earth's rotation.
>
> Like the other planets, Earth has enough mass to be in hydrostatic
> equilibrium. That means that Earth's shape is almost perfectly
> determined by the net forces acting on it, and the primary ones are
> Earth's self-gravity and its rotation.
in geology and didn't get past integrating a star in stellar evolution.
Ok hydrostatic equilibrium makes complete sense here. (I have to
wonder what that means for sunspots and volcanoes, though. Are they
insignificant?)
> Due to Earth's rotation, the "effective" gravitational force that you
> feel at Earth's equator is about 0.5% lower than at the poles, and
> Earth's shape reflects this.
> All planets have some amount of flattening around their rotational axis,
> and it has nothing to do with the Sun really. The tidal forces of the
> Sun on Earth are small; the moon would have a much stronger effect, and
> it is still secondary to Earth's rotation.
like a matter of philosophy, and I'm having trouble with the chicken and
the egg here.
I know I have fortran source for determining g on or above the earth,
but I think the input is in terms of the surface, and is has that
(b**2/a**2) correction which seemed sufficient for first order
approximations.
>
>> How oblate would the earth be if the moon were just booted away by some
>> celestial jokester?
>
> It would be basically the same that it is today. The moon's effect on
> Earth's shape is very tiny; much smaller than the effect of Earth's
> rotation.
will think blowing up the moon is a good idea. I'll worry less for them.
Peace.
--
Uno
Uno
and would conjecture that we're talking about ordinary temperatures,
ordinary substances, without a supernova happening nearby.
I'm impressed by how many objects look like they've been pummeled
forever, but let me ask you about helene, classified as a moon. Is
Helene in hydrostatic equilibrium? Is our moon?
--
Uno
Daniel Carrera
> On 10/3/2011 11:47 PM, Daniel Carrera wrote:
>> On 10/04/2011 03:04 AM, glen herrmannsfeldt wrote:
>>> The mantle is fluid, so centrifugal force from the rotation.
>>
>> Even planets whose mantles have cooled down will be in hydrostatic
>> equilibrium. There comes a point where an object has enough mass that
>> the only shape it can have is the one determined by self gravity and
>> rotation.
>
> When you speak of "a point," I assume you mean "a set of conditions,"
> and would conjecture that we're talking about ordinary temperatures,
> ordinary substances, without a supernova happening nearby.
require more mass for gravity to overcome stiffness, but for any
substance there will come a point where the object becomes round by self
gravity.
Supernovae do not destroy planets... Their radiation might kill all life
in a planet, but they do not "push" the planet. If the nearest star to
the Sun exploded as a supernova, Earth wouldn't notice (except for
animals dying from radiation).
> I'm impressed by how many objects look like they've been pummeled
> forever, but let me ask you about helene, classified as a moon. Is
> Helene in hydrostatic equilibrium? Is our moon?
is. You can see that Helene looks potato-shaped and our moon is round.
As chance would have it, we did an exercise about this last week (I'm
doing a masters in astrophysics). For ordinary materials things become
round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
while Helene is 43 km along its longest dimension.
Cheers,
Daniel.
Uno
> q1) Can you determine the inclination of the brooklyn bridge?
hand in a problem that you wanted credit for. One of the D's is Diagram:
http://i52.tinypic.com/2a4zivb.jpg
and one of the steps is just to estimate it, but I'm stuck in a variety
of places here:
$ gfortran -Wall -Wextra brook2.f90 -o out
pi is 3.1415927
w.r.t. CM planet, speed of any point in brooklyn is 774.17438
40.708096 73.999672
pretty close to 74 west
40.701622 73.991112
less W: a good thing
$ cat brook2.f90
implicit none
real :: latitude, speed1, speed2, pi
real :: degree, minute, second, north, west
real, dimension(2) :: ba, zinga
pi = 4.0 * atan(1.0)
print *, "pi is ", pi
speed1 = 456.1 !m/s
speed1 = speed1 * 3600 !m/hr
speed1 = speed1 * 3.28 / 5280 !mi/hr
latitude = 40.624722 ! north latitude for brooklyn in degrees
latitude = latitude * pi /180.0 !radians
speed2 = speed1 * cos(latitude)
print *, "w.r.t. CM planet, speed of any point in brooklyn is ", speed2
! data entry
degree = 40
minute = 42
second = 29.15
north = degree + minute/60.0 + second/3600.0
degree = 73
minute = 59
second = 58.83
west = degree + minute/60.0 + second/3600.0
ba(1) = north
ba(2) = west
print *, ba
print *, "pretty close to 74 west"
degree = 40
minute = 42
second = 5.84
north = degree + minute/60.0 + second/3600.0
degree = 73
minute = 59
second = 27.99
west = degree + minute/60.0 + second/3600.0
zinga(1) = north
zinga(2) = west
print *, zinga
print *, "less W: a good thing"
end program
! gfortran -Wall -Wextra brook2.f90 -o out
$
I'm looking for some better notation here (tried to find stuff in mrc),
a subroutine to convert degrees to real radians, and a means to compare
this vector to the meridian that bisects it.
--
Uno
Daniel Carrera
> Thx, daniel, forgive my ignorance here, as I've had no formal training
> in geology and didn't get past integrating a star in stellar evolution.
> Ok hydrostatic equilibrium makes complete sense here. (I have to wonder
> what that means for sunspots and volcanoes, though. Are they
> insignificant?)
mountains. Mountains happen on Earth, but as you said, they are pretty
insignificant. Everest is about 9 km high, while the radius of the Earth
is over 6,000 km. If you were a giant so to you the Earth was the size
of a beach ball, it would be the smoothest, most perfect sphere you had
ever seen.
>> Due to Earth's rotation, the "effective" gravitational force that you
>> feel at Earth's equator is about 0.5% lower than at the poles, and
>> Earth's shape reflects this.
>
> ok. do you have equations?
Earth's equatorial radius is R = 6378 km.
Earth rotates in 24h, so its equatorial speed is:
V = 2 pi R / 24h = 463 m/s.
Thus, the centripetal acceleration must be:
a = v^2 / R = 0.033730 m/s^2
Earth's gravity is g = 9.8 m/s^2. Divide those two and you get:
a/g = 0.00344
So it's more like 0.34% of the gravity of the Earth at the equator. So,
on the equator you are 0.34% lighter than if the Earth was not rotating.
>> All planets have some amount of flattening around their rotational axis,
>> and it has nothing to do with the Sun really. The tidal forces of the
>> Sun on Earth are small; the moon would have a much stronger effect, and
>> it is still secondary to Earth's rotation.
>
> Can you show me some of the math on this? Without equations, it is more
> like a matter of philosophy, and I'm having trouble with the chicken and
> the egg here.
difference in the gravitational pull (e.g. from the Sun) across the
Earth. Gravitational force is:
F = G M m / R^2
Earth's orbital radius is 150 million km, and Earth's radius is 6,370
km. So the force on the side closer to the Sun, written as an
acceleration is:
a1 = G M / (150 million km - 6,370 km)^2 = 0.0058997 m/s^2
On the side away from the Sun the acceleration is:
a2 = G M / (150 million km + 6,370 km)^2 = 0.0058987 m/s^2
The difference is:
a1 - a2 = 0.000001 m/s^2
Divide this by two, and you get the force that you feel pulling out away
from the centre of the Earth. This is 67,000 times smaller than the
R = 3.8 x 10^8 m
M = 7.3 x 10^22 kg
Do the same calculation and you get:
a1 - a2 = 0.00000226 m/s^2
In other words, the moon's tidal force is 2.26 times stronger than the
tidal force from the Sun, but it is still 30,000 times weaker than the
Daniel.
Daniel Carrera
>> On 10/04/2011 03:04 AM, glen herrmannsfeldt wrote:
>>> The mantle is fluid, so centrifugal force from the rotation.
>>
>> Even planets whose mantles have cooled down will be in hydrostatic
>> equilibrium. There comes a point where an object has enough mass that
>> the only shape it can have is the one determined by self gravity and
>> rotation.
>
> When you speak of "a point," I assume you mean "a set of conditions,"
Yes, by "point" I mean "situation".
> and would conjecture that we're talking about ordinary temperatures,
> ordinary substances, without a supernova happening nearby.
AFAICT what I said is true for any substance. Some substances will
require more mass for gravity to overcome stiffness, but for any
substance there will come a point where the object becomes round by self
gravity.
Supernovae do not destroy planets... Their radiation might kill all life
in a planet, but they do not "push" the planet. If the nearest star to
the Sun exploded as a supernova, Earth wouldn't notice (except for
animals dying from radiation).
> I'm impressed by how many objects look like they've been pummeled
> forever, but let me ask you about helene, classified as a moon. Is
> Helene in hydrostatic equilibrium? Is our moon?
Uno
> Helene is not in hydrostatic equilibrium, but our moon most definitely
> is. You can see that Helene looks potato-shaped and our moon is round.
>
> As chance would have it, we did an exercise about this last week (I'm
> doing a masters in astrophysics). For ordinary materials things become
> round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
> while Helene is 43 km along its longest dimension.
>
gravity and heat transfer? Looks pretty static to me. (Equations help.)
--
Uno
Daniel Carrera
>> As chance would have it, we did an exercise about this last week (I'm
>> doing a masters in astrophysics). For ordinary materials things become
>> round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
>> while Helene is 43 km along its longest dimension.
>
> How are the forces not resolved on Helene, in particular those due to
> gravity and heat transfer? Looks pretty static to me. (Equations help.)
are thinking. It means that the force of gravity exactly balanced out by
pressure. The term comes from fluid dynamics. I think the idea is that
if all the forces are balanced out in the fluid, then the fluid is
"static" in that it is not being accelerated.
Imagine that I buy a ring and throw it into space. The ring will feel a
tiny gravitational pull, trying to pull the ring material into the
centre. In practice the ring would not change shape because its gravity
is very tiny compared to the stiffness of the material. The ring would
not be in hydrostatic equilibrium because it doesn't have a pressure
force that is balancing out gravity. The ring has essentially no
pressure; it is the stiffness of the material that keeps its shape.
Now make the ring 6,000 km in size and even the stiffest metal will give
in to gravity and the material will fall and make a ball of mass. Now it
is in hydrostatic equilibrium, because now it's pressure that is
balancing gravity.
For planets, moons and stars, hydrostatic equilibrium means "enough mass
to be round".
Cheers,
Daniel.
Uno
I thought Helene looked like the elephant man, and I was disappointed
for my friend helene, who is an attractive new york actress, made all
the more so by her interest in science, which she may have gotten from
her dad, a mathematician. Anyways, I was thinking there could be a way
to photograph helene with some more sympathetic lighting or something.
Has anyone post-processed those images?
--
Uno
Uno
> Helene is not in hydrostatic equilibrium, but our moon most definitely
> is. You can see that Helene looks potato-shaped and our moon is round.
She has the face only an astrophysicist can love:
http://en.wikipedia.org/wiki/File:Leading_hemisphere_of_Helene_-_20110618.jpg
So, now, daniel, I wonder how big would we have to enlarge helene in
order for her to fall apart. As I look at it, that unseemly protrusion
that might be a snout or the leaning tower of pisa would be the first
thing to crumble.
http://nssdc.gsfc.nasa.gov/planetary/factsheet/saturniansatfact.html
It makes life tolerable as a handyman to have such resources available
for amusement. I'm so impressed with the parts of my government that
don't live fat and happy from lobbyist money, especially those from
foreign warporations.
>
> As chance would have it, we did an exercise about this last week (I'm
> doing a masters in astrophysics). For ordinary materials things become
> round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
> while Helene is 43 km along its longest dimension.
>
What would that mass be? I guessed 1.0e21 kg:
$ gfortran -Wall -Wextra brook3.f90 -o out
$ ./out
3.00000008E+16 1.00000002E+21 helene to M nought 33333.332
7.36000003E+22 1.00000002E+21 M moon to M nought 73.599998
$ cat brook3.f90
implicit none
real:: helene, ratio, M_nought, M_moon
M_moon = 7.36e22 !kg
helene = .0003e20 ! kg
M_nought = 1.0e21 ! minimum mass to imply hydrostatic equilibrium
ratio = M_nought / helene
print *, helene, M_nought, "helene to M nought", ratio
ratio = M_moon / M_nought
print *, M_moon, M_nought, "M moon to M nought", ratio
end program
! gfortran -Wall -Wextra brook3.f90 -o out
$
--
Uno
"It all started with Wisconsin."
Uno
> I'm looking for some better notation here (tried to find stuff in mrc),
> a subroutine to convert degrees to real radians, and a means to compare
> this vector to the meridian that bisects it.
$ ./out
pi is 3.1415926535897931
40.708097222116258 73.999675000508631
40.701622222264611 73.991108333269750
bisecting meridian is close to 73.995391666889191
8.56666723888110937E-003 6.47499985164756708E-003
a protestor and a cop, looking at each other from across
the median of the brooklyn bridge, would, if they gerrymandered the
other's
frame of reference to include the tangential velocity of the earth see
speeds of 466.80839661712156 or 617.60498683516869 miles
per hour.
theta and comp are 52.916680591221521 37.083319408778479
$ cat brook5.f90
implicit none
integer, parameter :: dp = selected_real_kind(12,30)
real(dp) :: pi, speed3, speed4
real(dp) :: latitude, speed1, speed2
real(dp) :: degree, minute, second, north, west
real(dp) :: bis_mer, delta_y, delta_x, theta, comp
real(dp), dimension(2) :: ba, zinga
pi = 4.0_dp * atan(1.0_dp)
speed1 = 456.1 !m/s
speed1 = speed1 * 3600 !m/hr
speed1 = speed1 * 3.28 / 5280 !mi/hr
latitude = latitude * pi /180.0 !radians
speed2 = speed1 * cos(latitude)
print *, "w.r.t. CM planet, speed of any point in brooklyn is ", speed2
! data entry
degree = 40
minute = 42
second = 29.15
minute = 59
second = 58.83
ba(2) = west
print *, ba
minute = 42
second = 5.84
minute = 59
second = 27.99
zinga(2) = west
print *, zinga
print *, "bisecting meridian is close to ", bis_mer
delta_y = abs(zinga(2)-ba(2))
delta_x = abs(zinga(1)-ba(1))
print *, delta_y, delta_x
delta_y= delta_y * pi /180.0_dp !radians
delta_x= delta_x * pi /180.0_dp !radians
theta = atan(delta_y/delta_x)
speed3 = speed2 * cos(theta)
speed4 = speed2 * sin(theta)
print *, "a protestor and a cop, looking at each other from across"
print *, "the median of the brooklyn bridge, would, if they
gerrymandered the other's"
print *, "frame of reference to include the tangential velocity of the
earth see "
print *, "speeds of", speed3, "or", speed4, "miles per hour."
theta = theta * 180.0_dp/pi !degrees
comp = 90.0_dp - theta
print *, "theta and comp are ", theta, comp
contains
function deg_to_dp(degree, minute, second)
real(dp) :: deg_to_dp
real(dp), intent(in) ::degree,minute, second
deg_to_dp = degree + minute/60.0 + second/3600.0
end function deg_to_dp
end program
! gfortran -Wall -Wextra brook5.f90 -o out
$
If the brooklyn bridge is straight, then is its slope just delta y over
delta x, where the points are the coordinates on the globe?
--
Uno
Daniel Carrera
> On 10/05/2011 06:14 AM, Daniel Carrera wrote:
>
>> Helene is not in hydrostatic equilibrium, but our moon most definitely
>> is. You can see that Helene looks potato-shaped and our moon is round.
>
> She has the face only an astrophysicist can love:
>
> http://en.wikipedia.org/wiki/File:Leading_hemisphere_of_Helene_-_20110618.jpg
solar system are always named after mythological characters in the life
of the Greek version of the Roman god after which the planet is named...
Ok, let me rewind... Saturn is a roman god. The Greek equivalent of that
is Cronus. Helen of Troy was the granddaughter of Cronus... The
exception is Uranus, whose moons are named after Shakespearean characters.
The high quality pictures that we have today were only obtained in the
last few years (2007 - 2011) thanks to the Cassini spacecraft currently
orbiting Saturn.
> So, now, daniel, I wonder how big would we have to enlarge helene in
> order for her to fall apart.
mostly rock.
>> As chance would have it, we did an exercise about this last week (I'm
>> doing a masters in astrophysics). For ordinary materials things become
>> round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
>> while Helene is 43 km along its longest dimension.
>>
>
> What would that mass be? I guessed 1.0e21 kg:
Helene is made of the same material and it needs to have a 300 km radius
that makes a volume of:
V = 4/3 pi (300,000 m)^3 = 1.13e17
Density = 2 g/cm^3 = 2000 kg/m^3
So that gives a mass of 2.26e20 kg. So your guess was pretty good. Just
off by one digit.
Cheers,
Daniel.
Robert Miles
You forgot the velocity of the Milky Way toward another galaxy
in the local group of galaxies.
Robert Miles
Robert Miles
> On 10/05/2011 08:37 AM, Uno wrote:
>> On 10/3/2011 11:47 PM, Daniel Carrera wrote:
>>> On 10/04/2011 03:04 AM, glen herrmannsfeldt wrote:
> As chance would have it, we did an exercise about this last week (I'm
> doing a masters in astrophysics). For ordinary materials things become
> round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
> while Helene is 43 km along its longest dimension.
>
> Cheers,
> Daniel.
A masters in astrophysics? Then you're likely to be interested
in the two BOINC projects I found that are related to
astrophysics:
Einstein@Home
http://einstein.phys.uwm.edu/
Milkyway@Home
http://milkyway.cs.rpi.edu/milkyway/
Robert Miles
Robert Miles
> On 10/3/2011 11:44 PM, Daniel Carrera wrote:
>>> How oblate would the earth be if the moon were just booted away by some
>>> celestial jokester?
>>
>> It would be basically the same that it is today. The moon's effect on
>> Earth's shape is very tiny; much smaller than the effect of Earth's
>> rotation.
>
> Well that's good to know. I think, in some future arms race, someone
> will think blowing up the moon is a good idea. I'll worry less for them.
>
> Peace.
How much difference will it make in how many pieces of the moon would
then impact the earth?
Robert Miles
Daniel Carrera
> A masters in astrophysics? Then you're likely to be interested
> in the two BOINC projects I found that are related to
> astrophysics:
>
> Einstein@Home
> http://einstein.phys.uwm.edu/
>
> Milkyway@Home
> http://milkyway.cs.rpi.edu/milkyway/
Their website seems to be down right now but I'll try again later.
Cheers,
Daniel.
Dick Hendrickson
Dick Hendrickson
Uno
> On 10/06/2011 12:53 AM, Uno wrote:
>> So, now, daniel, I wonder how big would we have to enlarge helene in
>> order for her to fall apart.
>
> About 250 - 350 km in radius, depending on whether it is mostly ice or
> mostly rock.
$ gfortran -Wall -Wextra brook6.f90 -o out
$ ./out
V_nought = 5.23598810E+17 m**3.
M_nought = 1.04719760E+21 kg.
surface gravity of minimal necessarily-round object is 0.27951801
m/s**2.
ratio of g_earth to g_nought is 35.096130
ratio of g_moon to g_nought is 5.7956910
at g_nought acceleration an objects covers 45.840954 feet in 10
seconds.
$ cat brook6.f90
implicit none
real :: pi, G, G_earth, G_moon, ratio, d_comp
real:: rho, R_nought, V_nought, M_nought, G_nought
pi = 4.0 * atan(1.0)
R_nought = 500.0 !km (400 - 700) source: Daniel in c.l.f.
R_nought = R_nought * 1000.0 !meters
V_nought = 4.0/3.0 * pi * R_nought**3 ! m**3
print *,"V_nought = ", V_nought, "m**3."
rho = 2.0 !g/cm**3 typical asteroid density
rho = rho * 1000.0 ! kg/m**3 (/1000 *100 * 100 *100)
M_nought = V_nought * rho ! kg
print *,"M_nought = ", M_nought, "kg."
!g = [G*M(earth)]/d^2.
! gravitational constant = 6.67300 � 10-11 m3 kg-1 s-2
G = 6.67300e-11
G_nought = G * M_nought/ (R_nought**2)
print *, "surface gravity of minimal necessarily-round object is ",
G_nought, "m/s**2."
G_earth = 9.81
G_moon = 1.62
ratio = G_earth / G_nought
print *, "ratio of g_earth to g_nought is ", ratio
ratio = G_moon / G_nought
print *, "ratio of g_moon to g_nought is ", ratio
! how much g is this?
d_comp = .5 * G_nought * 10**2 * 3.28 ! feet in ten seconds
print *, "at g_nought acceleration an objects covers ", d_comp, "feet in
10 seconds."
end program
! gfortran -Wall -Wextra brook6.f90 -o out
$
Dude, your work is fascinating. I'm not quite sure how you got that
estimate for a helenian object, but I think that others think it might
be made completely of ice, given that it has a density of 1000 kg/m**3
at the official site, and I'm not the guy who says, "oh, because it's
official it must really be that way, so your estimate would be off by a
binary order magnitude, making my guess off by a garden-variety order of
magnitude, which my instructor for stellar evolution, Roger Ptak, use to
call "pretty good."
>>> As chance would have it, we did an exercise about this last week (I'm
>>> doing a masters in astrophysics). For ordinary materials things become
>>> round when they are a ~400 - 700 km wide. Our moon is ~3480 km wide
>>> while Helene is 43 km along its longest dimension.
>>>
>>
>> What would that mass be? I guessed 1.0e21 kg:
>
> Let's see... a typical asteroid has a density of about 2 g / cm^3. If
> Helene is made of the same material and it needs to have a 300 km radius
> that makes a volume of:
>
> V = 4/3 pi (300,000 m)^3 = 1.13e17
>
> Density = 2 g/cm^3 = 2000 kg/m^3
>
> So that gives a mass of 2.26e20 kg. So your guess was pretty good. Just
> off by one digit.
How do you figure out what the pressure at the core is of this minimal
object?
--
Uno
Daniel Carrera
> Dude, your work is fascinating. I'm not quite sure how you got that
> estimate for a helenian object, but I think that others think it might
> be made completely of ice, given that it has a density of 1000 kg/m**3
> at the official site, and I'm not the guy who says, "oh, because it's
> official it must really be that way, so your estimate would be off by a
> binary order magnitude, making my guess off by a garden-variety order of
> magnitude, which my instructor for stellar evolution, Roger Ptak, use to
> call "pretty good."
>
>>
>> V = 4/3 pi (300,000 m)^3 = 1.13e17
>>
>> Density = 2 g/cm^3 = 2000 kg/m^3
>>
>> So that gives a mass of 2.26e20 kg. So your guess was pretty good. Just
>> off by one digit.
>
> How do you figure out what the pressure at the core is of this minimal
> object?
For this calculation I don't need the central pressure. Solids and
liquids are not very compressible at the pressures you find in planets.
So I can just take the densities that I know from Earth and use them. If
you say that Helene is all ice, I know that its density is 1 g/cm^3.
That said, we *can* estimate the central pressure of a planet (or a moon
in hydrostatic equilibrium) if we can estimate (or guess) what the
interior is made of. The planet is a sphere. Think of it as a series of
thin spherical shells. Pick one shell at radius R:
Radius of the shell == R
Mass contained inside the shell == M(R)
Gravitational field at this point == G M(R) / R^2
Thickness of this shell == dR
Area of this shell == A(R)
Volume of this shell == A(R) dR
Density of the planet at distance R == rho(R)
Mass of the shell == rho(R) A(R) dR
Total weight of this shell == G M(R) rho(R) A(R) dR / R^2
Total weight per unit area == G M(R) rho(R) dR / R^2
Now, remember that pressure is force per unit area. If the planet (or
large moon) is in hydrostatic equilibrium, that means that pressure
exactly balances gravity. Let "dP" be the difference in pressure between
the top and the bottom of the shell. That "dP" must be equal to the
weight per unit area of the shell (i.e. hydrostatic equilibrium):
dP = - G M(R) rho(R) dR / R^2
The "-" sign is because when radius increases (positive dR) pressure
decreases (negative dP). This leads to the differential equation:
dP / dR = - G M(R) rho(R) / R^2
Now, if you have a way of estimating the density profile of the planet -
rho(R) - you can obtain M(R) too, and you can solve the differential
equation to obtain the pressure. You just need a couple of boundary
conditions:
Mass at the very centre (R = 0) is 0.
Pressure at the surface of the planet (R = R_max) is 0.
To get the density profile you have to either use some theoretical model
of how planets work, or send a space probe to take gravity measurements
(or in Earth's case, you can use earthquakes too).
As a simple example, we can pretend that the planet has uniform density,
so the ODE becomes very easy.
rho(R) = rho = constant
M(R) = 4/3 * pi R^3 * rho
dP / dR = - G 4/3 * pi R * rho^2
P(R) = - G * 2/3 * pi * rho^2 * R^2 + Const
Naturally, the constant is the central pressure P(0):
P(R) = - G * 2/3 * pi * rho^2 * R^2 + P(0)
Now use the boundary condition: P(R_max) = 0
P(R_max) = 0 = - G * 2/3 * pi * rho^2 * R_max^2 + P(0)
Therefore, we obtain the central pressure:
P(0) = G * 2/3 * pi * rho^2 * R_max^2
So now we have an estimate of the central pressure of the planet based
on the simplifying assumption that the density is constant. If you use
the average planet density for "rho", the above value will give you an
overestimate, but it will be in the right ball park.
Cheers,
Daniel.