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Author Topic:   Sun-Earth-Moon Gravity
jar
Member (Idle past 394 days)
Posts: 34026
From: Texas!!
Joined: 04-20-2004


Message 16 of 119 (347817)
09-09-2006 5:03 PM
Reply to: Message 14 by Ragged
09-09-2006 4:34 PM


You asked about earth-moon. That is what I was answering.
Stop and think about the equations.
distance earth to moon = about 240,000 miles.
distance earth or moon to sun = 93,000,000 miles.
when you go back to the equations you will find
somevalue 93,000,0002
and
somevalue 240,0002
That is a big difference. Regardless of the somevalue, the attraction between the earth and moon will be greater than between either and the sun.
The second part involves the 'somevalue'.
That will be dependant on the two variable, the M (mass of first) and m (mass of second).
When the two masses are relatively the same (as is true with the earth -moon system), the individual masses become significant. But look at the relative masses of the earth or moon vs the sun.
The sun is about 333,000 times the mass of the earth and so it doesn't much matter wether we are talking about the mass of the earth or of the moon, the answer will be dominated by the Mass of the sun and the distance between the two objects. The sun is BIG and far away.
Edited by jar, : 0

Aslan is not a Tame Lion

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Taz
Member (Idle past 3291 days)
Posts: 5069
From: Zerus
Joined: 07-18-2006


Message 17 of 119 (347826)
09-09-2006 5:55 PM
Reply to: Message 14 by Ragged
09-09-2006 4:34 PM


Time for a little review.
Gravitational force between Earth and Sun: 3.52e22 N
Gravitational force between Moon and Sun: 4.33e20 N
Gravitational force between Earth and Moon: 1.98e20 N
Your calculation appears to be off.
Ragged writes:
Or is it even wrong to separate the two?
It is not wrong to seperate the two. It is, however, easier for you to understand if you don't seperate the two.
You are still new to physics, so it is fine not fully grasp the picture. Give it a little time and I'm sure all will be well
P.S. "Alot" means to distribute fractions of something to people while "A LOT" (with a space in between) is the same as "many".

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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 18 of 119 (347830)
09-09-2006 6:02 PM
Reply to: Message 17 by Taz
09-09-2006 5:55 PM


Alot
"Alot" is a town and a nagar panchayat in Ratlam district in the state of Madhya Pradesh, India.
allot
verb (allots, allotting, allotted) give or apportion (something) to someone.
pedantic pain in the ass eh?

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MangyTiger
Member (Idle past 6353 days)
Posts: 989
From: Leicester, UK
Joined: 07-30-2004


Message 19 of 119 (347831)
09-09-2006 6:05 PM
Reply to: Message 10 by Ragged
09-09-2006 2:46 PM


Let's plug values[1] in the formula gasby quotes in Message 13:
G*M*m
--------- = F
d*d
G = universal gravitational constant
M = mass of one object in Kg
m = mass of the other object
d = distance between them in Meters
F = force in Newton
G = 6.674210-11 Nm2kg-2
M = 5.97421024 kg (mass of Earth)
m = 7.3476731022 kg (mass of Moon)
d = 405,696 km = 4.05696108 m (Perigee)
For the Earth and Moon this gives:
F = (6.674210-11 x 5.97421024 x 7.3476731022) / (4.05696108 x 4.05696108) N
F = 2.91037 / 1.61017 N [approx.]
F = 1.81020 N [approx.]
For the Sun and Earth:
m = 1.9884351030 kg (mass of Sun)
d = 147,098,074 km = 1.470980741011 m (Perihelion)
F = (6.674210-11 x 5.97421024 x 1.9884351030) / (1.470980741011 x 1.470980741011) N
F = 7.91044 / 2.21022 N [approx.]
F = 3.61022 N [approx.]
For the Sun and Moon:
M = 1.9884351030 kg (mass of Sun)
m = 7.3476731022 kg (mass of Moon)
d = 147,098,074 km = 1.470980741011 m (Perihelion - approximately the same as for Earth)
F = (6.674210-11 x 1.9884351030 x 7.3476731022) / (1.470980741011 x 1.470980741011) N
F = 9.81042 / 2.21022 N [approx.]
F = 4.51020 N [approx.]
So we have:
Earth-Moon attraction = 1.81020 N
Earth-Sun attraction = 3.61022 N
Sun-Moon attraction = 4.51020 N
As Coragyps worked out (with much less pain! ) in Message 15 the Earth-Sun attraction is about 100 times bigger than either of the others and the Earth-Moon attraction is two to three times bigger than the Sun-Moon attraction.
Never underestimate the power of the inverse square law
[1] All values taken from Wikipedia.

Oops! Wrong Planet

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Ragged
Member (Idle past 3552 days)
Posts: 47
From: Purgatory
Joined: 10-26-2005


Message 20 of 119 (347834)
09-09-2006 6:18 PM


Ok, thanks guys. I need to find my graphing calculator. Kinda hard to work with an online scientific calculator. Messed up by a factor of 2 on the Sun-Moon gravitation.

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JustinC
Member (Idle past 4844 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 21 of 119 (347868)
09-09-2006 10:02 PM


It might be instructional if you calculate the accelerations of the the moon due to the sun's gravitational pull and the earth due to the sun's gravitational pull.
This is why they are in approximately the same orbit and it reveals an interesting feature of gravitation.

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42
Inactive Member


Message 22 of 119 (363011)
11-10-2006 1:19 AM
Reply to: Message 11 by Taz
09-09-2006 3:34 PM


The Earth and the Moon together makes up one system (EM system). Now, think of EM as a single object. EM and the Sun makes up another system.
I'm guessing the same applies to solar systems too, which is why planets are not pulled (any more han stars) towards the centres of galaxies.
PS do galaxies also orbit each-other? And why don't moons have their own satelites etc etc ?
All the best

Human Evolution in 42 Steps

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Taz
Member (Idle past 3291 days)
Posts: 5069
From: Zerus
Joined: 07-18-2006


Message 23 of 119 (363068)
11-10-2006 12:56 PM
Reply to: Message 22 by 42
11-10-2006 1:19 AM


42 writes:
I'm guessing the same applies to solar systems too, which is why planets are not pulled (any more han stars) towards the centres of galaxies.
Correct.
PS do galaxies also orbit each-other?
Some do.
And why don't moons have their own satelites etc etc ?
Some do, but in most cases the orbits had been too unstable to maintain itself for long... if they existed in the first place.
You have to understand that when we are including a massive object of some kind the closer to the object the more perturbations in the gravitational field there are.
For example, a small object might be able to orbit Ganymede. However, because of Ganymede's close proximity to Jupiter, pertubations in Jupiter's gravitational field would be tugging the small object until the orbit becomes unstable enough for the object to leave Ganymede. In fact, off the top of my head I can't think of any example of a sat orbiting a sat. Perhaps someone here could point out an example.

Place yourself on the map at http://www.frappr.com/evc
The thread about this map can be found here.

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Taz
Member (Idle past 3291 days)
Posts: 5069
From: Zerus
Joined: 07-18-2006


Message 24 of 119 (363070)
11-10-2006 1:01 PM
Reply to: Message 21 by JustinC
09-09-2006 10:02 PM


JustinC writes:
It might be instructional if you calculate the accelerations of the the moon due to the sun's gravitational pull and the earth due to the sun's gravitational pull.
This is why they are in approximately the same orbit and it reveals an interesting feature of gravitation.
This reminds me of the good ole days when I had to actually drop a bowling ball and a tennis ball at the same time from the top of a ladder in the lecture room to prove to some that heavier stuff don't fall faster. You'd think that even English majors would know this much about gravitation

Place yourself on the map at http://www.frappr.com/evc
The thread about this map can be found here.

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AnswersInGenitals
Member (Idle past 151 days)
Posts: 673
Joined: 07-20-2006


Message 25 of 119 (363126)
11-10-2006 7:29 PM
Reply to: Message 1 by Ragged
09-08-2006 12:09 AM


You might find that this all makes a lot more sense intuitively if you consider two artificial satellites orbiting the earth, an international space station and the space shuttle, for example. This is the identical situation to the earth and the earth's moon orbiting the sun, and in their similar orbit around the sun, then orbiting around each other. Lets say that the shuttle and the ISS are both 200 Kms above the earth and 1 Km from each other, and that the shuttle weighs 20,000 Kgms and the ISS 200,000 Kgms. The shuttle and the ISS both orbit around the earth about once every 90 minutes. But they also orbit around each other. Can you calculate the earth=shuttle, the earth-ISS, and the shuttle-ISS forces and from that, determine the orbital period of the shuttle and the ISS around each other? Do astronauts need to consider these motions when docking?

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RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 26 of 119 (363235)
11-11-2006 2:36 PM
Reply to: Message 20 by Ragged
09-09-2006 6:18 PM


something missing?
Any object can orbit any other object in an infinite number of orbits, so the moon can orbit the sun at the current distance and the earth can orbit the sun at the same distance -- the formula for the gravitational attraction between the sun and the moon or the sun and the earth only tells you what the force holding them IN the orbit is.
FIN = Gravitational force = GmM/R^2
The equilibrium of an orbit is established by the force holding the objects OUT from the center, and this is based on obital velocity:
FOUT = Centripetal force = ~mV^2/R
(as a first approximation especially where m << M)
In a circular orbit this is simple, in an elliptical orbit it gets a little more complicated, but it is easy to see that
  • if FIN > FOUT that the object will be pulled in, and
  • if FOUT > FIN that the object will tend to drift out, AND
  • that an object in an elliptical orbit will oscillate between these two conditions
So the question is NOT how big the forces are but whether they BALANCE for the orbit(s) in question.
From
Page not found: The Worlds of David Darling
For a circular orbits where FIN = FOUT, and where m is closer to M, a more accurate formula for orbital velocity is given:
V = {G(M+m)/R}^1/2
And using this as a simple approximation we can then compare the orbital velocities for the moon in orbit around the sun (neglecting the earth), for the earth in orbit around the sun (neglecting the moon), and for the {earth\moon} system in orbit around the sun.
Prediction: the calculated velocities will be closest to the actual orbital velocity with the {earth\moon} system.
Enjoy.

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we are limited in our ability to understand
by our ability to understand
RebelAAmericanOZen[Deist
... to learn ... to think ... to live ... to laugh ...
to share.

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42
Inactive Member


Message 27 of 119 (363352)
11-12-2006 7:01 AM
Reply to: Message 24 by Taz
11-10-2006 1:01 PM


gasby writes:
This reminds me of the good ole days when I had to actually drop a bowling ball and a tennis ball at the same time from the top of a ladder in the lecture room to prove to some that heavier stuff don't fall faster.
Hooray for that experiment, and the imaginary one with the feather and heavy object falling together on the moon.
Altogether, this topic makes me wonder - does every piece of matter continuously fall, towards the rest of the universe's centre of gravity?

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RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 28 of 119 (363357)
11-12-2006 7:52 AM
Reply to: Message 27 by 42
11-12-2006 7:01 AM


falling feathers
and the imaginary one with the feather and heavy object falling together on the moon.
Imaginary?
http://www1.jsc.nasa.gov/er/seh/feather.html
video of the drop
http://www1.jsc.nasa.gov/er/seh/feather.avi
Altogether, this topic makes me wonder - does every piece of matter continuously fall, towards the rest of the universe's centre of gravity?
Well christians are always telling us we're fallen ...
Enjoy.
Edited by RAZD, : added last, title

Join the effort to unravel {AIDS/HIV} {Protenes} and {Cancer} with Team EvC! (click)

we are limited in our ability to understand
by our ability to understand
RebelAAmericanOZen[Deist
... to learn ... to think ... to live ... to laugh ...
to share.

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cavediver
Member (Idle past 3643 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 29 of 119 (363358)
11-12-2006 8:11 AM
Reply to: Message 27 by 42
11-12-2006 7:01 AM


does every piece of matter continuously fall, towards the rest of the universe's centre of gravity?
The natural state of a piece of matter is to trace out a straight path through space-time, known as a geodesic. Because space-time is curved, these straight paths appear curved from our limited 3d viewpoint. Examples of the these straight paths are a stone dropped down a shaft which has been evacuated of air, the moon's orbit around the Earth, the Earth's orbit around the Sun, etc.
For most models of the Universe, there is no centre of gravity as there is no centre...

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42
Inactive Member


Message 30 of 119 (363365)
11-12-2006 8:43 AM
Reply to: Message 28 by RAZD
11-12-2006 7:52 AM


Re: falling feathers

Aha - I imagined it was imaginary. Hooray for this real experiment!
I just realised my centre of gravity question was dumb. So does every piece of matter accelerate along a resultant force vector, one wonders.
Edited by AdminJar, : fix link coding

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