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Accelaration Of Free Fall 
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#1
May208, 04:58 AM

P: 29

Hello.
Why is it so that the accelaration of free fall on a unit mass on equator = Gravitational force Centripetel Force? Answer ample for a level purposes is required... 


#2
May208, 05:36 AM

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Hello ay2k!
(I think you need to explain to our American friends what A levels are! ) If it's in free fall, then by definition of free fall, the only acceleration will be that due to gravity, won't it? 


#3
May208, 06:12 AM

P: 14

The fact is that due to Newton's third law, the sum of forces applying to a system equals it's mass*it's acceleration.
Here we have 2 forces : the gravitational and the centrifugal forces. So we have mg  mv²/R = m a (using a centripete axis) Which yields immediately : a = g  v²/R So, in a way, acceleration = gravitational force  centrifugal force I assume you meant centrifugal when you said "centripetel" :) This equation looks ok to me, even though I've never read or heard anywhere that the centrifugal force, being a virtually force, could be used in newton's equation. It's also very intuitive, you have Earth's rotation that tends to eject objects on it's surface while its gravity field tends to attract objects. The resulting force we sense is then the sum of both. However, as you can imagine, the centrifugal force is very tiny in comparison with the gravitational force, this is why it is often not used. 


#4
May208, 06:46 AM

P: 14

Accelaration Of Free Fall
It's wierd I can't edit my post anymore :S
I wanted to correct : I'm talking about newton's second law of motion and not his third, but I'm sure everybody corrected by themselves :) 


#5
May208, 04:46 PM

P: 19

If we're neglecting any drag forces (which we are) then here we have *one* force acting on the object  the gravitational pull of the Earth on the object, which acts towards the centre of the Earth.
The acceleration (a=F/m) is directed towards the centre of the Earth. It can be considered as the sum of two parts... the acceleration towards the point on the Earth's moving surface directly below its release point AND its centripetal acceleration (since it was moving sideways when it was released, due to the rotation of the equator). Take the familiar form of Newton's second law... F = ma gravitational force on object = m (free fall acceleration + centripetal acceleration) Divide both sides by m... gravitational field strength = "free fall" acceleration + centripetal acceleration Rearrange... "free fall" acceleration = gravitational field strength  centripetal acceleration ...which is the thing you were asking about. ============================= At the poles this reduces to "free fall" acceleration = gravitational field strength since there is no need to consider centripetal acceleration about the earth's axis at the poles since you're moving in "circles" of radius zero. =================== The socalled "centrifugal force" is a pseudoforce: something that looks like a force because you're looking at the situation from a rotating reference frame. Its like writing F = ma and then shifting some of the acceleration from the "ma" side over to the "F" side of the equation, then treating it as if it is a force. =================== A further point  note that the centripetal acceleration will change as the object gets closer to the centre of the Earth, but not by a significant amount if the drop height is much less than the radius of the Earth. 


#6
May208, 04:52 PM

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#7
May208, 04:56 PM

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#8
May208, 04:57 PM

P: 19

Put some numbers in to give an idea of the size of the effect:
gravitational field strength at Earth's surface = 9.81 m/s/s At the equator the velocity of the Earth's surface is about 465 m/s. The earth's radius is about 6400 km, so centripetal acceleration = [tex]v^{2}[/tex] / r = 0.03 m/s/s << complete solution deleted by berkeman >> 


#9
May208, 07:20 PM

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However, relative to an observer standing on Earth it's a different story. The observer is also accelerating due to their centripetal motion. So the "observed" acceleration of objects in free fall is the difference in the accelerations of the object and observer: [tex] a_{observerd} = a_{gravity}  a_{centripetal} [/tex] 


#10
May208, 07:29 PM

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P: 12,069

Scroll to the last sentence in the "Latitude" section at: http://en.wikipedia.org/wiki/Earth%27s_gravity 


#11
May308, 04:20 AM

P: 29

What im confused about is that both centripetel force and garvitational forces are directed towards the center.....then how com we subtract them when both act in the same direction.... Can someone please explain the quoted euwation fully? Or please give a helpful link... 


#12
May308, 02:23 PM

P: 290

Acceleration net= Acc. do to G  Acc. do to Centrifugal force 


#13
May308, 03:12 PM

P: 14

In this case the only centripetal acceleration is the acceleration due to the gravity, there is no centripetal acceleration similar to the centrifugal acceleration (due to a virtual force).



#14
May308, 03:28 PM

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P: 12,069

[tex] a_{relative} = a_{object}  a_{observer} [/tex] What is the acceleration of an observer "at rest" on the Earth's surface? 


#15
May408, 12:44 AM

P: 29

So if the question would have asked for the absolute value for the value for accelaration....then it would have been simply value for gravitational accelaration??
this is the only bug in this thing left. Kindly answer this last one as well.... 


#17
May408, 08:02 AM

P: 29

cool...
Thankyou everyone... 


#18
May408, 12:11 PM

P: 5

acceleration of the free fall wud be acc due to gravity only isnt it...if it is force then also it will be mass into acc ,mg..



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