Why Does the Acceleration of Free Fall Differ at the Equator?

[ay2k]

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...

 

[tiny-tim]

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?

 

[Celunas]

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.

 

[Celunas]

It's weird 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 :)

 

[JimChampion]

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 so-called "centrifugal force" is a pseudo-force: 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.

 

[berkeman]

(I think you need to explain to our American friends what A levels are! )

I'm assuming that A-Level means coursework. Thread moved to Homework Help forums.

 

[cristo]

I'm assuming that A-Level means coursework.

An A level student is at the equivalent level of a senior in high school (or thereabouts).

 

[JimChampion]

Put some numbers into 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 = $$v^{2}$$ / r = 0.03 m/s/s


<< complete solution deleted by berkeman >>

 

[Redbelly98]

If it's in free fall, then by definition of free fall, the only acceleration will be that due to gravity, won't it?

That's true for the actual acceleration: due only to gravity while in freefall.

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:

$$a_{observerd} = a_{gravity} - a_{centripetal}$$

 

[Redbelly98]

Put some numbers into 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 = $$v^{2}$$ / r = 0.03 m/s/s


<< complete solution deleted by berkeman >>

To add to the effect, one will be closer to the Earth's center at the poles than at the equator. As a result, the value of g is 0.5% more at the poles (9.78 vs. 9.83 m/s^2).

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

 

[ay2k]

gravitational force on object = m (free fall acceleration + centripetal acceleration)

Isnt it like net accelaration of free fall= accelaration due to gravitational force + centripetel accelaration ?

What I am 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...

 

[robertm]

What I am 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...

Centrifugal 'force' Is directed in the opposite direction of Centripetal Acceleration, i.e... opposite of force do to gravity. So:

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

 

[Celunas]

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).

 

[Redbelly98]

Can someone please explain the quoted euwation fully?

To get the apparent or relative acceleration, subtract the observer's acceleration from the object's acceleration (similar to how you would get the object's velocity as measured by an observer):

$$a_{relative} = a_{object} - a_{observer}$$

What is the acceleration of an observer "at rest" on the Earth's surface?

 

[ay2k]

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...

 

[Redbelly98]

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...

Yes. The actual or absolute acceleration is GM/r^2 .

 

[ay2k]

cool...

Thankyou everyone...

 

[nandu11]

acceleration of the free fall wud be acc due to gravity only isn't it...if it is force then also it will be mass into acc ,mg..