The acceleration due to gravity should be less or the same on the equator than on the poles?

[Science Student]

There is wording in my textbook that is frustrating me, or it is me who does not understand the concept.

 

After explaining the centripetal acceleration at the equator but not at the poles, my textbook says, "for a spherically symmetric earth the acceleration due to gravity should be about 0.03 m/s^2 less at the equator than at the poles".

 

The textbook says it this way multiple times, so I know it wasn't a typo.

 

Shouldn't it be due to gravity and the centripetal acceleration? Or shouldn't it have said "due to the apparent g"?

Edited May 9, 2014 by Science Student

[studiot]

Does this help?

 

http://curious.astro.cornell.edu/question.php?number=310

[Science Student]

Does this help?

 

http://curious.astro.cornell.edu/question.php?number=310

I guess I am really just wanting to know why the textbook says that g is less on the equator rather than saying that g - a_c is 0.03 m/s^2 less than g. Doesn't g stay the same no matter how fast the Earth spins?

Edited May 10, 2014 by Science Student

[Rahul08]

Earth, being a sphere, is bulged out at the equator.So the distance between the centre of mass and the surface of earth is higher at the equator than at the pole. We know,according to newton's law of gravity, that the gravitational force is inversely proportional to the square of distances between two bodies. So, the gravitational force on the surface of earth at the equator becomes less and hence the acceleration becomes less.

 

To say accurately, The gravity at the equator becomes 0.53% less than that at the poles and hence, the acceleration becomes 0.18% less,that is, 9.78 m/s^2.

 

Conclusion: Gravity is stronger at the poles than at the equator.

 

Hope this helps You.

Please correct me if i am wrong.

[pwagen]

Well, the text book specifically say "spherically symmetric Earth". Wouldn't that suggest you ignore the bulge and its effects?

[timo]

Contrary to popular misconception most words used in physics do not have that unique definition that is the same between all physics texts or even within a single textbook (despite Wikipedia giving you the impression - that is because they have to decide on one definition to use in their text). Physics texts should be read in context and with the attitude of a sentient being. Not with the attitude of a 20th century computer (21st century computers can read context ...). It is uncommon to count centrifugal effects into gravity. It would be much less uncommon to count them into "g" (in fact that is what I would expect from "g"). But both can or can not be done, as long as it is clear what is meant on the physical level (which may not have a 1-to-1 mapping onto human language).

In this particular example, for me the meaning of the sentence in question is pretty clear. In the sense that the physical scenario is apparent. If you had the full text, it would probably be even much more obvious, since the sentence quoted probably comes at the end of an explicit calculation of the correction number quoted. Both alternative texts; "gravity and the centripetal centrifugal acceleration" and "due to the apparent g", would be fine, too (assuming "g" being defined). If it is the first time this results is stated the two alternatives may in fact be more suitable than what the book writes.

Edited May 10, 2014 by timo

[studiot]

Yes the book is unclear and you have correctly deduced that for a non spinning spherical Earth the weight would be the same at the poles and the equator.

 

For a more complete derivation you need to think in three dimensions.

 

Firstly the geometry.

In fig1 any point of the has latitude [math]\lambda [/math] on the surface of a sphere of radius R.

If we take horizontal (East-West) sections as shown by the dashed line through the point on the surface we generate a smaller circle of radius [math]{R_\lambda }[/math].

This has centre [math]{C_\lambda }[/math], as shown in fig4 beneath.

It is important to remember that the centripetal/centrifugal action takes place in this plane and is not directed towards the cente of the Earth.

This is shown in fig3

 

In fig2 we see that the acceelation due to gravity is directed towards the centre of the Earth.

This acceleration is the same everywhere on the surface of a rotating or non rotating spherical Earth.

 

The accelerations are vector quantities and combine vectorially so the component of the acceleration due to the rotation is
[math]{\omega ^2}{R_\lambda }\cos \lambda [/math]

ie

[math]{\omega ^2}\left( {R\cos \lambda } \right)\cos \lambda [/math]

Now the object of mass m is resting on the Earth's surface, but it is not in absolute equilibrium since it is spinning with the Earth.

 

We can reduce it to euqilibrium by introducing the fictious centrifugal force which acts outwards and opposes the force due to gravity or the weight of the mass.

 

This is shown in fig 6.

 

This centrifugal force reduces the weight of the mass by the equation shown at the bottom.

 

So the apparent weight as measured by any device that does not balance weights eg a spring balance will be lower than that measured by a weight balance eg a beam balance since the weights in the beam balance are affected the same as the mass.

[swansont]

I guess I am really just wanting to know why the textbook says that g is less on the equator rather than saying that g - a_c is 0.03 m/s^2 less than g. Doesn't g stay the same no matter how fast the Earth spins?

 

This is a bit of sloppiness in how terms are being defined, I think. Two simple ways you can define g are: g = GM/r² at the surface of the earth, or W = mg = -N, i.e. your weight is what you read on a scale, and that defines g. If you use the former, g is nominally a constant (it will still vary with mass distribution and non-constant r), but then W ≠ mg. This problem appears to be using the latter definition.

[Science Student]

Yes the book is unclear and you have correctly deduced that for a non spinning spherical Earth the weight would be the same at the poles and the equator.

 

For a more complete derivation you need to think in three dimensions.

 

Firstly the geometry.

In fig1 any point of the has latitude [math]\lambda [/math] on the surface of a sphere of radius R.

If we take horizontal (East-West) sections as shown by the dashed line through the point on the surface we generate a smaller circle of radius [math]{R_\lambda }[/math].

This has centre [math]{C_\lambda }[/math], as shown in fig4 beneath.

It is important to remember that the centripetal/centrifugal action takes place in this plane and is not directed towards the cente of the Earth.

This is shown in fig3

 

In fig2 we see that the acceelation due to gravity is directed towards the centre of the Earth.

This acceleration is the same everywhere on the surface of a rotating or non rotating spherical Earth.

 

The accelerations are vector quantities and combine vectorially so the component of the acceleration due to the rotation is

[math]{\omega ^2}{R_\lambda }\cos \lambda [/math]

 

ie

[math]{\omega ^2}\left( {R\cos \lambda } \right)\cos \lambda [/math]

 

Now the object of mass m is resting on the Earth's surface, but it is not in absolute equilibrium since it is spinning with the Earth.

 

We can reduce it to euqilibrium by introducing the fictious centrifugal force which acts outwards and opposes the force due to gravity or the weight of the mass.

 

This is shown in fig 6.

 

This centrifugal force reduces the weight of the mass by the equation shown at the bottom.

 

So the apparent weight as measured by any device that does not balance weights eg a spring balance will be lower than that measured by a weight balance eg a beam balance since the weights in the beam balance are affected the same as the mass.

Thank-you so much, I understand it much better now.

Thank-you everyone!!!!

Edited May 10, 2014 by Science Student