Calculating Acceleration Due to Gravity for a 96m Building

[Bjarne]

How can i calculate the Acceleration Due to Gravity expected to be measured

• At the bottom of the building, - for example at Ground Floor or in the cellar.
• And at the top of the building.

The weight is about 55.000.000 KG.
The is height 96 meter
Width = 20 meter
Lenght = 20 meter

This is not homework

 

[berkeman]

How can i calculate the Acceleration Due to Gravity expected to be measured

• At the bottom of the building, - for example at Ground Floor or in the cellar.
• And at the top of the building.

The weight is about 55.000.000 KG.
The is height 96 meter
Width = 20 meter
Lenght = 20 meter

This is not homework

Why do you expect it to be different? To what accuracy do you intend to make this calculation? And what is the application?

 

[DaveC426913]

g at 96 metres is the same as g at 0 metres, to many decimal places.

 

[Bjarne]

Why do you expect it to be different? To what accuracy do you intend to make this calculation? And what is the application?

I don't except it to be different
Accuracy? - I guess it can be calculated very accurate.
I only want to calculate, not t to know how exact it can be measured.

 

[sophiecentaur]

What information do you intend to use i order to make the calculation?
g is something that must be inferred from other measurements, isn't it?
Or are you wanting to include the effect of the mass of the building? That will be very very small because the Earth is so bloomin' massive in comparison.

 

[DaveC426913]

Bjarne, please give us some context, so we can understand how to answer.

 

[Bjarne]

What information do you intend to use i order to make the calculation?
g is something that must be inferred from other measurements, isn't it?
Or are you wanting to include the effect of the mass of the building? That will be very very small because the Earth is so bloomin' massive in comparison.

In the bottom of the building you should be able to calculate a "pull upwards" and opposite at the top a pull downward, - relative to the Free Air gradient..

 

[Bjarne]

Bjarne, please give us some context, so we can understand how to answer.

Well imaging you would be measuring acceleration due to gravity (ADG) at the top and at the bottom of that building, and let us assume you could do it 100 % exact.
Now you should also be able to calculate the exact same.
The questions which equation to use.
If the building would have a ball shape, it would be easy, but this is not the case.

 

[sophiecentaur]

Calculate for your given mass at the centre of mass of your building, then, and subtract it from g. You will need many significant figures, though.

To get an idea of how small your force difference will be, they have managed to measure gravitational perturbations in the vicinity of large mountains of iron ore. Some famous geezer, several hundred years ago managed to use this effect (on a plumb bob near some Scottish mountain) to measure G and was fairly accurate. He would have done even better if he had not ignored the presence of some other nearby mountains, apparently.

 

[Bjarne]

Calculate for your given mass at the centre of mass of your building, then, and subtract it from g. You will need many significant figures, though.
.

And this is the question
With a ball shaped body it is pretty easy to do such calculation, e.g; the ADG of the Earth
But what about such structure as a building?

 

[DaveC426913]

And this is the question
With a ball shaped body it is pretty easy to do such calculation, e.g; the ADG of the Earth
But what about such structure as a building?

Shape is irrelevant, or at least its effect is another several orders of magnitude smaller. You can treat the building as if it is a point-source of gravity.

i.e.: The presence of the building on g will make a difference of maybe 10^-6. The shape of the building will make a difference on g of maybe 10^-9. (These are ballpark magnitudes, just to give you an idea.)

 

[sophiecentaur]

Actually, I was being over simple. The gravitational potential of a cuboid is not the same as that of a sphere but you would certainly get a good idea of the magnitude of the force if you just went for a point mass at the cm particularly as it is so tall and thin!

 

[D H]

Gravitational acceleration is subject to the superposition principle. Calculate the gravitational acceleration due to the Earth and the building separately and just add. The Earth is fairly easy: Use the free air correction for points above the ground floor or a double negative Bouguer correction for the basement.

Gravitation due to the building is an exercise left to the OP. A Bouguer-style correction might give a reasonable approximation.

 

[Bjarne]

The presence of the building on g will make a difference of maybe 10^-6. The shape of the building will make a difference on g of maybe 10^-9.

I don't understand the point…..
A ball shaped object will give you the maximum acceleration due to gravity (ADG)
The more an object deviate from a ball shape, - the weaker ADG would you get.
For example a 500 ton steel ball would cause more gravity as a 500 ton steel cable between UK and France.
So I believe the point is; how to calculate how much a body deviate from the perfect shape; - > a ball shape.

 

[DaveC426913]

I don't understand the point…..
A ball shaped object will give you the maximum acceleration due to gravity (ADG)

"...give you..."? Where are you standing such that you measure the force from this object?

The force you feel from a 500 ton steel that's sitting on the Moon is the same force - to many, many decimal places - that you feel from a 500 ton steel cable sitting on the Moon.
Do you agree?

The more an object deviate from a ball shape, - the weaker ADG would you get.

From what vantage point are you measuring this force?

For example a 500 ton steel ball would cause more gravity as a 500 ton steel cable between UK and France.

Cause more gravity where?


There are some elements to your scenarios, and your assumptions, that are going unstated. I want to root them out.

 

[Bjarne]

"...give you..."? Where are you standing such that you measure the force from this object?

At the surface of the steel ball, or at the surface at the cable.
The steel ball would cause stronger acceleration due to gravity at the surface.

The force you feel from a 500 ton steel that's sitting on the Moon is the same force - to many, many decimal places - that you feel from a 500 ton steel cable sitting on the Moon.
Do you agree?

If you would add 500 ton to the mass of the moon, yes then it wouldn’t matter if you would add it as a steel ball or as a cable, the effect measured from Earth would be the same. _ But this is not the question.

From what vantage point are you measuring this force?

From the surface of the bodies

Cause more gravity where?

A steel ball would cause more acceleration due to gravity (ADG) everywhere at the surface

There are some elements to your scenarios, and your assumptions, that are going unstated. I want to root them out.

I think you must have misunderstood the question.
The question is about ADG a certain place near a body, - not about whether gravity is " different"
The point is that a ball shape is the perfect shape that is causing stronger ADG than any other kind of shape.

 

[DaveC426913]

At the surface of the steel ball, or at the surface at the cable.
The steel ball would cause stronger acceleration due to gravity at the surface.

At the surface. Thank you. That's what I was looking for.

I think you must have misunderstood the question.

No I haven't. You kept using phrases that make assumptions.

The question is about ADG a certain place near a body, - not about whether gravity is " different"
The point is that a ball shape is the perfect shape that is causing stronger ADG than any other kind of shape.

Shape does not cause gravity to be strong.

Without stating that you are talking about being directly adjacent to the object (standing on the sphere or standing on the cable), the statement about "how strong ADG is" does not make sense. If you were far enough away that the shape did not affect it then the question is virtually moot.

I needed to hear that you are measuring g from the object's surface.

A building is pretty near spherical for the intents and purposes of measuring gravity. OK, yes when you're standing at one of its corner, you're effectively standing on a pointy mountain that is projecting off the sphere, and there are 7 other pointy mountains sticking out of this sphere. That'll get you pretty high degree of accuracy.

But you want to know exactly what the effect is. The issue here is: there is no "exact", since g is known only to a certian number of places. How many decimal places do you need? At some point, you may be measuring effects that are smaller than our current measurement of g.

 

[Bjarne]

To calculate ADG at the surface of a ball shape is the most easiest thing to do > MG/r^2
Some month ago I had some equations how to calculate ADG for example on a top of a container (cylinder). (But lost these) . -
This equation to calculate ADG on or near a cylinder is a bit more complicated (as the one to calculate the ADG near a ball shaped structure).
In the same way, there exist equations how to calculate ADG for example on the top, or near of a building.
The shape of the building (or any other object) makes a big difference, according to how strong ADG you can expect a certain place, near or on the top of the building.
So I am asking about which equation to use to calculate ADG near, or on the top of any kind of building structure..
For exsample , how would you calculate ADG of the building mentioned above...if you were standing at the roof.