If gravity is weaker the farther from Earth's core, is there a height beyond which additional stories exert no added stress to a building?

9 Answers

  • 9 months ago
    Favorite Answer

    Although Jim Moor is perfectly correct as far as he goes it does not fully answer the question.  If ADDITIONAL stories are added beyond synchronous orbit those stories will be under tension. ( the force needed to maintain them moving in a circle is greater than the force of gravity) Which is still a stress.

    Lower stories under compression and higher ones under tension.

    And this would only work if we eliminated the moon and the sun. Tidal effects add other stresses.

    Now if the building was NOT at the equator the direction of the forces are not equal and opposite so the building would "bend like a banana".  Which is a very large stress.

    Now consider people IN the building.  The RELATIVE gravity on them reduces.  To zero at synchronous orbit. But beyond that?  Negative relative gravity.  You would need to swap the positions of the floor and the ceiling at higher floors.

    And of course you could not use a cable to run an elevator because it would not be under tension in the area near the geosynchronous orbit.

  • Mr. P
    Lv 7
    9 months ago

    There was a design of a low earth orbit elevator that supported its own weight because of centrifugal forces. I think they dropped the idea because there was nowhere to go at the top, and there was more of a risk to shipping and aircraft.

    The main detail here is that it was based on cables, so any materials used would need to be just as strong in tension as compression.

    Plus in the top stories there would be negative gravity, so people would have  a hard time up there, and the lack of oxygen could cause problems if you wanted the window open.

  • oyubir
    Lv 6
    9 months ago

    Those who reply 'no' forget the centrifugal forces.

    If Earth was not revolving, then, the answer would be no. Whatever the height, the gravity is smaller and smaller, but never null

    But Earth is revolving. And therefore, a story is submitted (as every object on Earth) to both gravity and centrifugal force (except if you build your tour at one of the poles)

    Gravity acceleration (ie force divided by mass) is MG/(R+z)^2 

    Where M is Earth mass (5.972×10^24 kg)

    G the gravitational constant (6.675×10^-11 m^3/kg/s^2)

    R is Earth radius (6.371×10^6 m), and z the altitude, and therefore, R+z the distance to the center of Earth

    Centrifugal acceleration is, by definition of "acceleration" the magnitude of the second derivative of the position of the story. 

    So (R+z)w^2 × cos(lat), 

    Where R+z is still the distance from the Earth center. And w the angular speed of Earth rotation, that is 1 rotation every 24 hours, or 2π/86400 = 7.272×10^-5 radians/s

    lat is the latitude

    The two forces are opposed to each other.

    At Earth surface, gravity clearly wins.

    5.972×10^24*6.675×10^-11 / (6.371×10^6)^2 is 9.82 m/s^2

    At the equator, centrifugal force  is 6.371×10^6 * (7.272×10^-5)^2 = 0.03 m/s^2

    At Europeans latitudes, centrifugal force are somewhere between 0.01 and 0.02 m/s^2.

    Hence the 9.81 or 9.80 m/s^2 used for gravity (which is, in reality, gravity, minus centrifugal force)

    So, the further you get from Earth center (the higher), the smaller is gravity and the bigger is centrifugal force. So, indeed, at some point, there is an equilibrium.

    At the equator, that point is the one for which

    MG/(R+z)^2 = (R+z)w^2


    MG/w^2 = (R+z)^3


    R+z = cubic root of 5.972×10^24 * 6.675×10^-11 / (7.272×10^-5)^2

    = cubic root of 7.538×10^22 

    = 42.24×10^6 m

    So z = 42.24×10^6 - 6.371×10^6 = 35.872 × 10^6 m

    In other words, your building should be 35872 km (22290 miles) at the equator. Even higher, elsewhere.

  • Jim
    Lv 7
    9 months ago

    This would be the same height as geosynchronous orbit! Located at 22,236 miles (35,786 kilometers) above Earth's equator, above which the building would subtract from the weight.

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  • User
    Lv 7
    9 months ago

    1) If gravity is weaker the farther from Earth's core

    Well...in fact, it's not...

    Core -> surface of Earth

    - gravity is weaker the closer you get to the core and is zero at the core

    Above surface of Earth

    - gravity is weaker the farther you get from the core and is zero at infinite distance

    2), is there a height beyond which additional stories exert no added stress to a building?

    No because Earth's gravity even affects things very far away - for example: the Sun!

  • 9 months ago

    Yes, I think that's true!  But the buildings would have to be thousands of miles tall. 

    There's been talk of a 'space elevator'.  Imagine a satellite in geosynchronous orbit--that means it travels around the earth in one day, so it always appears to be over the same spot on the ground. (It would also have to be over the equator.)  MANY satellites are in this orbit, like for satellite TV (so you don't have to go outside and adjust your dish every few minutes.)  This is about 24,000 miles high.

    So we have this satellite and we dangle a string from it.  The string is 24,000 miles long and it will actually reach the ground.  Now we can put objects into orbit by sending them up this string!  The problem with this is that the whole weight of the string is on the satelilte, so you need really light string, but strong enough that you could climb it like a rope.

    Anyway, this is essentially a 'building' 24,000 miles tall.  If you could build a  tower that tall, you could take the elevator to the top floor, and just step off!  And you'd be in orbit!  And weightless!

  • Anonymous
    9 months ago

    No is the short answer

  • 9 months ago

    probably around 18,000 stories

  • Rick
    Lv 7
    9 months ago

    sure, but there'd be no air that high ......................

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