Acceleration toward Earth

[Zanarh]

Hi there, I'm new to this forum. Nice to meet you!

 

Well, basically, I'm studying under the IB course and I have Mathematics, Physics and Economics as my higher level subjects.

 

Now that I just finished my exams yesterday, I chanced upon MIT's OCW Courses on Physics by Professor Walter Lewin. After watching the lecture about centripetal acceleration, I had a random thought about the acceleration of objects on the surface of the Earth and wrote a short paper about it. However, I do not have any much knowledge beyond the basics, and I'm not entirely sure about the resultant acceleration. There isn't exactly much information about this on the web too.

 

Acceleration Towards Earth

 

From the lecture, I understand that for an object tethered to a string and swirled around in a perfect circle, the centripetal acceleration towards the centre of the circle results in a perceived (from the object's point of view) gravity opposite of the centripetal acceleration (ie outwards of the circle). However, if an object is not tethered in anyway to Earth, does this mean that the spinning of the Earth results in the object accelerating away from the Earth? Does the gravitational acceleration of the Earth affect that?

 

I'll be discussing this with my Physics teacher soon, but could any of you provide any insights? Thanks!

Edited June 9, 2009 by Zanarh

[Sisyphus]

In the context of classical physics, the Earth's gravity is the "tether." If it were to suddenly turn off, we would indeed all fly off on tangent trajectories, just like a rock released from a sling. As is, the Earth's spinning does create a centrifugal effect, which is why it bulges slightly at the equator, and why it's easier to launch rockets the closer to the equator you are.

[swansont]

As is, the Earth's spinning does create a centrifugal effect, which is why it bulges slightly at the equator, and why it's easier to launch rockets the closer to the equator you are.

 

That's also to take advantage of the rotational speed, so you need to launch less maneuvering propellant, which I believe is a larger issue.

[Zanarh]

From what I understand, it is easier to launch a rocket at the equator because of lower gravitational force (since the equator is furthest [21km] away from the centre of the Earth than any other place).

 

Does this mean that the calculations here (Acceleration Towards Earth) are correct since they are based on the equatorial radius?

 

But I found on wikipedia that

At the equator, the effective gravitational acceleration is 9.7805 m/s². This means that the true gravitational acceleration at the equator must be 9.8144 m/s² (9.7805 + 0.0339 = 9.8144)

http://en.wikipedia.org/wiki/Equatorial_bulge

 

In my calculations, I assumed that perceived gravity due to the rotation of the Earth on an object is directed outwards from the centre of the Earth.

 

In the context of classical physics, the Earth's gravity is the "tether." If it were to suddenly turn off, we would indeed all fly off on tangent trajectories, just like a rock released from a sling. As is, the Earth's spinning does create a centrifugal effect, which is why it bulges slightly at the equator, and why it's easier to launch rockets the closer to the equator you are.

If this is true for an object tethered to a string and spun around (as explained here), it should also be true for objects on the surface of the Earth.

 

 

So shouldn't the final acceleration at the equator be roughly 9.7805 - 0.0339 = 9.7466m/s² instead of 9.7805 + 0.0339 = 9.8144m/s²?

[swansont]

So shouldn't the final acceleration at the equator be roughly 9.7805 - 0.0339 = 9.7466m/s² instead of 9.7805 + 0.0339 = 9.8144m/s²?

 

True gravitational acceleration is 9.8144 m/s^2. Perceived gravity must be less (9.8144-0.0339) because of the rotation requiring there be a centripetal acceleration. If the planet were rotating fast enough to have the centripetal acceleration be 9.8144 m/s^2, one would be in orbit, and feel no acceleration (freefall)