Free-fall Acceleration with Centrifugal Force

In summary, the problem involves finding the free-fall acceleration on a spherically symmetric planet, given that it has a magnitude g = g_0 at the North Pole and g = \lambda g_0 at the equator. The solution involves using a vector equation and a trial-and-error approach to arrive at the equation g(\theta) = g_{0}(Sin(\theta) + \lambda Cos(\theta)), while the book provides the answer g(\theta) = g_{0}\sqrt{Sin^{2}(\theta) + \lambda^{2} Cos^{2}(\theta)}.
  • #1
CoreyJKelly
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0

Homework Statement



From Taylor's "Classical Mechanics", problem 9.15:

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude [tex]g = g_0[/tex] at the North Pole and [tex]g = \lambda[/tex] [tex]g_0[/tex] at the equator (with 0 [tex]\leq[/tex] [tex]\lambda[/tex] [tex]\leq[/tex] 1). Find g([tex]\theta[/tex]), the free-fall acceleration at colatitude [tex]\theta[/tex] as a function of [tex]\theta[/tex].


Homework Equations



[tex]g = g_0 + (\Omega \times R) \times \Omega[/tex]

The Attempt at a Solution



The north pole corresponds to a colatitude of 0 degrees, and the equator corresponds to 90 degrees. Using these values, and a trial-and-error approach, I arrived at the equation

[tex]g(\theta) = g_{0}(Sin(\theta) + \lambda Cos(\theta))[/tex]

the book gives the following answer:

[tex]g(\theta) = g_{0}\sqrt{Sin^{2}(\theta) + \lambda^{2} Cos^{2}(\theta)}[/tex]

And I'm not really sure where to start with this. The equation I've given is the only one given for free-fall acceleration, but I can't see any way to manipulate it into anything like this answer. Any suggestions would be appreciated.
 
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  • #2
You can compute [tex]g[/tex] as a function of theta with [tex]g = g_0 + (\Omega \times R) \times \Omega[/tex]

this is a vector equation and [tex]g_0[/tex] doesn't point in the same direction as [tex](\Omega \times R) \times \Omega[/tex]
Once you've done that you can find [tex]\Lambda[/tex] by looking at what happens at the equator.
 

1. What is free-fall acceleration?

Free-fall acceleration refers to the rate at which an object falls towards the ground under the influence of gravity. It is a constant value of 9.8 meters per second squared on Earth, meaning that objects will accelerate at this rate regardless of their mass.

2. How is free-fall acceleration related to centrifugal force?

Centrifugal force is a fictitious force that appears to act on objects in circular motion. It is actually a result of inertia, but it can be used to explain the force felt by objects in free-fall due to the rotation of the Earth. This force is equal and opposite to the force of gravity, cancelling it out and allowing objects to appear weightless.

3. Is free-fall acceleration affected by altitude?

Yes, free-fall acceleration is affected by altitude. The further an object is from the surface of the Earth, the weaker the force of gravity becomes. This means that the rate of free-fall acceleration will be slightly less at higher altitudes than at sea level.

4. How does air resistance affect free-fall acceleration?

Air resistance, also known as drag, is a force that opposes the motion of an object through air. This means that as an object falls, it will experience a force that opposes its downward motion. As a result, the object will not accelerate at a constant rate and will eventually reach a terminal velocity, where the forces of gravity and air resistance are balanced.

5. Can free-fall acceleration be measured accurately?

Yes, free-fall acceleration can be measured accurately using various methods such as timing the fall of an object or using specialized equipment such as accelerometers. However, external factors such as air resistance and altitude may affect the accuracy of the measurement.

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