What Rotation Period Causes 9.8 m/s² Equatorial Acceleration?

In summary, the question is asking for the rotation period of Earth that would result in objects on the equator having a centripetal acceleration of 9.8 m/s^2. Using the equation for centripetal acceleration and solving for T, the period in seconds, the solution is then divided by 60 to get the answer in minutes. However, this answer may not be correct as it assumes a circular orbit, and the actual period may be affected by other factors such as the shape and mass distribution of Earth. GM or μ represents the standard gravitational parameter for Earth, which is the product of its mass and the Universal Gravitational constant.
  • #1
lauriecherie
44
0

Homework Statement



What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration equal to 9.8 m/s^2?

____ min


Homework Equations



Centripetal acceleration is equal to velocity * (2PI/T),
where T is the period in seconds.




The Attempt at a Solution



I set centripetal acceleration equal to 9.8m/s^2 and solved for T. Then I took T and divided it by 60 so I could get the answer in minutes. I came out with 4.97 min which Webassign says is incorrect. Any ideas?
 
Last edited:
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  • #2
One way to look at it is that if the object is not to fall into the center then calculate the orbit about a mass of Earth at 1 Earth radius.

So what is the period of such an orbit?

V2/R = GM/R2 = ω2R

ω2 = GM/R3 = (2π /T)2

T = 2π*(R3/GM)1/2
 
  • #3
What does GM stand for?
 
  • #4
lauriecherie said:
What does GM stand for?

That's sometimes written as μ which is the standard gravitational parameter for earth.

It is the product of Earth's mass and the Universal Gravitational constant.
 

FAQ: What Rotation Period Causes 9.8 m/s² Equatorial Acceleration?

What is Centripetal Acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path at a constant speed. It is always directed towards the center of the circle and its magnitude is equal to v^2/r, where v is the object's speed and r is the radius of the circle.

How is Centripetal Acceleration different from Tangential Acceleration?

Centripetal acceleration and tangential acceleration are both components of the total acceleration of an object in circular motion. However, while centripetal acceleration is directed towards the center of the circle, tangential acceleration is directed tangentially to the circle. Centripetal acceleration is responsible for keeping an object in a circular path, while tangential acceleration is responsible for changing the speed of the object.

What is the relationship between Centripetal Acceleration and Centripetal Force?

According to Newton's second law of motion, the net force on an object is equal to its mass multiplied by its acceleration. In the case of circular motion, the net force is the centripetal force and the acceleration is the centripetal acceleration. Therefore, the relationship between centripetal force and centripetal acceleration is given by the equation F = ma, where F is the centripetal force, m is the mass of the object, and a is the centripetal acceleration.

Can Centripetal Acceleration be negative?

No, centripetal acceleration cannot be negative. It is always directed towards the center of the circle and its magnitude is always positive. However, the direction of the velocity of the object can change, which can result in a negative tangential acceleration.

How is Centripetal Acceleration related to the frequency and period of circular motion?

The frequency of circular motion is the number of complete revolutions an object makes in one second, while the period is the time taken for one complete revolution. The relationship between centripetal acceleration and the frequency and period is given by the equations a = (2πf)^2r and a = (2π)^2r/T^2, where f is the frequency, T is the period, and r is the radius of the circle.

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