Centripetal Acceleration of Earth

In summary: Just for your future reference, take a look at the source (a table in your text?) where you found this velocity figure. This is the *escape* velocity of the Earth, which won't be of help to you in the problem.
  • #1
shell4987
43
0

Homework Statement


(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator owing to the rotation of Earth? (b) What would the period of rotation of Earth (in minutes) have to be for objects on the equator to have a centripetal acceleration with a magnitude of 9.80 m/s2?


Homework Equations


a=v squared/r and T=(2(pi)(r))/v


The Attempt at a Solution


How do I find out the radius of the Earth? Let alone the velocity of the Earth? I think if I knew those two that maybe I could solve this problem.
 
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  • #2
shell4987 said:
How do I find out the radius of the Earth?
Look it up!

Let alone the velocity of the Earth?
Once you have the radius, you should be able to figure out the speed of a point on the equator.
 
  • #3
shell4987 said:

Homework Statement


(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator owing to the rotation of Earth? (b) What would the period of rotation of Earth (in minutes) have to be for objects on the equator to have a centripetal acceleration with a magnitude of 9.80 m/s2?


Homework Equations


a=v squared/r and T=(2(pi)(r))/v


The Attempt at a Solution


How do I find out the radius of the Earth? Let alone the velocity of the Earth? I think if I knew those two that maybe I could solve this problem.

They assume that you will look up the radius of the Earth in your book. You have to calculate the speed using the fact that you know the period of rotation!
 
  • #4
Okay, I looked up all of my information in the book and got the radius to be 6.37e6 and the velocity to be 11.2km/s, I attempted part (a) and got the answer to be 2.0e13m/s squared and that came out to be wrong, what did i do wrong for that, i used the a=v squared/r formula with 11200 m/s as velocity and 6.37e6m as the radius? Am I doing something wrong here?

And for part (b) I solved it and got it correct. Thank you.
 
  • #5
shell4987 said:
Okay, I looked up all of my information in the book and got the radius to be 6.37e6 and the velocity to be 11.2km/s,
Show how you calculated the velocity. Even if you assume that speed, how did you calculate an acceleration of 2.0e13!
 
  • #6
Doc Al said:
Show how you calculated the velocity. Even if you assume that speed, how did you calculate an acceleration of 2.0e13!

I converted 11.2km/s (the velocity of the Earth) into meters by multiplying it by 1000, therefore getting 11200 m/s, then i put that into the a= v squared/r formula and used r as 6.37e6m... ahh i don't know what I'm doing wrong!
 
  • #7
shell4987 said:
I converted 11.2km/s (the velocity of the Earth)
You want the rotational speed of the Earth's surface. Figure it out using period (one day) and the Earth's radius.

therefore getting 11200 m/s, then i put that into the a= v squared/r formula and used r as 6.37e6m
Even using your numbers you won't get anywhere near the answer you got. (Check your arithmetic!)
 
  • #8
shell4987 said:
Okay, I looked up all of my information in the book and got the radius to be 6.37e6 and the velocity to be 11.2km/s,

Just for your future reference, take a look at the source (a table in your text?) where you found this velocity figure. This is the *escape* velocity of the Earth, which won't be of help to you in the problem.
 

What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It always points towards the center of the circle and is responsible for keeping the object in its circular motion.

How is centripetal acceleration related to Earth?

The centripetal acceleration of Earth is the acceleration that keeps the Earth in its orbit around the Sun. This acceleration is caused by the gravitational force between the two objects.

What is the value of centripetal acceleration of Earth?

The value of centripetal acceleration of Earth is approximately 0.0059 meters per second squared. This means that every second, the Earth's velocity changes by 0.0059 meters per second towards the center of its orbit around the Sun.

How does the centripetal acceleration of Earth affect its orbit?

The centripetal acceleration of Earth is responsible for maintaining the curvature of its orbit around the Sun. If the centripetal acceleration were to suddenly disappear, the Earth would move in a straight line tangent to its orbit.

Can the centripetal acceleration of Earth change?

Yes, the centripetal acceleration of Earth can change if there are external forces acting on the Earth, such as gravitational forces from other planets or asteroids. However, the change is usually very small and does not significantly affect Earth's orbit around the Sun.

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