Gravitation/circular motion problem

  • Thread starter Moebius
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In summary: Earth's equator due to Earth's rotation. The conversation discussed using the equation v = (2r * 6.4 * 10^6) / (24 * 3600) to calculate the speed, and questioned why the equation a = v^2/r could not be used instead. The answer was that a is the centripetal acceleration experienced while standing still, but when taking into account Earth's rotation, the forces are no longer balanced and the normal force pushing up is less, resulting in a difference in apparent weight.
  • #1
Moebius
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Hello, I'm having a problem with the concept of one question, which goes like this:

Calculate the linear speed of a point on the Earth's equator due to the Earth's rotation.

They gave the radius of Earth on the equator as 6.4 * 10^6..

Therefore,
v = (2r * 6.4 * 10^6) / (24 * 3600)
= 465 m/s

My question is this:

Why can't we use the equation a = v^2/r to solve this question?

As in 9.8 = v^2/(6.4 * 10^6)
v^2 = (9.8)(6.4 * 10^6)
v = root of [(9.8)(6.4 * 10^6)]

The answer is different, and I don't really understand why. Would anyone explain this for me?

Thanks in advance. :)
 
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  • #2
Originally posted by Moebius
They gave the radius of Earth on the equator as 6.4 * 10^6..

Therefore,
v = (2r * 6.4 * 10^6) / (24 * 3600)
= 465 m/s

My question is this:

Why can't we use the equation a = v^2/r to solve this question?

As in 9.8 = v^2/(6.4 * 10^6)
v^2 = (9.8)(6.4 * 10^6)
v = root of [(9.8)(6.4 * 10^6)]

The answer is different, and I don't really understand why. Would anyone explain this for me?

Gravity pulls you towards earth; acceleration due to spinning actually tries to throw you away from the earth.
 
  • #3
How much do you weigh?


Not trying to be nosy- just making the point that IF the acceleration due to gravity were precisely what is needed to keep you moving at the speed of rotation of the earth, the "net" downward force on you would be 0- you would weight nothing!

(I don't know about you but that is definitely not true for me!)
 
  • #4
Originally posted by HallsofIvy
Not trying to be nosy- just making the point that IF the acceleration due to gravity were precisely what is needed to keep you moving at the speed of rotation of the earth, the "net" downward force on you would be 0- you would weight nothing!

That's actually how satelites, the space station, and space shuttles stay seemingly weightless.
 
  • #5
Originally posted by Moebius
Why can't we use the equation a = v^2/r to solve this question?

As in 9.8 = v^2/(6.4 * 10^6)
v^2 = (9.8)(6.4 * 10^6)
v = root of [(9.8)(6.4 * 10^6)]
That "a" is the centripetal acceleration that you experience just standing "still". As Halls and ShawnD point out, that is not the acceleration due to gravity, g.

If you ignore the Earth's rotation, you would presume that you are in equilibrium when you are standing still. That means downward force (gravity) is balanced by the upward force (Normal force of the ground pushing you up).

If you include the Earth's rotation, you are no longer in equilibrium. You are now centripetally accelerated. So the forces no longer balance: the normal force pushing you up is less. That's why your apparent weight is less. As Halls said, if the Earth were spinning fast enough so that the centripetal acceleration equaled g, then there would be zero normal force: You'd be floating around. (That's what is meant by the term "weightless".)
 
  • #6
Kwaze Moto
 
Last edited:

1. What is the difference between gravitational force and centripetal force?

The gravitational force is the attractive force between two objects with mass, while centripetal force is the force that keeps an object moving in a circular path.

2. How does the mass of an object affect its gravitational force?

The greater the mass of an object, the stronger its gravitational force. This means that larger objects have a stronger pull on other objects compared to smaller objects.

3. What is centripetal acceleration?

Centripetal acceleration is the acceleration of an object towards the center of a circular path. It is always perpendicular to the velocity of the object and directed towards the center of the circle.

4. How do you calculate the centripetal force?

The centripetal force can be calculated using the formula F = m*v^2/r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

5. What is the relationship between circular motion and gravity?

Circular motion is caused by a balance between the centripetal force and the gravitational force. The centripetal force keeps the object moving in a circular path, while the gravitational force pulls the object towards the center of the circle.

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