Gravitation/circular motion problem

[Moebius]

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. :)

 

[ShawnD]

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.

 

[HallsofIvy]

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!)

 

[ShawnD]

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.

 

[Doc Al]

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".)

 

[radmonovic]

Kwaze Moto