What Day Length Would Allow Objects to Float at the Equator?

[MPat]

Homework Statement

Earth is a spherical object, which completes a full rotation every 24 hours. How long should the day on Earth be so that an object at the equator is able to float freely above the ground?

Homework Equations

v=2πr/T
Fr=mv^2/r
v=velocity
r= radius
T = time for comleting 1 revolution in seconds
Fr= centripetal force/force in a radial direction

The Attempt at a Solution

1 revolution in 24 hours so T = 24x3600sec = 86400

Fr=mv^2/r
Fr at the Earth is equal to mg
mg=mv^2/r
mass cancels out
g=v^2/r Eq. 1

v=2πr/T
vT/2π = r Eq. 2

Sub Eq. 2 into 1

g=v^2/(vT/2π)
re arrange the equation to
v=gT/2π
v= 9.8*86400/2π
v=134 760m/s

Thats where I'm at so far...I'm just not sure if that's correct...and if so where to go from there.

 

[Doc Al]

1 revolution in 24 hours so T = 24x3600sec = 86400

This is where you're messing up. That value for T is what the period is now. You want to solve for the period when the Earth is sped up so that the object floats.

So, do just what you did except now solve for T.

 

[MPat]

This is where you're messing up. That value for T is what the period is now. You want to solve for the period when the Earth is sped up so that the object floats.

So, do just what you did except now solve for T.

So for the object to float

Fr should equal 0 correct?

 

[Doc Al]

So for the object to float

Fr should equal 0 correct?

No. Your thinking before was correct: Fr = mg.

Fr=mv^2/r
Fr at the Earth is equal to mg
mg=mv^2/r
mass cancels out
g=v^2/r Eq. 1

This is good. Keep going. (Just don't assume a value for T.)

 

[MPat]

No. Your thinking before was correct: Fr = mg.This is good. Keep going. (Just don't assume a value for T.)

Ahh I'm confused.

So if I take v=gT/2π and solve for T i get T=2πr/v Eq 2...what do I do with that? I don't have any of the values to help me solve for T.

You said g=v^2/r Eq. 1 was correct...My assumption would have been that Fr would have to equal 0 in order for the object to float freely above the ground

 

[Doc Al]

You said g=v^2/r Eq. 1 was correct...My assumption would have been that Fr would have to equal 0 in order for the object to float freely above the ground

No. You want the centripetal acceleration to equal the acceleration due to gravity. (Or, in other words, the centripetal force to equal the weight.) Which is what you did to get that equation.

So if I take v=gT/2π and solve for T i get T=2πr/v Eq 2...what do I do with that?

Substitute that expression for v in your first equation. Then you can solve for T. (You'll need the radius of the earth. Look it up!)

 

[MPat]

No. You want the centripetal acceleration to equal the acceleration due to gravity. (Or, in other words, the centripetal force to equal the weight.) Which is what you did to get that equation.Substitute that expression for v in your first equation. Then you can solve for T. (You'll need the radius of the earth. Look it up!)

Okay...

g=v^2/r
r = 6371 km = 6371000m

v=sqrt g*r
v= 7901m/s

v=2πr/T

T=2πr/v
T=2π*6371000/7901
T=5066 seconds
T= 84.4 mins

So a day on Earth should be 84.4 mins so that an object can float freely above the equator.

One thing I'm still confused about:
Fr=mv^2/r
Fr=mg - I came to that because as an object on Earth the only thing creating a centripetal force would be gravity therefore Fr = mg. For the object to float I still thing Fr should equal 0 so that there is no pull on the object towards to earth...What am I missing here?

 

[haruspex]

Fr=mg - I came to that because as an object on Earth the only thing creating a centripetal force would be gravity therefore Fr = mg. For the object to float I still thing Fr should equal 0 so that there is no pull on the object towards to earth...What am I missing here?

You are mixing up two frames of reference.
In an inertial frame, the object is accelerating towards the centre of the Earth. This is centripetal acceleration, v²/r, and requires a net force to produce it. If it is floating (orbiting) then the only force is mg, mg=mv²/r.
In the frame of reference of the object, there is no acceleration, so the net force is zero. But now there is the "fictitious" force, centrifugal, to balance the force of gravity: mg=mv²/r.

 

[Doc Al]

One thing I'm still confused about:
Fr=mv^2/r
Fr=mg - I came to that because as an object on Earth the only thing creating a centripetal force would be gravity therefore Fr = mg. For the object to float I still thing Fr should equal 0 so that there is no pull on the object towards to earth...What am I missing here?

What you're missing is that you must have a non-zero centripetal force in order for the object to move in a circle. For it to "float" the speed must be just right so that the required centripetal force equals the weight. If the speed is too slow (like when the period is 24 hours!) the object doesn't float--it sits on the ground. And if it's too fast, it goes flying off. Make sense?