Gravitational forces

What would the period of rotation of the Earth need to be if objects at the equator were to appear weightless?
r= 6.4x10^6

Am I not able to equate the centripetal force and the gravitational force, rearrange for v and then sub in t=2pi/v ? Can't get the asnwer

Thanks

Correct method. But t = 2pi. r / v

Correct method. But t = 2pi. r / v

What does it mean by appear weightless?

Thanks!

I still getting the wrong answer for some reason. Using the following:

(2pi x (6.4x10^6))
------------------------
sqrt( ((6.67x10^-11)(5.98x10^24))/(6.4x10^6))

When you equate the centripetal force to weight you write

mg = mv²/r

This gives

g = v²/r

and

v = √(gr)

Is this not the same as:

mv²/r=GmM/r²

v = √(GM/r)

As when I went to the next stage using v =√(gr) gave the same answer. I sub in v=2pi.r/T

to get T= 2pi.√(r/GM)

When i sub in r, g and m it gives 5074 as opposed to 1.4 which the answer should be

Is this not the same as:

mv²/r=GmM/r²

v = √(GM/r)

As when I went to the next stage using v =√(gr) gave the same answer. I sub in v=2pi.r/T

to get T= 2pi.√(r/GM)

When i sub in r, g and m it gives 5074 as opposed to 1.4 which the answer should be

5074 what?
1.4 what? seconds, hours, minutes, days, years?
5074 is in seconds!!!
Stop and think for just a second.
You need to be careful with units.

5074 what?
1.4 what? seconds, hours, minutes, days, years?
5074 is in seconds!!!
Stop and think for just a second.
You need to be careful with units.