How Long Must a Day Be for Weightlessness at the Equator?

AI Thread Summary
To achieve weightlessness at the equator, calculations indicate that a day would need to last approximately 84 minutes, based on the balance of gravitational and centrifugal forces. The formula used involves equating gravitational force to centrifugal force, leading to a velocity of 7920 m/s and a period of 5077 seconds. Participants in the discussion agree on this 84-minute figure, questioning the teacher's assertion of a five-minute duration. The apparent strength of gravity at the equator is expressed as g_a = g - rω² cos² φ, which confirms the longer duration needed for weightlessness. The conversation highlights a discrepancy in calculations and invites further clarification from the teacher.
madis
Messages
2
Reaction score
0
Hello, this is my first post here :)
The question I'm having problem with follows: how long would one day have to be so that there would be weightlessness on the equator? (the teacher said that the correct anwser would be roughly 5 minutes)

Solving this by myself I couldn't reach any other conclusion than:
gravitational force = centrifugal force => mg=mvv/r => g=vv/r => vv=gr
(v-velocity of any point on the equator, r- radius of the Earth, 6,4 million meters)
which means that: v=7920m/s => T=2*pi*r/v=5077s=84 minutes

What did I leave out of the account?
 
Physics news on Phys.org
Welcome madis.

An equation of the apparent strength of gravity is:
g_a = g - r\omega^2 \cos^2 \phi
Where \phi is latitude (which would be zero at equator) and g_a is apparent force of gravity, which in your case would be zero. I can show the derivation if you like.
 
Thanks! Using this equation I still got 84 minutes for the solution so I guess this is the real right anwser...
 
Yes, I get around 85 mintues also. I would be interested in seeing how your teacher calculated the five minutes, if you could post it here please:biggrin:
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanged mass'

Thread 'A cylinder connected to a hanged mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How Do You Calculate Centripetal Acceleration in a Blood Centrifuge?
Replies
8
Views
1K
Centrifugal and centripetal forces on a half pipe
Replies
8
Views
4K
Normal force in a curvilnear motion
Replies
12
Views
3K
Centrifugal Force, Centripetal Force, and Space
Replies
2
Views
4K
How Does Centrifugal Force Compare to Gravity in Car Cornering Dynamics?
Replies
10
Views
5K
Foucault Pendulum at the North Pole
Replies
9
Views
2K
Medical Is a small tidal force on the body harmful over the long term?
Replies
12
Views
2K
What Day Length Would Allow Objects to Float at the Equator?
Replies
8
Views
3K
Centrifugal Force: Are these equal?
Replies
2
Views
2K
How High Must a Rocket Travel for Gravity to Halve?
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top