Go to Home Page
You are here
Go to Reference Section
Go to Directories Section
Go to Community Section
Go to Fun Section
Go to Science Store
Go to About PhysLink.com
Club PhysLink
   Not a member yet?
   Get Free Membership
   Username:
   
   Password:
   
    Remember me
   
   Forgot your login?
Top Destinations Menu
 Ask the ExpertsAsk the
Experts

 Physics and Astronomy Departments DirectoryUniversity
Departments

 Discussion ForumsDiscussion
Forums

 Online Chat Online
Chat

 FREE Einstein eCardsEinstein
eGreetings

 PhysLink.com Science eStoreScience
eStore

Community

Chikrii Word2TeX Software

Click here for a free 2-week trial

Become a Sponsor


   Question

How fast would the Earth have to rotate so that it would neutralize gravity?

Asked by: Brad Nelson

Answer

In order to neutralise the acceleration due to gravity the centripetal acceleration needs to be equal to the acceleration due to gravity:

Centripetal acceleration = 9.81 m/s2

The centripetal acceleration is, a:

a=r x w2

Where r is the Earth’s radius (in our case the radius at the equator), and w is the angular velocity.

Let a = 9.81 m/s2 and r = 6.4 x 106 m

9.81 = 6.4 x 106 x w2

Therefore w = 0.00124 rad/s

This is how fast the Earth would need to rotate to get centripetal acceleration at the equator equal to 9.81 m/s2.

So if we use this value in this equation:

w = 2/T

Where w is the same as before, the numerator is constant, and T is the time for rotation or the period.

If we put our value of omega (angular velocity) into the equation we find that T = 5074.99 seconds or 1.409 hours. This means that the Earth would need to rotate with a period of 1 hour 24 minutes. This means it would need to rotate approx. 20 times faster than it does now!

Answered by: Dan Summons, Physics Undergrad Student, UOS, Souhampton


go to the top  
Advertisement:



All rights reserved. © Copyright '1995-'2004 PhysLink.com