MrBruce Posted February 11, 2006 Report Share Posted February 11, 2006 Hello all just signed up and was wondering if anyone can answer this question for me :) Right i was thinking the other day about this, i dont know why but i did, so if any brainbox can answer it would be interesting to know lol.. here goes.... Right as we all know we are held to the ground by a force called gravity, so i was thinking that in the way if you spin a roundabout (childs play thing in a park) the things on the round about will be forced to the outside and will eventualy fall off as the spinning forces push the object to the edge and off.. now the earth spins at a rate of 1 revolution a day, so how fast would the earth have to spin in order to overcome the effect of gravity and makes us start floating/moving away from the ground?? now i am drunk while typing this so i will read the wording tomoz but i think i got down what i mean :hihi: :eek: :eek: cheers :eek: Quote Link to comment Share on other sites More sharing options...
snark1100 Posted February 21, 2006 Report Share Posted February 21, 2006 the formula for the centrifugal force is w*w*r*sin(theta), w is the angular velocity in radians/sec, r the radius, and theta the longituder for the earth is 6.38*10**6 meters, and we want the centrifugal forceto equal the gravitational acceleration, 9.8 m/(sec*sec). if we are at the equator, so that theta=90 degrees, sin(theta)=1, and we must solve w*w = 9.8/(6.38*10**6), or w = sqrt(9.8/(6.38*10**6) = 1.24 * 10 **-3. the earth actual angular velocity, 1 rev. per day, gives 2*pi/(24*60*60) = 7.29*10**-4 radians per second, so, in order to balance gravity at the equator, it would have to be spinning about 580 times faster, that is, a day would be about 148 seconds long.. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 22, 2006 Report Share Posted February 22, 2006 Actually, that formula gives the magnitude of the acceleration but, if you're not on the equator, the direction is different so they would never total zero. The centrifugal term changes the direction of effective g from the exact radius and this is why Earth isn't exactly spherical. Quote Link to comment Share on other sites More sharing options...
arkain101 Posted February 22, 2006 Report Share Posted February 22, 2006 its pretty close to the speed of 18,000 mph to send you off into orbit. Quote Link to comment Share on other sites More sharing options...
snark1100 Posted February 22, 2006 Report Share Posted February 22, 2006 yeah, qf, there is that sin(theta) term, and, yeah, if you wander away from the equator, the force is no longer parallel to the radius so further calculaion would be necessary, and i am lazy. want to work it out for chicago? Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 23, 2006 Report Share Posted February 23, 2006 I'm lazy too! But before wasting time on calculations, the first thing that would happen as you increased the rotation, the Earth would begin to look like a bun and then more towards like a pancake but, as ground at the equator reached zero-g it would be ready to break away itself. Quote Link to comment Share on other sites More sharing options...
GAHD Posted February 25, 2006 Report Share Posted February 25, 2006 Not only that, increasing the earth's rotational speed to such a high speed would in of itself add to the gravity in the area, as the energy would add to earth's mass, Maby not that much of a difference but it would be there. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 25, 2006 Report Share Posted February 25, 2006 increasing the earth's rotational speed to such a high speed would in of itself add to the gravity in the area, as the energy would add to earth's mass:lol: Quote Link to comment Share on other sites More sharing options...
MrBruce Posted February 27, 2006 Author Report Share Posted February 27, 2006 hehe cheers for that :hyper: Quote Link to comment Share on other sites More sharing options...
Kayra Posted February 27, 2006 Report Share Posted February 27, 2006 But before wasting time on calculations, the first thing that would happen as you increased the rotation, the Earth would begin to look like a bun and then more towards like a pancake but, as ground at the equator reached zero-g it would be ready to break away itself. and would that not change the "Diameter" used in the formula.. reducing the required speed? Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 27, 2006 Report Share Posted February 27, 2006 the formula for the centrifugal force is w*w*r*sin(theta), w is the angular velocity in radians/sec, r the radius, and theta the longituder for the earth is 6.38*10**6 meters, and we want the centrifugal forceto equal the gravitational acceleration, 9.8 m/(sec*sec). if we are at the equator, so that theta=90 degrees, sin(theta)=1, and we must solve w*w = 9.8/(6.38*10**6), or w = sqrt(9.8/(6.38*10**6) = 1.24 * 10 **-3. the earth actual angular velocity, 1 rev. per day, gives 2*pi/(24*60*60) = 7.29*10**-4 radians per second, so, in order to balance gravity at the equator, it would have to be spinning about 580 times faster, that is, a day would be about 148 seconds long..Your approach is correct, snark, but you made a couple or arithmetic mistakes: 2*pi/(24*60*60) = 7.29*10**-5 So (proposed angular velocity)/(actual angular velocity)= 1.24*10**-3/7.29*10**-5 =~ 17. The earth must spin 17 times faster, for a day about 84.5 minutes long Quote Link to comment Share on other sites More sharing options...
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