Where Does Gravity Have the Most Impact?

[Creator]

Originally posted by climbhi
I'm sorry to bring this back up but I really must be missing something, wouldn't the spinning of the Earth make you heavier at the equator not lighter? F = mv^2/r ; v = ωr ; F = mω^2r , thus since ω is constant for anyone on the Earth the further you are away from the center of the Earth the heavier you are ... or am I doing something wrong here?

Climbhi,
No problem; Its an honest inquiring question, obviously with a willingness to find the truth.

You are correct again with the equation, but you are simply pointing the force in the wrong DIRECTION. The centrifugal force is OPPOSITE to the DIRECTION of acceleration due to gravity. Thus it must be subtracted from the gravitational force.

If the Earth were rotating fast enough, objects on the equator would have a 'effective' value of g = 0 and the objects would seem to be weightless.

...Which brings us around again to the OTHER reason why GRACE satellite measurements don't give you the effective g value equivalent to that meassured at a specific surface location;-- the satellite doesn't take into account the lower measured g value due to the object's ROTATION on the surface.

You and Rockzella both deserve credit for bringing that up.

Creator

 

[Andre]

Thanx, your deity, creator, for your sublime posts, how delicately and patiently put. I clearly see my omission now thanks to your elaborated masterpieces.

Perhaps it is permitted to return to my water droplet world again, let's build one, orbiting in space around the sun having an average radius of 6371km. Let's assume rather heavy water, with a mean density of 5515kg/m^3. Now let's rotate this droplet with one revolution per 86400 seconds. We see that results in flattening and the equatorial radius increases to 6378,1 km while the polar radius decreases to 6356,8 km. We measure the gravity to be 9,78 m/s^2 at the equator and 9,83 m/s^2 at the poles, a difference of 0,5%.

We calculate that the rotation gives us a centrifugal force of rω^2 =0,034 m/s^2 on the equator surface. We also calculate a delta in the gravity due to the difference of the pole and equator surface to the center of gravity to be a ratio of 0,67% or about 0,0677 m/s^2. Both values added together is just a bit more than the actual difference in the measured values (0,05 m/s^2). But we assumed a sphere here, which is a invalid simplification since the mass distribution is not spherical and mass is pulling down more effective at the equator, the basic message of the whole thread. This should make up for the difference.

Another question is: do we have a balanced situation now?

For balance we still require equilibrium, but –and here is my omission- the equilibrium is not dependable on force but on pressure. It is the total pressure from every angle in every position that needs to be balanced and of course, pressure is force per area. So for more area, less force is required to maintain equilibrium.

I would ramble that on every plane through the center of gravity the value of the sum of all pressures must be equal for all possible disks, while the vector sum of all forces should be zero. I figure this would require a great deal of calculus to catch this. But perhaps we can make some very simplified assumptions to get an approximation.

Lets concentrate on two disks, one disk cut in the plane of the equator and one disk cut in a polar plane. In two dimensions the total pressure of gravitational forces of the rims for instance should be equal. Hence the sum of the forces times the circumference. This proportionality equals to the ratio between the circumferences: More circumference, less force. And obviously, the equatorial rim is longer than the polar rim (67 kilometers).

But not only the rim is acting as a mass, the whole disk is. Integration for the disk would yield the same logic for the ratio of the areas of both disks, hence more area, less force. This should equal the difference in relation between both extreme gravity values or 0,5%.

Now the equatorial disk has a area of 127,80 million km square while the polar disk has a area of 127,37 million km square, a difference of only 0,33%, not enough to match the gravity difference. Hence it seems that the equatorial bulge of the water droplet is insufficient to form equilibrium. I see that I have to increase the equatorial bulge with some 4 kilometers and reducing the poles likewise, to having the difference in areas to match the 0,5% gravity difference, but this also would affect the ratio in gravity differences. The centrifugal force would only increase 0,07% for instance, the differential gravity however would increase 34%. It seems that we need to calculate the effect of the uneven mass distribution to find a more accurate result.

It seems that all the mentioned effects are working, including the effect of uneven mass distribution. But instead of forces, pressures should be in equilibrium. This leads to different values for gravitational forces but the sums don’t add up due to the mass distribution not being spherical. And in my example, the water droplet seemed not to be flattened enough to be in equilibrium. So I wonder about the Earth. http://www.gsfc.nasa.gov/topstory/20020801gravityfield.html But then again, Earth is much more complicated than a water droplet.