Assuming the Earth is a perfect sphere, how much less would you weigh on the equator than on either of the poles? How would your weights differ if you do take into account the non-spherical shape of the planet? Crazy...
I'm sure others will correct me if I am in error but as I understand it the bulge at the equator due to it's rotational affect is in exact balance. That is the added diameter (mass) completely compensates for the rotation velocity and hence you have no net change in weight.
Quite the opposite, MacM. Please Register or Log in to view the hidden image! The equatorial bulge places you slightly farther from the center of gravity plus the centrifugal force created by the rotation is a second factor that reduces your weight. (Your mass stays the same, of course, but your weight will be less.)
Good observation. As usual the font of all wisdom is Wikipedia Please Register or Log in to view the hidden image! (http://en.wikipedia.org/wiki/Gravity#Units_of_measurement_and_variations_in_gravity). They say about 0.5% variation between poles and equator. -Dale
Thanks, Dale. That's actually more difference than I was expecting. An entire pound difference, if you're around 200 lbs, just by changing location on the planet...that's pretty cool.
To even further complicate matters, gravitational acceleration is not constant even at the equator at different locations. Contrary to what might seem logical, the location with the LEAST observed gravity is near the Andes mountains in South America, far above sealevel. The area with the GREATEST gravitational acceleration is over the Indian ocean. Even though the mountains are closer to an orbiting object, as calculated as a radius from the center of the Earth, the density of the Earth is greater beneath the Indian ocean. A geostationary satellite located near the Andes tends to require a good bit of station-keeping to maintain its location. It tends to drift in a direction opposite the Andes because of the reduced gravity, like it was on a slight incline. Geostationary satellites near the Indian ocean are very stable, requiring little station-keeping. NASA uses the location over the Indian ocean to park obsolete satellites because they will stay in position well, with no additional maintance required.
So then how do you calculate gravitational attraction when one of the bodies isn't uniformly dense, such as with Earth? It would make sense to me that you would treat the center of gravity as the center of the object (hence the name "center of gravity" lol). In the case of the Earth, the center of gravity would be shifted from the center of the planet towards the Indian Ocean, thus a satellite over the Indian Ocean would be closer to the "center" of the planet, causing it to feel more gravitational pull. Is this how it works? I would assume that it is...simply the term "center of gravity" seems to explain it, but I really don't know much about gravitational attraction of non-uniform bodies.
Using the center of the Earth is an excellent model, but has to be modified for exacting needs, like the GPS satellites for instance, or errors in ranging occur. As mentioned above, the slight differences in gravitational pull has to be considered when placing geostationary in locations and the fuel required onboard the satellite for station-keeping maneuvers. Yes, a geostationary satellite is similar to be in a depression when over the Indian ocean, thus very stable.
******************** Extract ********************** http://www.metaresearch.org/cosmology/gps-relativity.asp General Relativity (GR) predicts that clocks in a stronger gravitational field will tick at a slower rate. Special Relativity (SR) predicts that moving clocks will appear to tick slower than non-moving ones.Remarkably, these two effects cancel each other for clocks located at sea level anywhere on Earth. So if a hypothetical clock at Earth’s north or south pole is used as a reference, a clock at Earth’s equator would tick slower because of its relative speed due to Earth’s spin, but faster because of its greater distance from Earth’s center of mass due to the flattening of the Earth. Because Earth’s spin rate determines its shape, these two effects are not independent, and it is therefore not entirely coincidental that the effects exactly cancel. ******************************************************* I am somewhat amazed that a highly trained professional such as yourself is unaware of this fact.
What we have here is the eternal conflict between grammar garblers versus syntax sticklers. Live and let live. I believe it are all good.
Regarding the centrifugal force "created by the rotation." Are we being pushed away from Earth or being pulled toward it? There is no such thing as centrifugal force, I've argued with my Physics teacher and I've lost...many times. Dont tell me he's an idiot either, he's the smartest teacher at my school.
Beavis, I guess I can see where you are coming from. http://scienceworld.wolfram.com/physics/CentrifugalForce.html This site explains centrifugal force briefly and explains why it is a "fictitious force." But nonetheless you cannot deny the "force" pushing you outwards from a rotating body, even if it is created by a linear force on a "rotating coordinate system." Cool stuff. What was your argument against your physics teacher and what was his counter-argument? Just curious.
First of all, we were talking about space stations in orbit around Earth. I believed that since you could feel something holding you down on the inside of the space station, it must be pushing outwards. Common assumption, correct? Then my teacher told me to draw vectors regarding the force and velocity of the system, and add up those vectors to create a Net Force. That Net Force didnt point outwards, it pointed diagonal. Let me upload a picture for you guys of what I had to draw. Please Register or Log in to view the hidden image! Where is the arrow for the "Centrifugal Force?" That force seems like its there, because at every single singularity, the Net Force vectors add up to a perfect circle! This makes it seem as though we are being pushed "outwards." EDIT: Instead of "Direction of space station" imagine that you see the words "Velocity of space station."
Hi Beavis, The centrifugal force is a "ficticious force". That means that it does not exist in a non-accelerating (aka inertial) reference frame. So, in the space station example, the centrifugal force does not exist in the proper frame of a person on a spacewalk outside the station. However, if you have an accelerating (aka non-inertial) reference frame you need the centrifugal force. So, in the space station example, the centrifugal force does exist in the proper frame of a person inside the station. On another thread we had a fairly involved argument about this recently: http://www.sciforums.com/showthread.php?t=50898 -Dale
