ideasrule said:Start by drawing a free-body diagram of a person standing at the equator. Remember that part of gravity goes into providing the centripetal force.
huybinhs said:Ok. I got:
a = v^2 / r = 465^2 / (6378100) = 0.0321 m/s^2 (I took 6378.1 km as radious. Is this still correct?)
If yes, then:
m = F / a = 96.9 / 0.0321 = 3020 kg. Correct?
huybinhs said:If the person is standing at the equator, then force of gravity will be equal 0. Is this correct?
ideasrule said:Do you really think there's no gravity for people at the equator? The people there don't just float around.
ideasrule said:Again, draw a free-body diagram, label all the forces, and write down Newton's second law. If you've done any of these steps, you haven't shown them here.
huybinhs said:There will be 2 forces in the diagram: gravity and the normal force. The normal force represents the apparent weight of the of the object = 96.9 N.
the acceleration of the object = v2 / r = 4652 / (1.274×10^7 /2) = 0.03394 m/s^2.
Then I got equation forces:
96.9 - m (0.03394) => m = 2855 kg. Then my normal force = 2855 * 9.780 = 27922 N = final answer. (Note that 9.780 is gravity at equator). Is this correct?
ideasrule said:The normal force is the apparent weight of the object, but it isn't 96.9N. That's the normal force at the pole, which is just equal to mg since Earth's rotation has no effect there. From this you can find m, and from there the normal force at the equator (which is what the question asks for).
Seems right.
Where's your Newton's second law? Newton's second law says Fnet=ma. Fnet is composed of two forces: the normal force and gravity. a is just the acceleration that you calculated earlier.
mparsons06 said:How would the equation be set up to get g = 9.77 m/s^2?
I have different numbers, but I was just curious how the equation would look on the other side of g.
Then where did you plug g back into?
huybinhs said:After I found out my g is 9.77 m/s^2. I use the equation F at equator = m*a = 9.77 * 9.88 = 96.5 N . Hope this helps!
mparsons06 said:How did you get the 9.77?
Is it like:
96.9 - (9.88) * g = (9.88)(0.03394)
g = (9.88)(0.03394)/96.9 - 9.88
I'm so confused??
mparsons06 said:I get it now. Thank you.
Do you know how to start out problem #2?
The Earth's rotation causes centrifugal force, which pushes objects towards the equator, making them weigh slightly less than they would at the poles.
No, the Earth's obliquity (tilt) does not have a significant effect on an object's weight at the equator.
The equatorial bulge of the Earth, caused by its rotation, results in a slightly greater distance from the center of the Earth to the equator. This can cause objects at the equator to weigh slightly less than they would at the poles.
Yes, the higher the altitude of an object at the equator, the less it will weigh due to the decrease in gravitational pull at higher altitudes.
The difference in weight at the equator compared to the poles is very small, usually only about 0.5% of an object's total weight.