OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread

Original post by Summerdays

Hmm, well I guess because the equator is further away from the centre of mass thus the elements at the equator eperience the greatest centripetal force (is this correct?)
I am really talking about gravity cancelling.

I'm not so sure to be honest. I've looked it up on google and it seems it cannot be explained in terms of centripetal acceleration alone, so I don't think they would go as far as asking why the earth bulges at the equator.

Reply 403

Birkatron

Original post by Mikkels88

yeh but i think it acts to that effect, and it seems to make sense...

is it worth talking about pressure, even if its damn hard to understand? (like in my teachers answers)
BASICALLY: (i think)
If the earth was just a sphere, non rotating, then the gravitational force on some element of its volume (just a tiny section of it) would equal the force due to the pressure difference across that volume, (as there is more mass above the lower part than higher? im not sure... similar to air pressure i think) ,so its in equilibrium. However, if the earth is spinning, there must be a resultant force to the centre of the earth, and so the gravitational force needs to be larger than this pressure difference force. The element of volume therefore moves outward to reduce the pressure difference force (and ensure there is a resultant force), and so the radius from centre to equator becomes bigger! (You can even say that the polar radius therefore shortens to conserve volume, so even more bulge)

that make any sort of sense? :colondollar:

This is what my answers for the prerelease say, basically its saying the bulge is the earth correcting itself?

Reply 404

hannahbeex

Original post by Unkempt_One

I'm not so sure to be honest. I've looked it up on google and it seems it cannot be explained in terms of centripetal acceleration alone, so I don't think they would go as far as asking why the earth bulges at the equator.

Well I certainly hope they don't. If they give us some numbers and ask us to work something out related to it I'm fine with that....if I have to explain something then I'm screwed!

Reply 405

Birkatron

conservation of volume!

Reply 406

Original post by Mikkels88

yeh but i think it acts to that effect, and it seems to make sense...

is it worth talking about pressure, even if its damn hard to understand? (like in my teachers answers)
BASICALLY: (i think)
If the earth was just a sphere, non rotating, then the gravitational force on some element of its volume (just a tiny section of it) would equal the force due to the pressure difference across that volume, (as there is more mass above the lower part than higher? im not sure... similar to air pressure i think) ,so its in equilibrium. However, if the earth is spinning, there must be a resultant force to the centre of the earth, and so the gravitational force needs to be larger than this pressure difference force. The element of volume therefore moves outward to reduce the pressure difference force (and ensure there is a resultant force), and so the radius from centre to equator becomes bigger! (You can even say that the polar radius therefore shortens to conserve volume, so even more bulge)

that make any sort of sense? :colondollar:

Thank you, that explains it nicely. And because the presure difference depends on the distance of the element from the centre of the mass, and the gravitational force also depends on distance (GMm/r^2) The two force always coincide and balance :smile:

But when there's rotation, because PV = constant, each of the elements increase their volume in order to reduce the force due to the pressure difference so that Mg, of a particular element, is grater than the upthrust due o the pressure differencefrom the Earth's mantle fluid. Thus the element's begin to rotate.

Reply 407

Birkatron

This is a highly specific line to go down, but we all know what OCR B are like!.

Reply 408

Original post by Birkatron

conservation of volume!

That's a consequence but not a reason :redface:

Reply 409

Original post by Birkatron

This is a highly specific line to go down, but we all know what OCR B are like!.

I can't wait to get OCR B out of my life. It's quite safe to say that OCR B doesn't cater to physics :mad:

(edited 12 years ago)

Reply 410

Rogercbinboy

11

I'm feeling good about the first 60% but dreading section C - don't even know how they'd stretch 40 marks out of the 4 pages.

Reply 411

http://www.sparknotes.com/testprep/books/sat2/physics/chapter19section3.rhtml

Original post by Rogercbinboy

I'm feeling good about the first 60% but dreading section C - don't even know how they'd stretch 40 marks out of the 4 pages.

Hopefully with really easy questions

Reply 412

arianex

9

Original post by Summerdays

Hmm, the electron vibrates at specific wavelengths where its wavelength is equal to an integer multiple of the circumference of the nucleus. The electron's wavelength forms standing waves "at the limits" of the nuclues. The wavelength of the electron determines the enrgy of the elctron (E = hc/lambda.) The ground state of the electron corresponds to the first harmonic of its standing wave.

This link might also help :smile:

Oh yes, SAT Physics on Sparknotes! XD (I actually took that yonks ago in November.) Cheers dude, you're a lifesaver. :biggrin:

Reply 413

Well, here's what I can gather about why the earth bulges at the equator. You have to consider both total gravitational potential energy and rotational kinetic energy. The bulge is the formation by which the sum of those two energies is at a minimum.

Reply 414

Rogercbinboy

11

Probably obvious but when it says "0.1 virga changes direction in 3959.2 times in half an hour" how do you get the period from that?

Is it 3959.2/2 (since changes direction twice per oscillation) then divide that by 30*60 or is it (30*60)/(3959.2/2)?

Reply 415

Original post by Rogercbinboy

Probably obvious but when it says "0.1 virga changes direction in 3959.2 times in half an hour" how do you get the period from that?

Is it 3959.2/2 (since changes direction twice per oscillation) then divide that by 30*60 or is it (30*60)/(3959.2/2)?

Frequency = 3959.2/2=1979.6per half hour = 1979.6/(30x60) = 1.099...per second
=>time period = 1/f = 0.91s

Reply 416

Rogercbinboy

11

Original post by Unkempt_One

Frequency = 3959.2/2=1979.6per half hour = 1979.6/(30x60) = 1.099...per second
=>time period = 1/f = 0.91s

Right can remember that, cheers.

EDIT: Also, on the last page does anybody know why the 2nd method is more accurate than the 1st?

(edited 12 years ago)

Reply 417

JoeCarr

Original post by Summerdays

Gotta love physics without mathematics :colone:

THIS

ahaha but yeh, I need to do well so I can justify not taking maths to help with physics. :biggrin:

Reply 418

Original post by Rogercbinboy

Right can remember that, cheers.

Thanks to you as well. I will now remember to divide that number by 2, which I probably wouldn't have done otherwise.

Reply 419