Physics Question from Nova Physics

[In Missourah]

If you have a copy of Nova's "The MCAT Physics Book," perhaps you could help me with this question. It's from the passage in Chapter 5, page 75. The question is question 5 following the passage, on page 76.

How can C be the correct option? According to the second-to-last paragraph, the only reason there is a difference between the distance from the center of the earth to the equator and from the center to the pole is because the earth is rotating. If one were to measure the force of gravity at the equator and poles of a nonrotating earth, their distances would be the same, and thus their gravitational force (and scale reading) would be the same. Wouldn't it?

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[Zona Pellucida]

Tag.

I just went back into my book and noticed I have this question starred to check with my Physics Prof as I don't understand their explanation as well.

 

[EyEnStein 07]

wouldnt it be better if you posted the question? and the choices? that way everyone who doesnt have the book can assist you as well...

 

[physics junkie]

wouldnt it be better if you posted the question? and the choices? that way everyone who doesnt have the book can assist you as well...

That would be too logical.

 

[physics junkie]

Oh, and I think you might be referencing a phenomenon that is the result of the centrifugal force generated by the earth's rotation. I can't really answer your question because I don't have that book.

http://en.wikipedia.org/wiki/Equatorial_bulge

http://en.wikipedia.org/wiki/Earth's_gravity#Latitude

http://www-istp.gsfc.nasa.gov/stargaze/Srotfram1.htm

It turns out that when we stand straight up we are not actually standing straight up but instead have an angle pointing us towards the equator. Why don't you notice this? Because everything stands at this angle(If I remember correctly its around 5 degrees). Bizarre, I know.

 

[In Missourah]

wouldnt it be better if you posted the question? and the choices? that way everyone who doesnt have the book can assist you as well...

The reason I didn't post the passage was that I did not want to unwittingly violate a copyright by posting copyrighted material. But here it is (and mods, if it is inappropriate please feel free to delete it):

Consider an object sitting on a scale at the surface of the Earth.The scale reading is the magnitude of the normal force which the scale exerts on the object. To a first approximation, there is force balance, and the magnitude of the scale's force is the magnitude of the gravitational force:

F = GMm/(R^2) (1)

Where F is the gravitational force, G is Newton's constant, M is the mass of the Earth, and R is the radius of the Earth. The simple result is that the force of gravity, and the reading of the scale, is proportional to the mass:

F = mg (2)

Where g has the value GM/(R^2) = 9.8 m/(s^2). We have made several idealizations, however, and if we want to calculate the scale reading, we need to be more careful.

For example, we have ignored the rotation of the Earth. Consider a man standing on a scale at the equator. Because he is moving in a circle, there is a centripetal acceleration. The result is that the scale will not give a reading equal to the force of gravity (equation [1]).

We have also assumed that the Earth is a perfect sphere. Because it is rotating, the distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1%.

A third effect we have ignored is that the Earth has local irregularities which make it necessary to measure g in the local laboratory, if we need an exact value of the effective acceleration due to gravity.


Now for the question...

If two identical men stood on scales at the south pole and at the equator of an Earth identical to this one but nonrotating, how would the reading of the polar scale compare to the equatorial one?
A. It would be less.
B. It would be the same.
C. It would be greater.
D. There is not enough information to answer this question.


Now I've posted the passage and the question, look at my reasoning in the first post and tell me how C can be the correct option.

 

[physics junkie]

The reason I didn't post the passage was that I did not want to unwittingly violate a copyright by posting copyrighted material. But here it is (and mods, if it is inappropriate please feel free to delete it):

Consider an object sitting on a scale at the surface of the Earth.The scale reading is the magnitude of the normal force which the scale exerts on the object. To a first approximation, there is force balance, and the magnitude of the scale's force is the magnitude of the gravitational force:

F = GMm/(R^2) (1)

Where F is the gravitational force, G is Newton's constant, M is the mass of the Earth, and R is the radius of the Earth. The simple result is that the force of gravity, and the reading of the scale, is proportional to the mass:

F = mg (2)

Where g has the value GM/(R^2) = 9.8 m/(s^2). We have made several idealizations, however, and if we want to calculate the scale reading, we need to be more careful.

For example, we have ignored the rotation of the Earth. Consider a man standing on a scale at the equator. Because he is moving in a circle, there is a centripetal acceleration. The result is that the scale will not give a reading equal to the force of gravity (equation [1]).

We have also assumed that the Earth is a perfect sphere. Because it is rotating, the distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1%.

A third effect we have ignored is that the Earth has local irregularities which make it necessary to measure g in the local laboratory, if we need an exact value of the effective acceleration due to gravity.


Now for the question...

If two identical men stood on scales at the south pole and at the equator of an Earth identical to this one but nonrotating, how would the reading of the polar scale compare to the equatorial one?
A. It would be less.
B. It would be the same.
C. It would be greater.
D. There is not enough information to answer this question.


Now I've posted the passage and the question, look at my reasoning in the first post and tell me how C can be the correct option.

It's pretty straight forward. The Earth isn't a perfect sphere. It is wider at the equator.

I THINK that you might have just misread the question. You assume the Earth is spherical when you use the F=mg equation. You are not supposed to keep making that assumption.

You can treat the earth as a point particle whose entire mass is concentrated at the center. If you want a mathematical proof of why you can check out Feynman's Physics Lecture's Vol 1. I suspect you don't so feel free to take my word on it.

The force of gravity is measured by the equation: F=G(m1)(m2)/(r^2)
(r^2) refers to the distance between the two objects. Let's say m1 is the mass of the earth and we say its all concentrated at the center of the earth and m2 is the person. Then the 'r' in r^2 refers to the distance between the center of the earth and the person on the surface. For someone at the equator, because it bulges, the distance from the center to the surface is larger than it is for someone on the south pole, where the distance between the center of the earth is shorter.

Look at the equation. If you decrease r^2 then you are dividing by a smaller number, thus making force increase. Force is weight.

All the information is in the passage. Reading comprehension.

 

[physics junkie]

Also, please post your questions in the correct forum. There is a sub-forum called Study Questions Q&A.

 

[In Missourah]

It's pretty straight forward. The Earth isn't a perfect sphere. It is wider at the equator.

I THINK that you might have just misread the question. You assume the Earth is spherical when you use the F=mg equation. You are not supposed to keep making that assumption.

You can treat the earth as a point particle whose entire mass is concentrated at the center. If you want a mathematical proof of why you can check out Feynman's Physics Lecture's Vol 1. I suspect you don't so feel free to take my word on it.

The force of gravity is measured by the equation: F=G(m1)(m2)/(r^2)
(r^2) refers to the distance between the two objects. Let's say m1 is the mass of the earth and we say its all concentrated at the center of the earth and m2 is the person. Then the 'r' in r^2 refers to the distance between the center of the earth and the person on the surface. For someone at the equator, because it bulges, the distance from the center to the surface is larger than it is for someone on the south pole, where the distance between the center of the earth is shorter.

Look at the equation. If you decrease r^2 then you are dividing by a smaller number, thus making force increase. Force is weight.

All the information is in the passage. Reading comprehension.

Our comprehension skills must be different, then. Look at the sentence you bolded. "Because it is rotating, the distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1%."

As I understand it, the sentence could be viewed as, "The distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1% because [the Earth] is rotating." Ergo, the earth's rotation is responsible for the difference in distances, and said difference would disappear if the Earth were not rotating. Hence, the Earth would be a perfect sphere if it didn't rotate, and gravitation at all parts would be the same.

That was what I got from trying to get all my information from the passage using my limited reading comprehension skills (as you implicitly state). Apparently, I needed the Feynman lecture series to grasp the concepts presented in it. But what do I know? I'm "illogical" and dumb.

 

[delempicka]

I'm not even going to comment on PJ (as I will from now on affectionately call him) posts...just know that you are not alone, and lots of people have been walloped...
Sometimes he has good stuff though. You just need to take a big red pen to 99% of it.
Watch.

It's pretty straight forward. The Earth isn't a perfect sphere. It is wider at the equator.

I THINK that you might have just misread the question. You assume the Earth is spherical when you use the F=mg equation. You are not supposed to keep making that assumption.

You can treat the earth as a point particle whose entire mass is concentrated at the center. If you want a mathematical proof of why you can check out Feynman's Physics Lecture's Vol 1. I suspect you don't so feel free to take my word on it.

The force of gravity is measured by the equation: F=G(m1)(m2)/(r^2)
(r^2) refers to the distance between the two objects. Let's say m1 is the mass of the earth and we say its all concentrated at the center of the earth and m2 is the person. Then the 'r' in r^2 refers to the distance between the center of the earth and the person on the surface. For someone at the equator, because it bulges, the distance from the center to the surface is larger than it is for someone on the south pole, where the distance between the center of the earth is shorter.

Look at the equation.If you decrease r^2 then you are dividing by a smaller number, thus making force increase. Force is weight.

All the information is in the passage. Reading comprehension.

Personally, I love the parting comments. I would love just a whole thread of them.
I will look at the quesiton tommorrow...
and also, who wants to write out a whole passage question. clearly it's a pain and if someone else has the book and knows the answer, it's easier.

sub-forums, schmub-forums...if you've only posted 5 times, perhaps it's difficult to know where to put it...

oh, but, wait....reading comprehension. (i think I might have read his (assumption) mind...)

 

[physics junkie]

Our comprehension skills must be different, then. Look at the sentence you bolded. "Because it is rotating, the distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1%."

As I understand it, the sentence could be viewed as, "The distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1% because [the Earth] is rotating." Ergo, the earth's rotation is responsible for the difference in distances, and said difference would disappear if the Earth were not rotating. Hence, the Earth would be a perfect sphere if it didn't rotate, and gravitation at all parts would be the same.

Ah, I see why you are confused now. More dirt/rocks/sand has been deposited at the equator over millenia because the Earth because of the Earth's rotation. This is the reason why the Earth is not a sphere. If you had a planet identical to Earth(same size, shape, mass) but not rotating the equator would still bulge, right? If you don't maintain the shape you can hardly consider the two to be identical. Once again, reading comprehension.

If you think about your explanation it doesn't really make sense on any intuitive level that because you are spinning you are further away from the center. Think about being a merry-go-round. Does it make any sense that you would be further away from the center when someone comes and spins it? Nope, not one bit.

As an aside, abruptly stopping the Earth from spinning isn't going to magically redistribute all the dirt/rocks/sand around the globe. In fact, the only way to do that would be to either wait a few million years for equilibrium to set in or start spinning the Earth on a north-south direction instead of an east-west one for just enough time for that bulge to redistribute itself uniformly around the globe. Luckily, the question bypassed any of this line of inquiry by just assuming you are on a planet identical to Earth...otherwise you would have a different answer if the question asked you about the force of gravity on Earth if it were to gradually slow down(it would equilibrate the bulge as it slowed). The Earth has actually been equilibrating the bulge out of existence since it was first formed as its angular velocity was greater in the past than it is today.

Where I come from you typically offer a thank you when someone takes the time to help you. Having an ego about advice to read a textbook will get you nowhere.

We can piss our pants but it will only keep us warm for so long, gentleman.

How's that for an end comment?

 

[physics junkie]

I'm not even going to comment on PJ (as I will from now on affectionately call him) posts...just know that you are not alone, and lots of people have been walloped...

Sometimes he has good stuff though. You just need to take a big red pen to 99% of it.

Let me fix that for you.

99% of the time he has good stuff though. Sometimes you need to take a big red pen to 1% of it.

I'm glad that you enjoy my posts though.

Edit: Oh wait, I just realized you're the guy from the PV=NkT thread! I only treated you like a 5 year old because you asked me to do so! That makes much more sense now. That was one of my meaner responses, though. Your line "I said treat me like a 5 year old, not like I was just released from Gitmo!" made me laugh.

 

[In Missourah]

Where I come from you typically offer a thank you when someone takes the time to help you. Having an ego about advice to read a textbook will get you nowhere.

How's that for an end comment?

Thank you.