# Thread: Piel: How much does the Earth weigh?

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## Piel: How much does the Earth weigh?

Glad to hear that your Greek/Olympics Girl Scout camp went so well, PL. Here is a game for your next one. (Who says book larnin' has no practical uses?)

Your task is to figure out how much the Earth weighs. You are allowed to select two tools to help you. Naturally, you choose a pair of scissors and a shovel.

Point one blade of the scissors at the North Star, the other blade at the horizon and measure the angle between them. This is your latitude. (I just did this in my back yard. The latitude of Detroit is 42 and a quarter degrees.)

Now go a few miles to the South and do it again. (I drove 52 miles to Toledo. According to my scissors, Toledo is located at 41 and a half degrees North latitude.)

OK, this is a difference of three-quarters of a degree. Now we can find the circumference (C) of the Earth by the ratio

.75 degrees / 360 degrees = 52 miles / C

So the circumference of the Earth is 25,000 miles.

Step 2. Now take your shovel and dig a tunnel through the center of the Earth, all the way to the other side. When I did this last night, the length of my tunnel (= the diameter of the Earth) was

D = 25000/pi miles plus 3 millimeters.

But according to Euclid (C = pi D), this is three millimeters too long! This excess is due to the local warping of space by the Earth's gravitational field. Einstein says the extra length should be about .00000000000000000000005 cm. per metric ton of mass. So the Earth weighs

.3 cm / .00000000000000000000005 cm/ton = 6,000,000,000,000,000,000,000 tons.

(I can't believe I wrote that.)

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Oh my!

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Hmmmm....after hearing how fat the earth is, I'm feeling almost ice-dancer thin in relation. Thanks, Mathman!

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Ok, I got lost after the shovel and scissors part.

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Originally Posted by Mathman
Glad to hear that your Greek/Olympics Girl Scout camp went so well, PL. Here is a game for your next one. (Who says book larnin' has no practical uses?)

Your task is to figure out how much the Earth weighs. You are allowed to select two tools to help you. Naturally, you choose a pair of scissors and a shovel.

Point one blade of the scissors at the North Star, the other blade at the horizon and measure the angle between them. This is your latitude. (I just did this in my back yard. The latitude of Detroit is 42 and a quarter degrees.)
There is no way a pair of scissors can give you a 42 degree reading, give me a break, I have never seen a pair of scissors equipped with that. You need more than a pr of scissors.

Now go a few miles to the South and do it again. (I drove 52 miles to Toledo. According to my scissors, Toledo is located at 41 and a half degrees North latitude.)
Again, scissors can give the accuracy of half a degree, maybe if you have superman's eyes

OK, this is a difference of three-quarters of a degree. Now we can find the circumference (C) of the Earth by the ratio

.75 degrees / 360 degrees = 52 miles / C

So the circumference of the Earth is 25,000 miles.

Step 2. Now take your shovel and dig a tunnel through the center of the Earth, all the way to the other side. When I did this last night, the length of my tunnel (= the diameter of the Earth) was

D = 25000/pi miles plus 3 millimeters.
How did you measure the 25000/pi miles + 3 mm? by grandma's measuring tape? Even if you can do it with grandma's measuring tape, you need more than just a pr of scissors and a shovel.

But according to Euclid (C = pi D), this is three millimeters too long! This excess is due to the local warping of space by the Earth's gravitational field. Einstein says the extra length should be about .00000000000000000000005 cm. per metric ton of mass. So the Earth weighs

.3 cm / .00000000000000000000005 cm/ton = 6,000,000,000,000,000,000,000 tons.

(I can't believe I wrote that.)
oh well. Now how does all this relate to the black holes which btw was one of the favorite topics here at GS?

Then how does spin effect the radiation recoil, when gravitational waves come off how much momentum and energy come with it? So when one black hole plunges into another beaming the gws and momentum with it how does the larger black hole accomodate or conserve the momentume?

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Originally Posted by gezando
There is no way a pair of scissors can give you a 42 degree reading, give me a break, I have never seen a pair of scissors equipped with that. You need more than a pr of scissors.

How did you measure the 25000/pi miles + 3 mm? by grandma's measuring tape? Even if you can do it with grandma's measuring tape, you need more than just a pr of scissors and a shovel.
You caught me. We need a pair of scissors, a shovel, a ruler and a protractor.

Again, scissors can give the accuracy of half a degree, maybe if you have superman's eyes.
OK, we need a shovel, a ruler, a protractor and a mariner's astrolabe. All aspiring romantic poets -- rbp take note -- should be acquainted with the word "astrolabe," because it is the only word in the English language that rhymes with "babe."

You're my sweetie, you're my Babe,
My compass, sextant and astrolabe.

oh well. Now how does all this relate to the black holes which btw was one of the favorite topics here at GS?
Ahem. Cough cough. Funny you should ask. The estimate of .00000000000000000000005 cm per ton of mass is valid only in the special relativistic approximation. If the object is very massive, like a black hole, or rotating at a measurable fraction of the speed of light, like a neutron star, then we have to use the full force of general relativity. Then the problem becomes, how to distinguish space from time.

Then how does spin effect the radiation recoil, when gravitational waves come off how much momentum and energy come with it? So when one black hole plunges into another beaming the gws and momentum with it how does the larger black hole accomodate or conserve the momentum?
Since I believe in Einstein but not so much in quantum theory, I think that the momentum that is lost by gravity waves is exactly compensated for by the decay of the orbit of the smaller object.

This is such a cool theory. It only has one tiny flaw. Neither gravity waves nor gravitons have ever been observed. But we can't let that detail stand in our way. All we need is for grandma's tape measure to be graduated in units of 10^(-21) centimeters. Then we could measure how many gravitons Michelle is giving off in this spin, knowing her mass of 105 pounds and estimating her angular velocity from the angle of her skirt.

http://www.freep.com/art/1998/jan/18/0118_kwan.jpg

Mathman

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Originally Posted by Mathman

The estimate of .00000000000000000000005 cm per ton of mass is valid only in the special relativistic approximation. If the object is very massive, like a black hole, or rotating at a measurable fraction of the speed of light, like a neutron star, then we have to use the full force of general relativity. Then the problem becomes, how to distinguish space from time.
OK, MM you rule, I am not worthy. .

This is such a cool theory. It only has one tiny flaw. Neither gravity waves nor gravitons have ever been observed. But we can't let that detail stand in our way. All we need is for grandma's tape measure to be graduated in units of 10^(-21) centimeters. Then we could measure how many gravitons Michelle is giving off in this spin, knowing her mass of 105 pounds and estimating her angular velocity from the angle of her skirt.

http://www.freep.com/art/1998/jan/18/0118_kwan.jpg

Mathman
A MK picture and you can measure her angular vel by the angle of her skirt.

Now I really have a question for you math men, women, persons. Recently I bought a laptop computer. I paid for 30GB. When I received the computer and checked on the C drive property, it stated that I have 24.8 GB capacity. I don't have other drives that eats up GB. I called the tech support people, they told me that the manufacturer in this case Dell considers 1 GB as 1 billion bytes, but microsoft XP considers 1 GB as 1 billion, 73 million, 741 thousand and 824 bytes. When I divide 24.8 by 30 = 0.83. I can't divide 1 billiion by 1 bill 73 mil 741 thousand 824 because my \$5 walmart calculator does not go that far. But I believe even if it does I won't get 0.83. So what kind of fuzzy math is Dell trying to pull?

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Well, I can explain one part of it. There are two ways to compute a "kilobyte." You might think that means 1000 bytes. But since a computer works internally in binary, it is very convenient to use powers of 2 as the basic units. 2^10 = 1024. So for most computer applications,

one kilobyte = 2^10 = 1024 bytes,

one megabyte = 2^20 = 1048576 bytes (instead of 1,000,000), and

one gigabyte = 2^30 = 1,073,741,824 bytes (instead of 1,000,000,000).

This is about 93%. I don't know wher the other 10% comes from (83% instead of 93%), but I assume that the computer hardware itself uses up some of its own capacity just by existing and being compatible with various operating systems, etc. On my computer I am supposed to have 100 GB on my hard drive, but I only have 57.2.

I think that's...let's see now, take the cube root of the hypontenuse...I think it's about 57.2%.

Not that I could ever use of all that. So far I have only used 4.1 GB, and that counts my whole MK collection.

Mathman

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Originally Posted by Mathman
This is about 93%. I don't know wher the other 10% comes from (83% instead of 93%), but I assume that the computer hardware itself uses up some of its own capacity just by existing and being compatible with various operating systems, etc. On my computer I am supposed to have 100 GB on my hard drive, but I only have 57.2.

I think that's...let's see now, take the cube root of the hypontenuse...I think it's about 57.2%.

Mathman
Ok, I buy that the computer hardware itself uses up some of itw own capacity. BTW are you serious about the cube root?

Right now my laptop has 17.5 (binary GB free space) I guess that should be enough since I have already loaded microsoft office, window media, real one, antiviral, firewall, the dvd/ cd burning programs.

In addition, the flash cards (memory sticks) are so convenient. For some crazy reason I have three PDAs, and a desktop at home, so I don't need this laptop to store any MK pictures.

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