Physics Problem, some lateral thinking required :(

[Ragnarak]

This isn't homework but is a question on a past paper i'm trying to figure out.

Calculate the ratio of the mean density of the Earth to that of the Sun from the following data only. Give full details of your method.

Now the question is easy except all you are allowed to use to get the answer are:

One year: 3 x 10^7 seconds
Acceleration due to gravity at Earth's surface: 10 m/s²
Angular diameter of the Sun from the Earth: 1/2 degree
Length of 1 degree latitude on Earth's surface: 10^5 m

I've been playing around for a while and can't see how to get the answer

[at0mic.c0w]

Length of 1 degree latitude on Earth's surface: 10^5 m

use this to calculate the earth's radius, diameter etc

Acceleration due to gravity at Earth's surface: 10 m/s²

use this to calculate the earth's mass

Angular diameter of the Sun from the Earth: 1/2 degree

use this to calculate the radius at which the earth orbits the sun

One year: 3 x 10^7 seconds

use this to calculate the earth's speed and using it's mass u can calculate the centripetal force and hence the sun's mass

[DMZ]

For the Earth:

Given the length of 1 degree of latitude, you can work out the circumference
From the circumference, you can determine the radius ( r = c / 2.PI )
From the acceleration due to gravity on the surface and the radius, you can determine the mass ( a = GM / r^2)
From the radius and the mass, you can determine the mean density ( M = d. 4/3 PI r^3 )

Presuming the length of 1 degree latitude is on the equator, that is, otherwise there is no solution. As usual, a badly set question.

Coming soon... the sun!

[MrL_JaKiri]

What he said.

[DMZ]

Originally Posted by at0mic.c0w

Length of 1 degree latitude on Earth's surface: 10^5 m

And to think I rushed to get the first reply in. I hate you.

[at0mic.c0w]

you're welcome

[DMZ]

Originally Posted by at0mic.c0w

you're welcome

You actually have a mistake - you cannot determine the radius the earth orbits the sun from the sun's angular dimension, since you are not given the sun's dimensions to work with.

I believe the next step is to use the mass of the earth, orbital period and angular dimension of the sun to determine the ratio of mean densities between the two bodies - you cannot determine the mass or radius of the sun from the initial values provided, surely?

[at0mic.c0w]

Originally Posted by DMZ

You actually have a mistake - you cannot determine the radius the earth orbits the sun from the sun's angular dimension, since you are not given the sun's dimensions to work with.

that's down to my understanding of the english language tho. i'm not even quite sure what a mean density is

[Ragnarak]

Originally Posted by DMZ

For the Earth:

Given the length of 1 degree of latitude, you can work out the circumference
From the circumference, you can determine the radius ( r = c / 2.PI )
From the acceleration due to gravity on the surface and the radius, you can determine the mass ( a = GM / r^2)
From the radius and the mass, you can determine the mean density ( M = d. 4/3 PI r^3 )

Presuming the length of 1 degree latitude is on the equator, that is, otherwise there is no solution. As usual, a badly set question.

Coming soon... the sun!

I was with you up to the end of this but then i couldn't manage to do the same for the Sun but at0mic.c0w sorted that out

Thanks guys

[Ragnarak]

I'm about ready to give up on this as i seem to have forgotten how to do just about everything

Seurat was a pointillist paper, his pictures were made up of thousands of dots, typically separated by 0.5 mm. About how far away must one stand from a painting by Seurat to gain the impression of continuous colours?

To do this i used
d sin theta = 1.22 lambda

d = aperture diameter which i took to be around 5mm for the eye.
lambda = wavelength of light which i took to be 500 nm

For 2 dots to be resolved you need to be close enough so that the angular distance theta is less than 1.22 lamba / d , therefore we need theta greater than this value so dots aren't resolved.

Can get angular distance in terms of the distance away from painting by using:

theta/360 * 2*Pi*x = 0.5mm

then you set this less than the 1.22 lambda / d above and solve for x.


However i've ****ed up somewhere as i get answers that seem far to big.

If anyone spots a glaring mistake in the above i'd appreciate it if you let me know, otherwise i'll try again to figure out where i went wrong tonorrow.

[Stew]

Can't you just ring up/meet someone from your course?

[Ragnarak]

Originally Posted by Stew

Can't you just ring up/meet someone from your course?

These are just two of like a million past problems. Most are ok and it's obvious what to do but a couple are like

No one else is really bothering to do any of them so can't really ask anyone.

[DMZ]

Originally Posted by Ragnarak

I'm about ready to give up on this as i seem to have forgotten how to do just about everything

You think that's bad - last time I had to do anything like that was 10 years ago.. horrifying to find out I've gone from being able to do high order / partial differential equations in multiple dimensions to not remembering what a tangent is.

[MrL_JaKiri]

It's from the latin, tangere, to touch.

[DMZ]

Originally Posted by MrL_JaKiri

It's from the latin, tangere, to touch.

Is that a sine of something?

[Dead_Meat]

It's a sine of Mark having nothing to do on a Sunday but post here.


OMFGALOLMARMITEONTOASTALOLOLMFG

[Jennifer]

You can't approximate the dots to point sources is basically your answer. The whole painting will be emitting radiation, and each colour will be a broad section of the spectrum. Consider the effect of different colours on the appearance of your sinc function, and then consider that all 'points' on the background will also have such sinc functions, just with different wavelengths. I think you need to reconsider what 'counts' as being blurred out enough for the picture to appear continuous. You certainly wouldn't have to wait until the Rayleigh Criterion was satisfied.

[HAL-9000]

Originally Posted by DMZ

Presuming the length of 1 degree latitude is on the equator, that is, otherwise there is no solution. As usual, a badly set question.

Coming soon... the sun!

Degrees of latitude run parallel to each other. They are all of the same distance apart and measured from the angle between the centre of the earth to the equator and the position north or south of the equator that you wish to measure (by centre of the earth i mean the very middle, deep in the core!) . For instance, at 50 degrees of latitude the angle from the 3 points Equator, Centre of Earth, 50 degrees latitude is, believe it or not, 50 degrees! As one degree is equal to 60 nautical miles, you will see that every line of latitude is the same distance apart: 60nm. I believe you have confused latitude with longitude, in which case the distance would be dependant on the position relative to the equator.