Hi, I really got major problems with any piece of mathematical work. I took an a course in my university not knowing that it includes maths so it looks pretty bad now. Would be great if you could help me with this question:

1. Earth’s oceans cover roughly 70 percent of our planet, to an average depth of about 3.5 km. The density of ice is 0.9
that of water (that is why ice floats). Over a recent seven year period, several ma jor sections of Antarctic ice have broken off.
Among these are the Larsen A ice shelf, which measured 1,500 sq km, and which broke off in 1995. The 1,200 sq km Wilkins
ice shelf fell off in 1998 and the 14,500 sq km Larsen B dropped away in 2002. If each of these was 2.0 km thick (the average
ice thickness in Antarctica) and melted completely, how much would they have contributed to the global sea level rise over this
period? Note that the Antarctic ice sheet sits on a land mass so that if it melted the oceans would rise. By contrast, in the
Arctic, the ice floats on the Arctic Ocean so if it melted sea level would not be affected very much. [5 marks]

First off lets calculate how many cubic kilometers of ice that is. (1500x2)+(1200x2)+(14500x2)=34400 cubic kilometers of ice

34400x0.9=30960 cubic kilometers of fresh water

Fresh water has a density of 1, salt water has a density of 1.025

30960 * 1/1.025 = 30204 cubic kilometers of salt water

Maybe I'll do more later or someone can run with this.

This post was edited by Azrad on Nov 7 2009 08:35am

Thanks a lot so far its definately useful!

Would be great if someone could add to it.

ok question 1 is completed:


Q1

Larsen A: 1500 km²
Wilkins ice shelf: 1200 km²
Larsen B: 14500 km²

1500 km² + 1200 km² + 14500 km² = 17200 km²

17200km² x 2,0 km (thickness) = 34400 km³

34400 km³ x 0,9 (to get water) = 3096 km³

Earth surface is ~510 million km²
510 000 000 km² x 0,7 (to get the water area) = 357 000 000 km²


3096 km³ : 357 000 000 km² = 8,67e^(-6) km

8,67e^(-6) km x 1000 = 0,00867 m

0,00867 m x 100 = 0,867 cm

water would rise 0,867 cm worldwide



thats right or??


pls help me with 2 and 3 I might be able to pay some fg if its good.

This post was edited by MyAccountIsOsterHues on Nov 7 2009 08:03pm

forgot 0 there

For question 2 I got that:

2)

a) How long is the round-trip-time for a radio-signal from earth to saturn?

closest distance between saturn and earth: 1192,3 million km
radiowaves move at lightspeed: lightspeed = 300000 km/s

(1192300000 km) / (300000 km/s) = 3974,33 s

round-trip-time: 3974,33 s * 2 = 7948,66 s

latency is 7948,66 s thats like 2,2h


b.) How far would a spacecraft in 200000km orbit around Saturn travel in that time?

At first u have to get the speed of the spacecraft. I found this on the interwebz:

The speed (v) of a satellite in circular orbit is:

v = SQRT(G * M / r) "SQRT" means "square root"

G is the universal gravitational constant (6,6726 E-11 N m^2 kg^(-2))
M is the mass of Saturn (5,l85 E26 kg)
r is the radius from the center of the planet. (200000 km)
// a little note: I think he means from the center, cause everything is always measured from the center of the
// planets as far as i saw it. And it would match with the the part of the question where it says "just outside
// the ringsystem".

Now the maths:


v = SQRT(G * M / r)

G = 6,6726 E-11 N * m^2 * kg^(-2) //info: 1N = 1 kg * m * s^(-2)
= 6,6726 E-11 m^3 * kg^(-1) * s^(-2)
M = 5,685 E26 kg
r = 200000000 m

put all in

v = SQRT(189668655 m^2 * s^(-2))

v = 13772 m/s = 13772 m/s * 1km/1000m * 3600s/1h = 49579,2 km/h

This means the spacecraft would travel
7948,66 s * 13772 m/s = 109468945,52m = 109468,94552 km
in this time.


c) Do you think that mission control could maneuver the spacecraft in real time - that is,
control all its functions directly from Earth?

Control a object that moves with 49579,2 km/h directly woth a latency of 2,2 h?
Thats impossible, except u learned anything realy cool, of which i didn´t think here.



d) In particular, describe the sequence of events you would have to go through to
take a picture of Titan.

Puh thats a rough one. I think its only possible by pure math. U have to calculate everything through,
when and where the spaceship/satelite will be. Taking tons of picture with a awesome resolution and the fact
that Titan is with 5150 km a real big moon will increase the chances for good pictures.

.
Thats all i got. read it over and think about it for urself. Maybe u learned something, which I dont know.
Also as u see is v now different to the past PM. Was a lil lazy and hit wrong button at calc. Was / instead of *.
But this time calculation are right. Checked them over.

Ok to cover 357,000,000 km² with 0.001 kilometers of water (1 meter):
357,000,000 x 0.001 = 357,000 cubic kilometers, we have 30,204 cubic kilometers.

30204/357000=0.0846 so we can raise it about 8.46% of a meter, or 8.46 centimeters.

This post was edited by Azrad on Nov 8 2009 02:50am

why multiply with 0.001 km's ?

Basically it was an attempt to make the numbers look prettier.

I had a volume of the water the ice would make (a cubic number), and the surface area of the Earth's oceans (a square number). To be honest I wasn't sure how to handle the difference in dimensions so I did the following:

I decided to see how much water it would take to raise the ocean a certain height (which would be a cubic number), then compare that to how much we had from the ice (also a cubic number). I originally was going to raise it 1 kilometer, but immediately realized that the resulting number would have been hundreds of times larger than the amount of water we had from the ice. This method would have worked and gotten the same answer but it would not have given us the pretty fraction that reduced to 8.46% of a meter, it would have been 0.00846% of a kilometer. Does this explanation help? Or did I take a wrong turn somewhere in my calculation?

thanks man. nah didnt find any mistakes with the calculations.