# Homework Help? Science

Discussion in 'Community Discussion' started by obey908, Oct 6, 2009.

1. ### obey908 macrumors 6502a

Joined:
Sep 21, 2008
Location:
San Francisco
#1
I am not sure if this is the right spot to post this, but I was hoping someone could help me out with a homework assignment for my Earth class.

1. earth orbits the sun at a distance of 1AU. Determine its orbital velocity, Ve, in units of m/s. Assume circular orbit, for simplicity. Recall that the orbital period of the Earth is 1 year.

2. In a frame of reference in which the Earth and the Sun are at rest, the Earth's momentum (MeVe) must always be equal and opposite that of the Sun (MsVs). This implies that the Sun must also be moving around the center of mass of system. What is the Sun's orbital velocity around the center of mass?

3. The current detection limit for radial velcity, or Doppler, method of planetary detection is about 1 m/s. in this method, one is measuring the velocity of the star by using the Doppler shift. suppose that one was looking at the Earth-Sun system with this technique from a point within the orbital plane of the Earth, so that one would observe maximum velocity. Could you detect the Earth with currently available equipment?

4. Now, do the same calculation for Jupiter, which orbits the Sun at 5.2 AU. To do so, first find Jupiter's orbital velocity using Keplers 3rd law, then repeat steps 1-3. could we detect a planet like jupiter orbiting a star like the sun? how long would it take to do so?

Data:

1 au = 1.496x10^8 km
mass of earth: Me= 5.97x10^24 kig
mass of sun: Ms= 1.99x10^30 kg
mass of jupiter: Mj= 1.9x10^27 kg

I am pretty lost, and appreciate any help

Thanks

2. ### jake.f macrumors 6502

Joined:
Sep 4, 2008
Location:
NSW, Australia
#2
Note to self: NEVER take one of these "Earth" classes.

3. ### obey908 thread starter macrumors 6502a

Joined:
Sep 21, 2008
Location:
San Francisco
#3
lol, it is terrible. We are required to take 9 credits of science (3 classes)

I figured out the first one, but still working on the second

4. ### Chundles macrumors G4

Joined:
Jul 4, 2005
#4
OK, so one C half life is given as 5730 years.

Therefore, after 5730 years the ratio will be 1/2, another 5730 years it will be 1/4, then 1/8, then 1/16 and finally 1/32. So that's 5 half lives or 28,650 years.

Same sort deal for the second one, if it's a 1:1 reaction whereby one atom of U235 decays producing one atom of Pb207 and the ratio of U235b207 is 1:15 then it's just a matter of calculating how many half lives (or fractions of half lives it would take to go from 1:0 down to 1:15.

5. ### Xapphire13 macrumors regular

Joined:
Jan 14, 2009
Location:
South Australia
#5
1. 28,650 years (5 half lifes)
2. 2816 million years (4 half lifes)

6. Oct 7, 2009
Last edited: Oct 14, 2011

Joined:
Oct 7, 2009
7. ### obey908 thread starter macrumors 6502a

Joined:
Sep 21, 2008
Location:
San Francisco
8. ### Xapphire13 macrumors regular

Joined:
Jan 14, 2009
Location:
South Australia
#8

9. ### Ttownbeast macrumors 65816

Joined:
May 10, 2009
#9
concerning the carbon 14 to carbon 12 ratio I don't know the standards sorry I don't know the standards just the concepts for carbon dating but good luck and I hope you do well with your studies.

10. ### obey908 thread starter macrumors 6502a

Joined:
Sep 21, 2008
Location:
San Francisco
#10
i put a new problem up, if anyone would be able to help

11. ### SLC Flyfishing Suspended

Joined:
Nov 19, 2007
Location:
Portland, OR
#11
I night have gone about this overly simplistically but here you go, maybe it will get you thinking about a better way of doing it:

Simple geometry can give you a pretty good way of finding the earth's orbital velocity.

pi*r^2 is the formula for the circumference of a circle: 1 AU is the distance of the earth to the sun. But the sun has a diameter of 1,391,000 km. We divide that in half: 695,500 km radius of sun. Then convert to meters by multiplying by 1,000. 695,500,000 meters is the radius of the sun.

Then add that number to the AU figure you were supplied with (converted to meters as well: That gives us 150,295,500,000 meters is the distance from the earth to the center of the sun (or r in the pi*r^2).

That circle circumference formula then gives us the circumference of the earth's orbit. which I calculated to be: 7.0964611219167x10^22 meters

Then calculate the number of seconds in a year by taking 365 days * 24 hours * 60 minutes * 60 Seconds: 31,536,000 seconds per year.

Then take the circumference in meters divided by the seconds in the year and you get 2,250,273,059,968,512 m/s. or 2.25*10^15 m/s roughly estimated. You may need to re run this calculation using solely the AU figure (ignoring the radius of the sun as this may be included in the AU figure) or you may need to include the radius of the sun and the radius of the earth. I'm not an astronomer so I'm not familiar with exactly what the AU represents.

But the basic idea is still there.

SLC

12. ### SLC Flyfishing Suspended

Joined:
Nov 19, 2007
Location:
Portland, OR
#12
For number 2: once you get a solid answer to #1; #2 is easy! Simple algebra solves this.

Me*Ve=Ms*Vs

You already have Me given to you, you solved for Ve, you also have Ms given to you, so you solve algrbraically for Vs such that MeVe/Ms=Vs

I'm lost on #3, I know nothing about doppler shift etc so can't help there.

Hopefully that gives you a good start though.

SLC

13. ### obey908 thread starter macrumors 6502a

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Sep 21, 2008
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San Francisco
14. ### Ttownbeast macrumors 65816

Joined:
May 10, 2009
#14
Yah then the answer would just be in your pocket reference book like any good engineer has LOL

15. ### obey908 thread starter macrumors 6502a

Joined:
Sep 21, 2008
Location:
San Francisco
#15
i figured out the first two, stuck on the last two.

i hate being a business major and required to take science