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Physics Problem

post #1 of 10
Thread Starter 
Given that The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 \\times 10^{8}\\; {\
m km}, and the earth travels around this orbit in 365 days, what is the orbital velosity of the earth?

I tried to do the following

x = 1/2 vt
1.5x10^11 meters = (.5)(v)(31536000 seconds)
1.5x10^11 meters = 1.5768x10^7 seconds * v
9512.94 meters/second = velocity

but this isn't the right answer, what am I doing wrong?
post #2 of 10
You have a radius but didnt convert it to the circumference of the circular orbit?
2*pi*1.5*10^8 meters?

Why did you use the equation x = 1/2*v*t? Where does the 1/2 come from? Distance = velocity * time.
post #3 of 10
Thread Starter 
it comes from the equation

x(final) - x(initial) = Time(v(initial)+v(final))/2

well all inital values can be considered 0 so this simplifies down to

x = 1/2vt
post #4 of 10
I'm sorry, but I just have to ask - do you think the folks here should be doing your homework for you?

And folks who are doing the homework - do you think that's fair?

If you need assistance you should be asking your prof.
post #5 of 10
Orbital velocity as I know it is calculated by

v= sqrt (GM/r)

where G is gravitational constant (6.674 x 10^-11 m^3 kg^-1 s^-2)
and M is central mass of object (Earth) (5.9742 x 10^24 kg)

Try these numbers and see if it matches with your "right" answer.

Edit: Your method calculates linear velocity. With orbital, velocity increases inversely as the square root of the orbital radius... hence my earlier equation.
post #6 of 10
Well apparently they don't mind doing your homework then. I hope they get some of the credit when you hand in your papers.
post #7 of 10
Quote:
Originally Posted by algebrapro18 View Post
it comes from the equation

x(final) - x(initial) = Time(v(initial)+v(final))/2

well all inital values can be considered 0 so this simplifies down to

x = 1/2vt
V initial is equal to V final... so this would give a 2v on the top. So 2 in the numerator and 2 in the denominator would cancel out... so your equation has a 1/2 that doesn't need to be there.

Re: Februa's equation- though useful, you don't really need it to solve the problem. You're making way too much work out of simple algebra. You only really need x = vt

Quote:
Originally Posted by Yosemite View Post
I'm sorry, but I just have to ask - do you think the folks here should be doing your homework for you?

And folks who are doing the homework - do you think that's fair?

If you need assistance you should be asking your prof.
I didn't actually do the work, but am trying to explain how he ought to do it. Although, to be honest, it's probably more important to know how to get an answer than to actually get one. I do agree that he should ask the professor, and it couldn't hurt to work on the homework with classmates...
post #8 of 10
Gee, that's what I thought. Calculate the distance around the orbit, and divide it by the number of hours it takes in order to get a velocity. Unless I'm seriously overlooking something (not impossible), this is a pretty good first semester high school math problem (and no algebra needed).
post #9 of 10
If its that simple a problem, isnt the answer just v=distance/time
where distance = 2pi(1.5 x 10^11 m)

and time is= 31536000 seconds (per year)

=29885.8 m/s
post #10 of 10
Quote:
Originally Posted by Februa View Post
If its that simple a problem, isnt the answer just v=distance/time
where distance = 2pi(1.5 x 10^11 m)

and time is= 31536000 seconds (per year)

=29885.8 m/s
I was thinking that it was something like that since the only given he mentioned was the radius of the orbit. If so, I find it a little bit silly because it isn't much of a physics problem and should be in his realm of comfort.
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