# Another kibble problem

1. Oct 31, 2008

### haplo

Hi everybody, thank for helping with previous problem. I have another quesion:
This is not an actual problem. While the problem is actually solved in the book one of the steps there is not entierly clear as a result I cannot understand simmilar problem.
So here it is
1. The problem statement, all variables and given/known data
IF the earth orbit is divided in to by it's minor axis, how much does it spent in one half than another.

2. Relevant equations

Here is how is is solved in text. Eccentricity of earth orbit is 0.0167. It is clear that area of swept are is 1/2 area of elipse plus/minus the area of triangle with base 2*b and height ae. so A=1/2*Pi*a*b plus/minus a*e*b. Thus the time tame n are (1/2 plus/minus e/Pi) Years.

Thats what I have difficult understanding. How did they make a jump from area to years. Apparently the second evauation was obtined by dividing A over area of elipse a*b*Pi. If this is the case, how did author extracted years?

2. Oct 31, 2008

### Dick

The text used Kepler's second law. Is that the question?

3. Oct 31, 2008

### haplo

It was one of my guesses. If the second keplers law is used than dt=constant/da= J/(2*m*da) . I just fail to see why suddenly Pi appears in the denominator, as if they divided area by a*b*Pi

4. Oct 31, 2008

### Dick

The ratio of the areas swept is proportional to the ratio of the times. I've got to say I'm having problems following the rest of your post. But the ratio of the areas of the triangles is also proportional to the areas of the corresponding parts of the ellipse.

5. Nov 1, 2008

### wdednam

Hi there again,

I really struggled to figure out how they got the final answer for the difference between the times in this example.

But here's how I eventually got the answer (using what Dick suggested you use in his reply above):

The position vector sweeps out a total area of pi*a*b every year (period of the earth's orbit around the sun). So if you divide 0.5*pi*a*b +/- a*e*b by pi*a*b / year, it will give you the final answer 0.5 +/- e/pi years, the difference between the times spent by the position vector in the two halves of the Earth's orbit.

Hope this helps,

Wynand.

6. Nov 1, 2008

### haplo

thakns!, it is just Pi*a*b/year, dam I am retarded..

7. Nov 1, 2008

### wdednam

Don't mention it, I'm glad I could be of help:)

I don't think you're a retard, otherwise I'm one too:

I was actually completely bewildered when I first read this example. I didn't even realise that the origin of the position vector was one of the foci of the ellipse. I was still thinking in terms of the centre of a "circle". Haha - how's that for forgetting high school analytic geometry?