- #1

- 655

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*(in days) of 7, and*

**P***(height of the peaks above zero) of 4*

**a**Now, here is what I am having trouble doing:

"Convert P and a to years and AU. Then, convert a to KM"

Can anyone help me with this?

I know that one J.D = 0.00096 = 142,984 km

?

- Thread starter nukeman
- Start date

- #1

- 655

- 0

Now, here is what I am having trouble doing:

"Convert P and a to years and AU. Then, convert a to KM"

Can anyone help me with this?

I know that one J.D = 0.00096 = 142,984 km

?

- #2

- 655

- 0

Thanks

(in days) of 7, andP(height of the peaks above zero) of 4a

Now, here is what I am having trouble doing:

"Convert P and a to years and AU. Then, convert a to KM"

Can anyone help me with this?

I know that one J.D = 0.00096 = 142,984 km

?

- #3

- 782

- 3

Strange question. Convert to years and AU, then to km. Why? You already have it in JD. So multiply by that conversion factor you've quoted to get km. Though diameters are a bit odd, since it's usually in radii as that's what plugs into the equations.Anyone? :)

Thanks

As for conversion to 'years', which year do they mean? A standard Julian year is 365.25 days, but there's several other years that get used to, such as the Gaussian.

An AU is presently defined as 149,597,870.7 km. You'll see 149,597,870.691 km get quoted, but that's a bit too precise for the error bars. Hopefully someone has seen fit to tell you what AU value to use.

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