you knew it was coming...keplers third law.

I don't even know how to ask the question.

This law relates a planets orbital period (p) to the semimajor axis
of it's elliptical orbit (a) astronomical units.

(p) squared=(a) cubed

orbital period would be the time it takes to orbit the sun?

what is (a)? the distance between the two focus points of the ellipse?

If anyone could help with this...dumbed down please..
thanks.
jojo

"jojo" wrote in message
. ..
I don't even know how to ask the question.

This law relates a planets orbital period (p) to the semimajor axis
of it's elliptical orbit (a) astronomical units.

(p) squared=(a) cubed

orbital period would be the time it takes to orbit the sun?

what is (a)? the distance between the two focus points of the ellipse?

If anyone could help with this...dumbed down please..
thanks.
jojo

The square of the period is proportional to the cube of
the semimajor axis:

T^2 ~ a^3

What's the major axis? That's the longer of the two
axes of the ellipse. For a planetary ellipse, that
would be the straight line segment from perihelion to
aphelion.

Now, what's the semimajor axis? "Semi" means half, so
the semimajor axis is half of the distance between
perihelion and aphelion. For a circular orbit, the
semimajor axis is equal to the semiminor axis and equal
to the radius.

For an elliptical orbit, the distance between the two
foci is related to the lengths of the major and minor
axes. If a is the length of the major axis, b the minor
axis, and c the distance between the foci, then

a^2 = b^2 + c^2

"jojo" wrote in message
. ..
I don't even know how to ask the question.

This law relates a planets orbital period (p) to the semimajor axis
of it's elliptical orbit (a) astronomical units.

(p) squared=(a) cubed

orbital period would be the time it takes to orbit the sun?

what is (a)? the distance between the two focus points of the ellipse?

If anyone could help with this...dumbed down please..
thanks.
jojo

The square of the period is proportional to the cube of
the semimajor axis:

T^2 ~ a^3

What's the major axis? That's the longer of the two
axes of the ellipse. For a planetary ellipse, that
would be the straight line segment from perihelion to
aphelion.

Now, what's the semimajor axis? "Semi" means half, so
the semimajor axis is half of the distance between
perihelion and aphelion. For a circular orbit, the
semimajor axis is equal to the semiminor axis and equal
to the radius.

For an elliptical orbit, the distance between the two
foci is related to the lengths of the major and minor
axes. If a is the length of the major axis, b the minor
axis, and c the distance between the foci, then

a^2 = b^2 + c^2

"jojo" wrote in message
. ..
I don't even know how to ask the question.

This law relates a planets orbital period (p) to the semimajor axis
of it's elliptical orbit (a) astronomical units.

(p) squared=(a) cubed

Someone else already answered your questions, I am just going to say that
the law DO NOT states that "(p) squared=(a) cubed", rather the law states
that "(p) squared IS PROPORTIONAL to (a) cubed". In other words, to write
the law as an equality you'd have to add a constant (let's call it "K"):

p^2 = K * a^3

The only time when p^2 = a^3 is when "p" is given in earth years and "a"
is given in astronomical units (AU).

Guillermo

"jojo" wrote in message
. ..
I don't even know how to ask the question.

This law relates a planets orbital period (p) to the semimajor axis
of it's elliptical orbit (a) astronomical units.

(p) squared=(a) cubed

Someone else already answered your questions, I am just going to say that
the law DO NOT states that "(p) squared=(a) cubed", rather the law states
that "(p) squared IS PROPORTIONAL to (a) cubed". In other words, to write
the law as an equality you'd have to add a constant (let's call it "K"):

p^2 = K * a^3

The only time when p^2 = a^3 is when "p" is given in earth years and "a"
is given in astronomical units (AU).

Guillermo