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11 answers

As the planet moves farther from the sun its velocity through space slows slightly. As it moves in towards the sun it speeds up. These velocity variations average out to give the planet one fixed time needed for one orbit of the sun.

2007-04-12 02:09:51 · answer #1 · answered by Chug-a-Lug 7 · 0 0

If one sphere orbits another it takes a given amount of time to do so, depending on orbital velocity or the speed with which it is orbiting and the distance between the objects.

The orbit may be irregular or uneven and the orbital velocity may be greater or slower. So while it is reasonable to assume that Mercury orbits the Sun in less time than the Earth for instance, as it is much closer, it may not be the case if the orbital velocity of Mercury is vastly slower than Earths.

2007-04-12 01:07:14 · answer #2 · answered by Nexus6 6 · 0 1

The square of the orbital period is proportional to the cube of the semi-major axis (Kepler's third law). Not only does the planet have further to travel to complete an orbit, it also travels slower because the gravitational attraction that's pulling the planet is less. It's called a Keplerian decay.

2007-04-12 01:40:35 · answer #3 · answered by Iridflare 7 · 0 0

What is your constant? The speed at which it travels?

If speed is constant then the further away it is from the sun he further it has to travel to orbit once.

Otherwise, it could just be travelling faster.

the circumference is the important factor here in simple terms assuming the orbit is circular. If speed is constant then distance from the sun multiplied by pi = circumference (or distance in the same units as the distance from the sun).

2007-04-12 00:59:47 · answer #4 · answered by PollyPocket 4 · 0 1

Newton's law of gravitation+ equation for cetripetal acceleration.
F(grav)=G*M*m/r^2: F(cent)=m*w^2*r
M=mass of sun, m=mass of planet, r=mean orbital distance,w=angular vel=2*pi/T: (T=orbital period)
>G*M*m/r^2=m*w^2*r
>G*M/w^2=r^3
>G*M*T^2/4*pi^2=r^3
>T^2 propnl to r^3
..or orbital period (squared) varies as the mean orbital dist (cubed)
works where speed of planet <

2007-04-12 13:23:47 · answer #5 · answered by troothskr 4 · 0 0

According to the formula p^2 = r ^3, developed by Kepler using observational data obtained by Tycho Brahe. As the (average) radius of a planet increases from the sun, its orbital period increases as (the radius ^ 3) ^ .5.

2007-04-12 01:09:23 · answer #6 · answered by Anonymous · 0 1

I recall a little equation that used the earths distance from the sun and one earth year to do the math. I can't remember what the equation looked like.

2007-04-12 01:16:14 · answer #7 · answered by Anonymous · 0 0

The farther from the sun the bigger the circle it has to make.

2007-04-12 00:59:33 · answer #8 · answered by Anonymous · 0 0

this rule is just for solar system and kepler find this rule which the period^2=distance^3 . period has the dimensions of the years and the distance has the dimensions of AU

2007-04-12 01:32:49 · answer #9 · answered by suerena 2 · 0 2

The farther out, the longer it takes.

2007-04-12 00:58:55 · answer #10 · answered by Gene 7 · 0 0

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