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8 answers

Kepler's third law:
The square of the period is proportional to the cube of the mean distance (or semi-major axis).

Mathematically, it works as:

(tau) = sqrt (4 pi^2 a^3/mu)
where mu is the heliocentric gravitational constant.

The heliocentric gravitational constant is equal to the mass of the Sun times the Universal Gravitational Constant, but has already been worked out as: 1.327 x 10^20 m^3/sec^2.

Your semi-major axis needs to be in meters and your result is in seconds. If you want something more convenient, like days, divide the seconds by 86400 (there's 86400 seconds in a day).

2007-06-05 08:40:27 · answer #1 · answered by Bob G 6 · 0 1

A day is typically determined by the rotation of a planet or moon. One rotation on its axis equals one day.

A year is typically determined by the orbit of the planet around the sun. One orbit around the sun equals one year.

On a planet where a day is shorter than a year, the number of days would be calculated by dividing the length of the day into the length of the year. So for example, a Martian day is 1.026 Earth days long (24.62 hours). A Martian year is 686.93 Earth days long. So a Martian year has 669.52 Martian days (686.93/1.026)

However, not all planets have years that are longer than their days. For example, some planets found in other solar systems have years as short as 4 or 5 Earth days. It is very likely that these planets are rotationaly locked to their star so that the same side always faces the star. So on that planet a year and a day are the same length.

Hope this helps.

2007-06-05 14:43:18 · answer #2 · answered by pdq 3 · 0 0

The proper formula would be | ( orbital period / rotational period ) - 1 | if the planet rotates in the same direction as it's orbit, or you add one at the end if it orbits in the opposite direction. If the result is 0, then one side of the planet is always in sunlight and the other in darkness. If the result is greater than 0 and less than 1, then it's day is longer than it's year. Take the reciprocal to find out how many years per day.

For example, Earth rotates once on it's axis every 23.934 hours, and it's tropical year (the year we use for our calendar) is about 8765.813 hours, so, 8765.813 / 23.934 = 366.249, subtract 1 (because earth rotates the same direction it orbits), and you get 365.25 approximately. This isn't exact as the numbers were rounded off a bit, but it's fairly close.

2007-06-05 15:59:57 · answer #3 · answered by Arkalius 5 · 1 0

A day is the length of time it takes a planet to rotate once on its axis, a year is the length of time it takes a planet to orbit it's star once. Divide the first number of units of time you used to determine the length of a day into the number of units you used up in the trip around the star.

2007-06-09 10:42:27 · answer #4 · answered by johnandeileen2000 7 · 0 0

A Solar year is defined by the time it takes for a planet to orbit the Sun. A Solar day is defined by the time it takes for a planet to make one complete rotation on its axis.

Divide Solar year time by Solar day time to get the number of "days" in any given planetary year.

Don't forget to catch tonight's one-hour episode of "The Universe" on the History Channel at 9 PM EST. The topic is Mars.

2007-06-05 14:34:49 · answer #5 · answered by Anonymous · 1 0

Figure out how many times it does a complete spin in the amount of time that it does one complete orbit.

Just think of the earth. One year is how long it takes to go around the sun. How many days in a year is how many spins it does within that year.

2007-06-05 14:30:01 · answer #6 · answered by Anonymous · 1 0

You observe the angular motion of some "landmark" on the planet, such as the red spot on Jupiter. If you measure the Earth time from when you see its center on Jupiter to when you see the center on the other side of Jupiter, you can compute the rotation of a planet.

2007-06-05 14:36:34 · answer #7 · answered by cattbarf 7 · 0 2

humans invented time keeping therefor we made it what ever we felt we wanted.

2007-06-05 14:30:39 · answer #8 · answered by Anonymous · 0 2

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