you need

- - stopwatch;
- - Calculator;
- - reference data on the orbits of the planets.

Guide

1

Measure with a stopwatch the time required rotating body to arrive at the starting point.This will be a period of its rotation.If the rotation of the body is difficult to measure, measure the time t, N complete reversal.Find the ratio of these values, this will be the period of rotation of the body T (T = t / N).The period measured in the same magnitude as the time.In the international system of measurement is second.

2

If you know the frequency of rotation of a body, find the period by dividing the number 1 on the frequency ν (T = 1 / ν).

3

If the body rotates in a circular path and is known for its linear speed, calculate its rotation period.To do this, measure the radius R of the path along which the body rotates.Verify that the rate does not change with time.Then, make a payment.For this divide the circumference along which the body moves, which is 2 ∙ π ∙ R (π≈3,14), the speed of rotation v.The result will be a period of rotation of the body around the circumference of T = 2 ∙ π ∙ R / v.

4

If you want to calculate the orbital period of the planet, which moves around a star using Kepler's third law.If two planets orbit a single star, the squares of the periods of their treatment are treated like cubes of semi-major axes of their orbits.If we denote the periods of the two planets T1 and T2, semi-major axis (they ellipticity), respectively, a1 and a2, the T1² / T2² = a1³ / a2³.These calculations are correct in that case, if the masses of the planets is significantly inferior to the mass of the star.

5

Example: Determine the orbital period of the planet Mars.To calculate this value, find the length of the major axis of the orbit of Mars, a1 and Earth, a2 (like planets, which also revolves around the sun).They are a1 = 227,92 ∙ 10 ^ 6 km and a2 = 149,6 ∙ 10 ^ 6 km.Earth's rotation period T2 = 365,25 days (one Earth year).Then get orbital period of Mars, transforming formula from Kepler's third law, to determine the period of rotation of Mars T1 = √ (T2² ∙ a1³ / a2³) = √ (365,25² ∙ (227,92 ∙ 10 ^ 6) ³ / (149,6∙ 10 ^ 6) ³) ≈686,86 days.

Sources:

- how to find treatment in the text