The cycle time of a body which moves on the closed trajectory can be measured by means of hours. If the address happens too quickly, it becomes after change of some number of full addresses. If the body rotates on a circle, and its linear speed is known, this size is calculated by a formula. The cycle time of the planet pays off under the third law of Kepler.
It is required to you
- — stop watch;
- — calculator;
- — help data on orbits of planets.
Instruction
1. Measure time which is required for the rotating body to come to a starting point by means of a stop watch. It will also be the period of its rotation. If it is difficult to measure rotations of a body, then measure t time, N full addresses. Find the relation of these sizes, it also will be the period of rotation of this body of T (T=t/N). The period is measured in the same sizes, as time. In the international system of measurement it is second.
2. If the frequency of rotation of a body is known, then find the period, having divided number 1 into value of frequency ν (T=1/ν).
3. If the body rotates on a circular trajectory and its linear speed is known, calculate the period of its rotation. For this purpose measure radius R trajectories on which the body rotates. Make sure that the module of speed does not change over time. Then make calculation. For this purpose divide length of a circle on which the body which is equal 2 ∙π ∙ to R (π ≈ 3.14), on the speed of its rotation of v moves. The period of rotation of this body on a circle T=2 ∙π ∙ R/v will be result.
4. If it is necessary to calculate the period addressesof planets which moves around a star, use the third law of Kepler. If two planets rotate around one star, then squares of the periods of their address belong as cubes of big half shafts of their orbits. If to designate cycle times of two planets of T1 and T2, big half shafts of orbits (they elliptical), respectively, a1 and a2, then T1²/by T2² = a1³/a2³. These calculations are right in case the mass of planets considerably concede to the mass of a star.
5. Example: Define a planet Mars cycle time. To calculate this size, find length of a bigger half shaft of an orbit of Mars, a1 and Earth, a2 (as planets which rotate around the Sun too). They are equal to a1=227.92∙10^6 of km and a2=149.6∙10^6 of km. Period of rotation of the earth of T2=365.25 days (1 terrestrial year). Then find a cycle time of Mars, having transformed a formula from the third law of Kepler, for definition of the period of rotation of Mars of T1= √ (T2² ∙ a1³/a2³)= √ (365.25² ∙ (227.92∙10^6)³ / (149.6∙10^6)³) ≈ 686.86 days.