You will need

- — stopwatch;
- calculator;
- — reference data on the orbits of the planets.

Instruction

1

Measure with a stopwatch the time it takes a rotating body to arrive at the original point. This is the period of rotation. If you measure the body rotation constraint, measure the time t, N is full of references. Find the ratio of these quantities, this is the period of rotation of the body T (T=t/N). The period is measured in the same units, and time. In the international system of measurement is second.

2

If we know the speed of the body, then you can find the period by dividing the number 1 by the value of the frequency ν (T=1/ν).

3

If the body is rotating on a circular path and is known for its linear velocity, calculate the period of rotation. To do this, measure the radius R of the trajectory on which the body rotates. Make sure that the module of the velocity does not change with time. Then make the calculation. To do this divide the circumference on which the moving body, which is equal to 2∙π∙R (π≈3,14), the rotation speed v. The result is the period of rotation of the body around the circumference T=2∙π∙R/v.

4

If you need to calculate the orbital period of a planet that moves around a star, use Kepler's third law. If two planets revolve around one star, the squares of the periods of their treatment are as cubes of semi-major axes of their orbits. If we denote the periods of revolution of two planets, T1 and T2, the semi-major axes of the orbits (they are the ellipticity), respectively, a1 and a2, then T12/ T22= a13/a23. These calculations are correct in case if the masses of the planets are much 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 greater semiaxis of the orbit of Mars, and the Earth a1, a2 (as planets, which also revolve around the Sun). They are equal a1=227,92∙10^6 km and a2=149,6∙10^6 km and the rotation Period of the earth's T2=exactly 365.25 days (1 earth year). Then find the period of revolution of Mars, by converting the formula from the third Kepler 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.

# Advice 2: How to find Mars

**Mars**is the fourth farthest the Sun and seventh in size of the planets of the Solar system. It got its name after the Roman God of war. Sometimes

**Mars**is called the red planet, reddish tint of the surface gives the oxide of iron contained in the soil.

You will need

- Amateur telescope or powerful binoculars

Instruction

1

The opposition of Earth and

When the Earth is exactly between the Sun and

**Mars**andWhen the Earth is exactly between the Sun and

**Mars**ω, i.e. at a minimum distance equal to 55.75 million km, the ratio of the planets is called confrontation. At the same time he**Mars**is in the direction opposite the Sun. Such violence occur every 26 months at different points of the orbits of Earth and**Mars**. This is the most favorable points for observation of the red planet in Amateur telescopes. Once in 15-17 years, there are the great confrontation: the distance to**Mars**and the minimum, and the planet reaches its greatest angular size and brightness. The last great confrontation was 29 January 2010. The next one will be July 27, 2018.2

The conditions of observation

In the presence of the Amateur telescope you should look for

In the presence of the Amateur telescope you should look for

**Mars**in the sky in times of struggles. The surface detail available for observation only during those periods when the angular diameter of the planet reaches the maximum value. Large Amateur telescopes are available many interesting features on the planet's surface, seasonal evolution of the polar caps**of Mars**and the signs of Martian dust storms. In a small telescope you can see the "dark spots" on the surface of the planet. You can see the polar caps, but only during the great confrontations. A lot depends on experience, observations, and atmospheric conditions. So, the more observation experience, the less can be a telescope to "capture"**Mars**and parts of its surface. Lack of experience is not always kompensiruet expensive and powerful telescope.3

Where to look

The evening and morning

The evening and morning

**Mars**is visible in the red-orange light, and in the middle of the night in yellow. In 2011,**Mars**can be seen in the sky during the summer and until the end of November. Until August planet can be seen in the constellation Gemini, in the Northern hemisphere of the sky. Since September,**Mars**is visible in the constellation of Cancer. It is located between the constellations of Leo and Gemini.Note

If the experience of observation is small, should wait a period of confrontation.

# Advice 3: How to find linear velocity

Call the linear

**velocity**with which the body moves along an arbitrary trajectory. At a known length of the trajectory and the time in which it was passed, find the linear**speed**by the ratio of the length of time. Linear**speed**motion in a circle is equal to the product of the angular velocity, not its radius. Also use other formulas to determine linear velocity. It can be measured by the speedometer.You will need

- stopwatch, protractor, tape measure or range finder, speedometer

Instruction

1

In the most General case, to determine the linear velocity of a body in uniform motion, measure the length of the path (the lines on which the moving body) and divide by the time it took to pass this way v=S/t. However, non-uniform linear motion

**speed**in a given time, determine by means of a speedometer or a special radar.2

When a body moves on a circle it has an angular and linear velocity. To measure the angular velocity, measure the Central angle which describes the body around the circumference for a certain period of time. For example, measure the time during which the body describes half of a circle, in this case the Central angle is equal to π radians (180º). Divide this angle in the time it took the body to undergo a semi-circle, and will receive the angular

**velocity**. If you know the angular**velocity**of the body, then its linear**speed is**equal to the product of the angular velocity, the radius of the circle along which the body moves, which measure a tape measure or rangefinder v=ω•R.3

Another method of determining the linear speed of a body moving in a circle. Using the stopwatch measure the time of a full revolution of the body around the circumference. This time is called the period of rotation. Rangefinder or a tape measure, measure the radius of the circular path on which the moving body. Calculate the linear

**speed**by dividing the product of the radius of the circle and the number of 6.28 (circumference) at the time of her passing v=6,28•R/t.4

If you know the centripetal acceleration, which acts on a body moving on a circle with constant

**speed**u, additionally measure its radius. In this case, the linear**speed**of a body moving on a circle is equal to the square root of the product of the centripetal acceleration at the radius of the circle.