Kepler's laws explanation, exercises, experiment

The Kepler's laws on planetary motion were formulated by the German astronomer Johannes Kepler (1571-1630). Kepler deduced them based on the work of his teacher the Danish astronomer Tycho Brahe (1546-1601).

Brahe carefully compiled data on planetary movements over more than 20 years, with surprising precision and accuracy, considering that the telescope had not yet been invented at the time. The validity of your data is still valid today.

Article index

• 1 Kepler's 3 Laws
• 2 The law of universal gravitation and Kepler's third law
• 3 Exercises
  • 3.1 - Exercise 1
  • 3.2 - Exercise 2
• 4 experiment
  • 4.1 Materials 
  • 4.2 Procedure
• 5 References

Kepler's 3 Laws

Kepler's laws state:

-First law: all planets describe elliptical orbits with the Sun in one of the foci.

-Second law or law of equal areas: a line directed from the Sun to any planet (focal radius), sweeps equal areas in equal times.

-Third law: the square of the time it takes for any planet to orbit the Sun is proportional to the cube of its average distance from the Sun.

Be T said time, called orbital period, Y r the average distance, then:

T^two is proportional to r³

T = k r³

This means that the quotient T^two/ r³ is the same for all planets, which makes it possible to calculate the orbital radius, if the orbital period is known.

When T is expressed in years and r in astronomical units AU *, the constant of proportionality is k = 1:

T^two= r³

* An astronomical unit equals 150 million kilometers, which is the average distance between the Earth and the Sun. The orbital period of the Earth is 1 year.

The law of universal gravitation and Kepler's third law

The universal law of gravitation states that the magnitude of the gravitational force of attraction between two objects of masses M Y m respectively, whose centers are separated by a distance r, is given by:

F = G mM / r^two

G is the universal gravitation constant and its value is G = 6.674 x 10 ^-eleven N.m^two/ kg^two .

Now, the orbits of the planets are elliptical with a very small eccentricity.

This means that the orbit is not very far from a circumference, except in some cases like the dwarf planet Pluto. If we approximate the orbits to the circular shape, the acceleration of the planet's motion is:

to_c = v^two/ r

Given the F = ma, have:

G mM / r^two = m.v^two/ r

Here v is the linear velocity of the planet around the Sun, assumed static and of mass M, while that of the planet is m. Then:

This explains that the planets farther from the Sun have a lower orbital speed, since this depends on 1 / √r.

Since the distance the planet travels is approximately the length of the circumference: L = 2πr and it takes a time equal to T, the orbital period, we obtain:

v = 2πr / T

Equating both expressions for v gives a valid expression for T^two, the square of the orbital period:

And this is precisely Kepler's third law, since in this expression the parenthesis 4π^two / GM is constant, therefore T^two is proportional to the distance r cubed.

The definitive equation for the orbital period is obtained by taking the square root:

How much is the mass of the Sun worth? It is possible to find out by this equation. We know that the orbital period of the Earth is one year and the orbital radius is 1 AU, equivalent to 150 million kilometers, so we have all the necessary data.

In our previous equation, we solve for M, not without first converting all the values ​​to the International System of Units SI:

1 year = 3.16 x 10⁷ seconds.

1 AU = 150 million km = 1.5 x10^eleven m.

Training

Although Kepler had only the planets in mind when he derived his famous laws, these are also valid for the motion of satellites and other bodies in the solar system, as we will see below..

- Exercise 1

Knowing that the orbit of Jupiter is 5.19 times that of the Earth, find the orbital period of Jupiter.

Solution

According to the definition of the Astronomical Unit, Jupiter is distant from the Sun 5.19 AU, therefore, according to Kepler's third law:

T^two= r³= (5.19)³ years

Therefore T = (5.19)^3/2  years = 11.8 years

- Exercise 2

Comet Halley visits the Sun every 75.3 years. Find:

a) The semi-major axis of its orbit.

b) The aphelion measure, if the perihelion measures 0.568 AU.

Solution

Comet Halley visits the Sun every 75.3 years. Find:

a) The semi-major axis of its orbit.

b) The aphelion measure, if the perihelion measures 0.568 AU.

Solution to

When a planet or any other star is at its closest point to the Sun, it is said to be in the perihelion, and when it is further away, in aphelion. In the special case of a circular orbit, r in Kepler's third law is the radius of the orbit.

However, in the elliptical orbit the celestial body is more or less far from the Sun, the semi-major axis "a" being the average between aphelion and perihelion:

Therefore, we substitute r for a in Kepler's third law, which results for Halley in:

T^two= a³→ a = (T)^2/3 → a = (75.3) ^2/3 UA = 17,832 UA

Solution b

a = ½ (Perihelion + Aphelion)

17.832 = ½ (0.568+ Aphelion) → Aphelion = 2 x 17.832 - 0.568 AU = 35.10 AU.

Experiment

Analyzing the motion of the planets requires weeks, months, and even years of careful observation and recording. But in the laboratory a very simple scale experiment can be carried out to prove that Kepler's law of equal areas holds..

For this, a physical system is required in which the force that governs the movement is central, a sufficient condition for the law of areas to be fulfilled. Such a system consists of a mass tied to a long rope, with the other end of the thread fixed to a support..

The mass is moved a small angle from its equilibrium position and a slight impulse is given to it, so that it executes an oval (almost elliptical) movement in the horizontal plane, as if it were a planet around the Sun..

On the curve described by the pendulum, we can prove that it sweeps equal areas in equal times, if:

-We consider vector radii that go from the center of attraction (initial point of equilibrium) to the position of the mass.

-And we sweep between two consecutive instants of equal duration, in two different areas of the movement.

The longer the pendulum string and the smaller the angle that departs from the vertical, the net restoring force will be more horizontal and the simulation resembles the case of movement with central force in a plane..

Then the described oval approaches an ellipse, such as the one that the planets travel.

Materials

-Inextensible thread

-1 dough or metallic ball painted white that acts as a pendulum bob

-Ruler

-Conveyor

-Photographic camera with automatic strobe disk

-Brackets

-Two light sources

-A sheet of black paper or cardboard

Process

Assembling the figure is needed to take photos of multiple flashes of the pendulum as it follows its path. To do this, you have to put the camera just above the pendulum and the automatic strobe disk in front of the lens..

In this way, images are obtained at regular time intervals of the pendulum, for example every 0.1 or every 0.2 seconds, which allows to know the time it took to move from one point to another..

You also have to illuminate the mass of the pendulum properly, putting the lights on both sides. The lentil should be painted white to improve the contrast on the background, which consists of a black paper spread on the ground.

Now you have to check that the pendulum sweeps equal areas in equal times. To do this, a time interval is chosen and the points occupied by the pendulum in said interval are marked on the paper..

On the image a line is drawn from the center of the oval to these points and thus we will have the first of the areas swept by the pendulum, which is approximately an elliptical sector like the one shown below:

Calculation of the area of ​​the elliptical section

Angles are measured with the protractor θ_or Y θ₁, and this formula is used to find S, the area of ​​the elliptical sector:

S = F (θ₁) - F (θ_or)

With F (θ) given by:

Note that to Y b are the semi-major and minor axes respectively. The reader only has to worry about carefully measuring the semi-axes and the angles, since there are calculators online to evaluate this expression easily..

However, if you insist on doing the calculation by hand, remember that the angle θ is measured in degrees, but when entering the data into the calculator, the values ​​must be expressed in radians.

Then you have to mark another pair of points in which the pendulum has inverted the same time interval, and draw the corresponding area, calculating its value with the same procedure.

Verification of the law of equal areas

Finally, it remains to verify that the law of areas is fulfilled, that is, that equal areas are swept in equal times.

Are the results deviating a bit from what was expected? It must always be borne in mind that all measurements are accompanied by their respective experimental error.

References

1. Keisan Online Calculator. Area of ​​an elliptical sector calculator. Recovered from: keisan.casio.com.
2. Openstax. Kepler's Law of Planetary Motion. Recovered from: openstax.org.
3. PSSC. Laboratory Physics. Editorial Reverté. Recovered from: books.google.co.
4. Palen, S. 2002. Astronomy. Schaum Series. Mcgraw hill.
5. Pérez R. Simple system with central force. Recovered from: francesphysics.blogspot.com
6. Stern, D. Kepler's three laws of planetary motion. Recovered from: phy6.org.