Using Kepler's laws to calculate elliptical planetary motion

  • #1
kirito
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TL;DR Summary: orbital speed laws

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I would appreciate a bit of explanation on how did we find e1 and e2 and if there are any useful references to learn about Kepler laws since I am lost for the most part, and would like to gain understanding and solving ability
,and if you can go into some details on how to know when there is a conservation of momentum and energy
 
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  • #2
kirito said:
TL;DR Summary: orbital speed laws

how did we find e1 and e2 and if there are any useful references to learn about Kepler laws
Kepler laws will not give you the equations for ##E_1## and ##E_2##. You need the Newton law of gravity for that.
 
  • #3
Hill said:
Kepler laws will not give you the equations for E1 and E2. You need the Newton law of gravity for that.
so I should I think that for it to be in orbit the the gravitational force os the central force so
$$ f= m_ac = mv^2/r= G m M/r^2 $$ so I so I get $$v^2= GM/r$$ , how should I go on from there if I may ask
 
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  • #4
Please, use LaTeX.
kirito said:
gravitational force os the central force so f= m_ac = mv^2/r= G m M/r^2 so I so I get v^2= GM/r
You cannot assume that ##a=v^2/r## because the orbit here is not circular but rather elliptical.
You need to move from the gravitational force to the energy. Do you know the formula for gravitational potential energy?
 
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  • #5
Hill said:
Please, use LaTeX.

You cannot assume that ##a=v^2/r## because the orbit here is not circular but rather elliptical.
You need to move from the gravitational force to the energy. Do you know the formula for gravitational potential energy?
I don't I will try to search it up , I will also try to add latex now , thank you
 
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  • #6
Simply put, Orbital energy is the sum of kinetic energy and gravitational potential energy.
KE = 1/2mv^2
GPE = -GMm/r which is derived by applying calculus to Fg= GMm/r^2
Now, for a circular orbit, v=sqrt(GM/r)
Plugging this in for v in the KE+GPE equation(E2), and simplifying gives you -GMm/2r
It can be proven that an elliptical orbit with a semi-major axis of a has the same total orbital energy as a circular one of radius r (a circle being a special case of an ellipse where a=r)
thus E1= GMm/2a
 
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  • #7
Your reference quotes the vis-viva equation. The link I provided shows its derivation which I think is what you want.
 
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  • #8
Janus said:
Simply put, Orbital energy is the sum of kinetic energy and gravitational potential energy.
KE = 1/2mv^2
GPE = -GMm/r which is derived by applying calculus to Fg= GMm/r^2
Now, for a circular orbit, v=sqrt(GM/r)
Plugging this in for v in the KE+GPE equation(E2), and simplifying gives you -GMm/2r
It can be proven that an elliptical orbit with a semi-major axis of a has the same total orbital energy as a circular one of radius r (a circle being a special case of an ellipse where a=r)
thus E1= GMm/2a
thank you for the explanation , I do admit that I have a lot of confusion in the subject that's why I asked for some resource your comment is direct and organised surely a bit more organised
 
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FAQ: Using Kepler's laws to calculate elliptical planetary motion

What are Kepler's laws of planetary motion?

Kepler's laws of planetary motion consist of three fundamental principles that describe the motion of planets around the Sun. The first law, the Law of Ellipses, states that planets move in elliptical orbits with the Sun at one focus. The second law, the Law of Equal Areas, indicates that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. The third law, the Law of Harmonies, establishes a relationship between the squares of the orbital periods of planets and the cubes of the semi-major axes of their orbits.

How can I use Kepler's laws to calculate the orbital period of a planet?

You can calculate the orbital period of a planet using Kepler's Third Law, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit. The formula is T² = k * a³, where k is a constant that depends on the mass of the Sun and the gravitational constant. For planets in our solar system, k is approximately 1 when T is measured in Earth years and a in astronomical units (AU).

What is the significance of the semi-major axis in elliptical orbits?

The semi-major axis is a crucial parameter in defining the size of an elliptical orbit. It represents half of the longest diameter of the ellipse and is directly related to the orbital period of the planet. A larger semi-major axis indicates a longer orbital period, meaning the planet takes more time to complete one full orbit around the Sun.

How do I determine the eccentricity of a planet's orbit?

The eccentricity (e) of an elliptical orbit can be calculated using the formula e = (distance between foci) / (length of the major axis). It quantifies the deviation of the orbit from being circular, with values ranging from 0 (a perfect circle) to just below 1 (more elongated). The eccentricity can also be derived from the semi-major axis (a) and semi-minor axis (b) using the formula e = √(1 - (b²/a²)).

Can Kepler's laws be applied to exoplanets?

Yes, Kepler's laws can be applied to exoplanets, as they describe the motion of any celestial body under the influence of gravity. By observing the transit of exoplanets across their host stars and measuring their orbital characteristics, astronomers can use Kepler's laws to calculate their orbital periods, distances from the star, and other parameters, similar to how they are used for planets in our solar system.

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