Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Kepler’s laws and the shape of the orbit of the celestial bodies
Description: In this document I wrote a deduction for the shape of the orbit of the celestial bodies and I showed how Kepler's Laws can be demonstrated. I avoided to solve a difficult differential equation by using a shortcut. I used a special constant vector which gives the shape of the orbit.
Description: In this document I wrote a deduction for the shape of the orbit of the celestial bodies and I showed how Kepler's Laws can be demonstrated. I avoided to solve a difficult differential equation by using a shortcut. I used a special constant vector which gives the shape of the orbit.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Kepler’s laws and the shape of the orbit of
the celestial bodies
Figure 1: The change of coordinate system
1
Consider a cartesian system with it’s unit vectors e⃗x and e⃗y
...
Because m << M we can consider
that the star is fixed in O
...
Let’s consider that the position vector
of the celestial body makes the angle θ with Ox axis
...
Figure 2: The new unit vectors
By using Figure 2, we can write the unit vectors e⃗r and e⃗θ depending
on the unit vectors e⃗x and e⃗y :
e⃗r = e⃗x · cos θ + e⃗y · sin θ
(1)
e⃗θ = −e⃗x · sin θ + e⃗y · cos θ
(2)
We can write the position of celestial body relative to he star as
r · e⃗r
...
The
·
celestial body is attracted by the star with the force: F⃗ = − G·m·M
r2
e⃗r
...
We make the notation: C = −G · m · M
...
The angular momentum of
⃗ = ⃗r × P⃗ , where ⃗r is the position vector of the
the celestial body is: L
celestial body with respect to the star and P⃗ is the momentum of the
⃗
celestial body
...
⃗ = m · r · e⃗r × (⃗
L
eθ · dθ
⃗r · dr
dt · r + e
dt )
Because e⃗r ⊥⃗
eθ we can write: e⃗r × e⃗θ = e⃗z ¸si e⃗r × e⃗r = 0
Title: Kepler’s laws and the shape of the orbit of the celestial bodies
Description: In this document I wrote a deduction for the shape of the orbit of the celestial bodies and I showed how Kepler's Laws can be demonstrated. I avoided to solve a difficult differential equation by using a shortcut. I used a special constant vector which gives the shape of the orbit.
Description: In this document I wrote a deduction for the shape of the orbit of the celestial bodies and I showed how Kepler's Laws can be demonstrated. I avoided to solve a difficult differential equation by using a shortcut. I used a special constant vector which gives the shape of the orbit.