The Ratio of Velocities at Perigee and Apogee Proof

[thursday]

I need to know how to use kepler's 2nd law,(An object in an elliptic orbit will map out the same area in a certain time) to show that the ratio of the speeds of a planet at its near and far points from the sun is equal to the inverse ration of the far and near distances.

i.e. Vn = Df
Vf Dn

Where Vn= Velocity at perigee, Vf=Velocity at apogee
Dn= distacnce at perigee, Vf=distance at apogee

I found a good site that uses triangles to approximate the area of the ellipse for the path taken but i need a more accurate way for this proof.
http://www.phy6.org/stargaze/Skepl2A.htm


The following is from the site: it may help
The area A1 of such a triangle, by the formula for areas of right-angled triangles, is one half base time height, or

A1 = (1/2) V1r1
Similarly, the area A2 covered in one second after passing apogee A equals

A2 = (1/2) V2r2
However, by Kepler's 2nd law A1 = A2 so


(1/2) V1r1 = (1/2) V2r2
or, multiplying everything by 2

V1r1 = V2r2
A more useful form of that relation appears if both sides are divided by V2r1 :

V1 / V2 = r2 / r1
The ratio of velocities equals the inverse of the ratio of distances. The smaller the distance, the faster the motion. If perigee distance is half of the apogee distance, the velocity there is twice as large. (But please remember--this proportionality only holds with P and A, not with other points along the orbit).

 

[HallsofIvy]

What was your purpose in posting this? I see no question here.

 

[tacman]

To use Kepler's 2nd law to prove the ratio of velocities at perigee and apogee, we can start by looking at the definition of the law. It states that an object in an elliptical orbit will map out the same area in a certain time. This means that the area swept by the object's orbit in a specific time period will be the same, regardless of where the object is in its orbit.

Using this information, we can set up a proportion between the areas swept by the object at perigee and apogee. Let's call the areas A1 and A2 respectively.

A1/A2 = (Vn*T1)/(Vf*T2)

Where Vn and Vf are the velocities at perigee and apogee respectively, and T1 and T2 are the time periods for the object to sweep those areas.

Since Kepler's 2nd law states that the areas must be equal, we can set A1 equal to A2 and solve for the velocity ratio.

(Vn*T1)/(Vf*T2) = 1

Simplifying, we get:

Vn/Vf = T2/T1

Now, let's use Kepler's 3rd law which states that the square of an object's orbital period is proportional to the cube of its semi-major axis. In other words:

T2^2/T1^2 = r2^3/r1^3

Rearranging this equation, we get:

T2/T1 = (r2/r1)^3/2

Substituting this into our previous equation, we get:

Vn/Vf = (r2/r1)^3/2

Since we know that the ratio of distances at perigee and apogee is the inverse of the ratio of velocities, we can rewrite the equation as:

Vn/Vf = (r1/r2)^3/2

Which can also be written as:

Vn/Vf = (1/r2)^3/(1/r1)^3

Simplifying, we get:

Vn/Vf = r2^3/r1^3

This proves that the ratio of velocities at perigee and apogee is equal to the inverse ratio of the distances. So, for example, if the distance at perigee is half of the distance at apogee, the velocity at perigee will be twice the velocity