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## Main Question or Discussion Point

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).

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).