Logarythmic Messages 277 Reaction score 0 Thread starter Nov 22, 2006 #1 If I have the eccentricity and the perihelion of an orbit given, how can I compute the aphelion? Last edited: Nov 22, 2006
Tomsk Messages 227 Reaction score 0 Nov 22, 2006 #2 You need to think of what angles correspond the the perihelion and aphelion. Then use the relation between r, e, theta and l to get the aphelion.
You need to think of what angles correspond the the perihelion and aphelion. Then use the relation between r, e, theta and l to get the aphelion.
Logarythmic Messages 277 Reaction score 0 Nov 22, 2006 #3 Can I not just use the relation for an ellipse: [tex]e = \frac{d}{a}[/tex] where d is the distance from the focal point to the center and a is the semi major axis?
Can I not just use the relation for an ellipse: [tex]e = \frac{d}{a}[/tex] where d is the distance from the focal point to the center and a is the semi major axis?
Tomsk Messages 227 Reaction score 0 Nov 22, 2006 #4 Well yes if you know a and d. You'd get the same result. You might need J as well though.
Logarythmic Messages 277 Reaction score 0 Nov 22, 2006 #5 I just used the radial equation [tex]r_a = \frac{a(1 - e^2)}{1 + e \cos \pi} = a(1+e)[/tex] This leads to [tex]r_a = \frac{r_p(1+e)}{1-e}[/tex] where I have used that [tex]a = \frac{r_a + r_p}{2}[/tex] But for [tex]r_p = 0,2301 AU[/tex] I get [tex]r_a = 2988 AU[/tex] and this is wrong. I should get [tex]r_a =4699 AU[/tex]. Am I too tired or what is this? I have that [tex]e = 0,999846[/tex]
I just used the radial equation [tex]r_a = \frac{a(1 - e^2)}{1 + e \cos \pi} = a(1+e)[/tex] This leads to [tex]r_a = \frac{r_p(1+e)}{1-e}[/tex] where I have used that [tex]a = \frac{r_a + r_p}{2}[/tex] But for [tex]r_p = 0,2301 AU[/tex] I get [tex]r_a = 2988 AU[/tex] and this is wrong. I should get [tex]r_a =4699 AU[/tex]. Am I too tired or what is this? I have that [tex]e = 0,999846[/tex]