# Help finding the eccentricity of an orbit

I am trying to find the eccentricity of a planet in order to be able to calculate perihelion and aphelion distances. I can get a lot of the equation, but am having issues with the angular momentum, and the coefficient of the inverse-square law central force such as gravity or electrostatics in classical physics; that second one especially leaves me confused.

The characteristics are 3.296×1024 Kg, orbital period of 34713964.7399 seconds, rotation period of 69480 Seconds. Semi-major axis is 159425506264.96738 metres if I calculated that part correctly.

Assume the sun is the same as ours for simplicities sake. The star is supposed to be somewhat bigger, but comparable. I just need the sun as a assumption to make an upper/lower bound.

So what would the eccentricity be for this? And thank you for your time, if you choose to help. I apologize for this if it is a bother.

• The data you have is insufficient to define an eccentricity. You could have multiple valid orbits with different eccentricities that have the same period and semi-major axis. Jan 12, 2022 at 23:26

## 1 Answer

Given your provided requirements, eccentricity is essentially a free parameter here. Any ellipse with the same semi-major axis will have the same orbital period, regardless of eccentricity (in the range $$0 \le e \le 1$$), under the Keplerian-Newtonian two-body assumptions.

Pick one close to 0 if you don't want the eccentricity to noticeably affect temperatures, solar day length, how quickly the sun moves along the ecliptic, etc.

• I see, from the books it doesn't seem it has too big of a difference in temperature over the year. I guess I can assume Earth-like eccentricity. Thanks.
– Zoey
Jan 12, 2022 at 23:38
• Eccentricity can make a noticeable difference in the overall climate of an Earth-like planet, making winters more extreme in one hemisphere and summers more extreme in the other. Look into Milankovitch cycles for more information. Jan 13, 2022 at 15:20