What is the planet's farthest orbital distance from its star?

A newly discovered planet orbits a distant star with the same mass as the Sun at an average distance of 126 million kilometers. Its orbital eccentricity is 0.3.



Also, what is the planet's nearest orbital distance from its star?What is the planet's farthest orbital distance from its star?Eccentricity= (A-P) / (A+P)

A+P= 2 x average distance

126 x 2= 252

A+P= 252

A-P= Eccentricity x 2 x average distance

A-P= .3 X 2 X 126

A-P= 75.6

Then it just becomes a 2 step equation: 75.6+2P=252

solve the equation by subtracting 75.6 from both sides, so now the equation is: 2P= 176.4

divide 176.4 to get 88.2

Perihelion= 88.2 (88,200,000 km)

Aphelion= 252-88.2

Aphelion= 163.8 (163,000,000 km)

to check if this is right use the formula I wrote at the very top: (163.8-88.2) / (163.8+88.2) = 0.3

the answer is correct.

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Now to find the orbital period you have to use keplers third law: A^3=P^2

where A is the semi-major axis.

and P is the orbital period.

to find the semi-major axis, do the average of aphelion and perihelion: (163.8+88.2)/2=

so A=126,000,000 km

to find how many AU's are in 126,000,000, simply do 126,000,000/149,000,000

the answer comes out to about .845 AU's

So plug that into the equation: .845^3=P^2

You can do .845^3 and then find the square root of the answer or you can do .845^1.5. Either way you'll get the same answer.

.845^1.5= 0.776 years

orbital period (P)= 0.776 yearsWhat is the planet's farthest orbital distance from its star?p = r(1 - e)

a = r(1 + e)



where p = perihelion, a = aphelion, r = average distance and e = eccentricityWhat is the planet's farthest orbital distance from its star?If you use the average distance as the semi-major axis then you would have 88,200,000 for your short distance and 129,780,000 for your long distance. Hope this helps. Good luck.What is the planet's farthest orbital distance from its star?
This question goes above head hyperbolically

miata