Posted: November 20th, 2022

A newly discovered planet n orbit

QUESTION
A newly discovered planet n orbit a distant star with the same mass as the sun at an average distance of 120million kilometer s. Its orbital eccentricity is 0.1.Determine
1)The planet orbital period
2)Find the planets nearest orbital distance from the star.
SOLUTION
Kepler’s 3rd law relates,
p2∝d3
Where p=orbital period
And,
d= distance of the semi major axis.
In this relation,
T^2=(4π^2 a^3)/G(M1+M2)

Where ,T=period
M1 and M2=is the mass of the solar system and mass of the planet consequtively.
According to keplers gravitational law:
While,a3 is the distance of the semi major axis.
Taking a=120000000000 M
M1=M2=I.9891*10^30KG
And applying the formula

T^2=(4π^2 a^3)/G(M1+M2)

4π^2/G=1
a^3=1.728×〖10〗^33 m^3
M1+M2=3.9782〖×10〗^30

T2=1×1.728×〖10〗^33 m^3/3.9782〖×10〗^30
T2=434.3673

T=20.84YEARS

QUESTION2
Nearest orbital distance:
Kepler’s 1nd law relate;
e^2=1-b^2/a^2
Where :e=eccentricity
b=distance of the semiminor axis
a=distance of semimajor axis=120000000000m
Thus applying the formula:

e^2=1-b^2/a^2

b^2=(1-e^2 ) a^2
a^2=1.44×〖10〗^22
e^2=0.01
b^2=

Semi minor axis=344.67km

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