Question

Kepler-62e is a planet similar in size to the Earth with an orbital period of 122 days. The star it orbits has a mass of 1.4 x 1030 kg. Convert the period to hours and use Newton's version of Kepler's 3rd law to calculate the semimajor axis of this planet's orbit. Your answer will not match any of these choices, instead choose the answer that is closest to your calculated value.

Answer 1

Answer:

6.41 x 10^10 m

Explanation:

Mass, M = 1.4 x 10^30 kg

Period, T = 122 days = 122 x 24 = 2928 hours = 1.054 x 10^7 second

According to the Kepler's third law

$T^{2}=\frac{4\pi ^{2}}{GM}r^{3}$

where, r is the length of semi major axis

$r^{3}=\frac{1.054\times 1.054\times 10^{14}\times 6.67\times 10^{-11}\times 1.4\times 10^{30}}{4\times 3.13\times 3.14}$

r = 6.41 x 10^10 m

Answer 2

The semi major axis of this planet's orbit is $6.41*10^{-11}$ m.

Given that,  Time period, $T=122days$ , mass $m=1.4*10^{30} kg$

$T=122days=122*24*3600=1.054*10^{7}seconds$

From Newton's version of Kepler's 3rd law ,

               $T^{2}=\frac{4\pi^{2} }{Gm}r^{3}$

Where G is gravitational constant and r is semi major axis.

Value of G $=6.67*10^{-11}$ and $\pi=3.14$

Substituting all values in above formula.

             $(1.054*10^{7} )^{2}=\frac{4*(3.14)^{2} }{6.67*10^{-11}*1.4*10^{30} } *r^{3} \\\\r^{3}=2.63*10^{32} \\\\r=\sqrt[3]{2.63*10^{32}} \\\\r=6.41*10^{10}m$

Hence, The semi major axis of this planet's orbit is $6.41*10^{-11}$ m.

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