Answer:
Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m
Step-by-step explanation:
From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month
Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds
Now, the square of the period of any planet is given by the formula;
T² = (4π²/GM)α³
Where;
G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²
M is mass of sun which has a value of 1.99 x 10^(30) kg
α is the semi major axis of the comets orbit.
Thus, making α the subject, we have;
α = ∛(GMT²/4π²)
So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)
α = 1.89 x 10^(13) m