(hrw8c13p51) A comet that was seen in April 574 by Chinese astronomers on a day known by them as the Woo Woo day was spotted again in May 1994. Assume the time between observations is the period of the Woo Woo day comet and take its eccentricity as 0.11. What is the semimajor axis of the comet's orbit

Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m

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

change tension