Question

Suppose the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the samerotation speed. How much longer would a solar day be than a sidereal day? Mercury’s siderealday is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?

Answer 1

Answer:

A) 0.1955 days

B) 35.18 days

Explanation: Given that the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the same rotation speed. That is the orbital period will be

365.24/12 = P/4

P = 121.75 days

We have three periods, the orbital period Po = 121.75 days,

the sidereal period Ps = 23.93419 hours and

The mean solar day Pd =24 hours.

Since they had exactly the same rotation speed, the number of solar days in a year is Po/Pd.

The number of sidereal days in a year would be Po/Pd + 1 = 121.75 + 1 = 122.75 days. Since there are more sidereal days in a year this means that the sidereal day is a little bit shorter, 121.75/122.75 = 0.99185 times shorter.

0.99185 × 24 = 23.80448 hours.

24 - 23.80448 = 0.1955 days

Therefore, solar day is 0.99185 days longer than a sidereal day

Given that the Mercury’s side real day is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?

Using the formula

Ld = PoPs/(Po + Ps)

Ld = Length of Mercury's solar day

Po = orbital period

Ps = Mercury's side real day

Ld = (88 × 58.6)/ (88 + 58.6)

Ld = 5156.8/146.6

Ld = 35.18 days