If a planet has a semimajor axis of 2 AU (astronomical units)
a) what would the semimajor axis of a superior planet be if the angle of greatest elongation for the inferior planet is 20 degrees as viewed from the superior planet?
b) assuming a day on the superior planet is 30 hours, how long would the inferior planet be visible in the sky after sunset if it is at greatest elongation (answer must be in HR MIN SEC)
a) To find the semimajor axis of the superior planet, we can use the concept of greatest elongation. The angle of greatest elongation is the maximum angle formed between the superior planet, Earth, and the Sun as viewed from the inferior planet. It occurs when the inferior planet is at its maximum angular distance from the Sun.
In this case, the angle of greatest elongation for the inferior planet is given as 20 degrees. Since the superior planet is viewing this from its position, the angle of greatest elongation for the inferior planet as seen from the superior planet will also be 20 degrees.
Now, let's use the relationship between the semimajor axis (a) and the angle of greatest elongation (θ):
a = D / (1 + sin(θ))
where D represents the distance between the Sun and Earth.
Since the semimajor axis of the planet is given as 2 AU, we can assume this is the distance between the Sun and the planet (D = 2 AU).
Plugging in the values, we can calculate the semimajor axis (a) of the superior planet:
a = 2 / (1 + sin(20)) ≈ 1.843 AU
Therefore, the semimajor axis of the superior planet would be approximately 1.843 AU.
b) To determine how long the inferior planet would be visible in the sky after sunset, we need to consider the time it takes for the inferior planet to complete its apparent motion from the greatest elongation position to the sunset point.
Given that the day on the superior planet is 30 hours, we can assume that the planet takes 15 hours to go from the greatest elongation position to the sunset point (since greatest elongation occurs halfway between sunrise and sunset).
The next step is to convert this time into hours, minutes, and seconds. Since there are 60 minutes in an hour and 60 seconds in a minute, we can perform the conversion as follows:
15 hours = 15 x 60 minutes = 900 minutes
900 minutes = 900 x 60 seconds = 54,000 seconds
Hence, the inferior planet would be visible in the sky after sunset for approximately 15 hours, 900 minutes, or 54,000 seconds.