- #1
andreea_a
- 26
- 0
Hello!
I've been trying to solve this problem for some time but I can't seem to find the answer.
Can you help me?
On the night of 5th april an observer watches the sunset(η = 3.5 minutes,η = correction of time). The solar disk touched the horizon at 20h 02 m 02 s and then the Sun's disk disappears below the horizon at 20h 05m 10s.
Given the distance Earth-Sun= 149,6 * 10 ^6 km and the solar radius = 696 000 km find the latitude and longitude of the place.
What I have tried so far was to write the formula for angular diameter θ = 206265d/D =>θ = 0.53 degrees.
I have thought about finding the hour angle so after that I can use this formula:
sin φ = tan(90 - H)*tan h,where φ=latitude of the observer,H = hour angle,h = altitude of the Sun which is 0 at sunset
Now my question is:how do I calculate the hour angle using that angular diameter?
Thanks in advance!
I've been trying to solve this problem for some time but I can't seem to find the answer.
Can you help me?
Homework Statement
On the night of 5th april an observer watches the sunset(η = 3.5 minutes,η = correction of time). The solar disk touched the horizon at 20h 02 m 02 s and then the Sun's disk disappears below the horizon at 20h 05m 10s.
Given the distance Earth-Sun= 149,6 * 10 ^6 km and the solar radius = 696 000 km find the latitude and longitude of the place.
The Attempt at a Solution
What I have tried so far was to write the formula for angular diameter θ = 206265d/D =>θ = 0.53 degrees.
I have thought about finding the hour angle so after that I can use this formula:
sin φ = tan(90 - H)*tan h,where φ=latitude of the observer,H = hour angle,h = altitude of the Sun which is 0 at sunset
Now my question is:how do I calculate the hour angle using that angular diameter?
Thanks in advance!