- #1
mnmakrets
- 3
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An astronaut arrives on the planet Oceania and climbs to the top of a cliff overlooking the sea. The astronaut’s eye is 100 m above the sea level and he observes that the horizon in all directions appears to be at angle of 5 mrad below the local horizontal.
What is the radius of the planet Oceania at sea level?
How far away is the horizon from the astronaut?
[Hint: the line of sight from the astronaut to the horizon is tangential to surface of the planet at sea level.]
Having drawn out the problem I have two equations:
Pythagoras: x^2 + R^2 = (R+100)^2
Sine Law: x= (R+100)sinA
x is the distance between the horizon and the astronaut
R is the radius of the planet
A is the angle between the two lengths of radius of the planet (sorry might sound a bit confusing, it makes more sense if you draw it out!)
I then tried to combine them and solve for the radius of the planet R but I am not having any luck getting the right answer :(
Any steps, answer, and/or hints are all greatly appreciated! :)
What is the radius of the planet Oceania at sea level?
How far away is the horizon from the astronaut?
[Hint: the line of sight from the astronaut to the horizon is tangential to surface of the planet at sea level.]
Having drawn out the problem I have two equations:
Pythagoras: x^2 + R^2 = (R+100)^2
Sine Law: x= (R+100)sinA
x is the distance between the horizon and the astronaut
R is the radius of the planet
A is the angle between the two lengths of radius of the planet (sorry might sound a bit confusing, it makes more sense if you draw it out!)
I then tried to combine them and solve for the radius of the planet R but I am not having any luck getting the right answer :(
Any steps, answer, and/or hints are all greatly appreciated! :)