Substended angle, diameter question

[jehan4141]

We are on lesson 1 and 2, which are vectors and kinematics only. We haven't covered anything else, yet the professor assigned us this problem. Can anybody help me? I have searched google and have yet to understand this problem. Thank you!

To an observer on earth, the moon, which is 3.8x10^5 km away, substends an angle of 0.52 degrees. What is the moons diameter?

Homework Equations

I found on google that S=R x theta

The Attempt at a Solution

The answer is 3500 km, but no matter what configuration I put these values into the equation, I still don't get the right answer.

 

[Spinnor]

I think you have to convert the angle in degrees to an angle in radians for the formula to work? Try that.

 

[jehan4141]

thank you! yes that works!

 

[Ouabache]

alternatively (keeping theta in degrees),
recall from trigonometry, how to determine the length of a http://www.csgnetwork.com/trigtriformulatables.html".

 

[rwisz]

Hello there,

I understand that you are currently working on vectors and kinematics, and this problem may seem confusing to you. However, it is important to remember that in science, problems often require us to apply concepts from multiple topics in order to find a solution. In this case, we can use our knowledge of vectors and kinematics to solve this problem.

First, let's break down the given information. We know that the moon is 3.8x10^5 km away from Earth, and it subtends an angle of 0.52 degrees as seen from Earth. This means that if we draw a line from the observer on Earth to the moon, and another line from the observer to the edge of the moon, the angle between these two lines is 0.52 degrees.

Now, let's think about what we know about vectors. We know that vectors have both magnitude (size) and direction. In this case, we are interested in the magnitude of the vector that represents the distance from the observer to the edge of the moon. This will give us the diameter of the moon.

We can use the equation S = R x theta, where S is the length of the vector, R is the magnitude of the vector, and theta is the angle between the vector and a reference line. In this case, our reference line can be the line from the observer to the moon.

Plugging in our values, we get S = (3.8x10^5 km) x (0.52 degrees) = 1976 km. However, this is only the length of one half of the diameter. To get the full diameter, we need to multiply this by 2, giving us a diameter of approximately 3952 km.

I hope this helps you understand the problem better. Remember, in science, it is important to think critically and use concepts from different topics to solve problems. Keep up the good work!