Diameter of a planet, known the angular size and distance

In summary, using trigonometry, the diameter of the sun can be calculated by first finding the distance between the planet and the sun (3x10^5 km) and using it to create a right triangle with the angular size of the sun (1 degree) as one of the angles. By cutting the triangle in half and using the trigonometric ratio of sine, the value of the hypotenuse (diameter of the sun) can be found to be 300011 km. Using the Pythagorean theorem, the radius of the sun is calculated to be 2569 km, which when multiplied by 2 gives a diameter of 5138 km. The only uncertainty in this solution is whether the triangle can be
  • #1
subopolois
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Homework Statement


a planet is 5x10^8 km from its sun, from the planet, the sun has an angular size of 1 degree. what is the diameter of the sun?


Homework Equations





The Attempt at a Solution


ive drawn a diagram with the sun and the planet, with the distance between written in and the 1 degree size of the sun on the planets surface. i was thinking of using trignometry, since i know one angle and the length going through the center of the imaginary triangle, but I am not sure, i haven't used trig in a while
 
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  • #2
alright so here's what i worked out. since from the planet the angular distance is 1 degree and the distance to the sun to the sun is 3x10^5, i draw a triangle with one degree at the planet and the wider part of the triangle is defined by the base of the triangle. i cut the triangle down the center (cutting the 1 degree in half) making 2 right angle triangles, since one angle is 90 and i cut the 1 degree in half, that's 0.5 degrees and the third is 89.5. one side of the triangle is 300000 km. here's my work using trig:
sin(89.5)= 300000/x (x being the hypotenuse)
x= 300011

now i use Pythagorus to get the length of the base of the right triangle, which is the radius of the sun.
300000^2+b^2= 30011^2
b= 2569

since the radius is needed, i multply this by 2 to get a diameter of 5138 km.is this right, the only thing I am uncertain of in my solution is the i cut the one triangle to make 2. can i cut it down the middle like that making two triangles with 0.5 degrees from one triangle with 1 degree?
 
  • #3
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I would first start by converting the distance between the planet and its sun from kilometers to meters, as this is the standard unit used in scientific calculations. This would give a distance of 5x10^11 meters.

Next, I would use the formula for angular size, which is given by θ = d/D, where θ is the angular size, d is the diameter of the object, and D is the distance to the object. Rearranging this formula to solve for the diameter, we get d = θD.

Substituting in the known values, we get d = (1 degree)(5x10^11 meters) = 8.73x10^9 meters.

Therefore, the diameter of the sun is approximately 8.73x10^9 meters or 8,730,000 kilometers. This is a very large diameter, which is expected for a massive object like the sun.
 

1. What is the formula for calculating the diameter of a planet?

The formula for calculating the diameter of a planet is diameter = angular size * distance / 206265. This formula is based on the small angle formula, which calculates the size of an object based on its angular size and distance from the observer.

2. How do we measure the angular size of a planet?

The angular size of a planet can be measured using a telescope or other optical instrument. By measuring the apparent size of the planet in degrees, minutes, and seconds, we can plug this value into the formula for calculating the diameter.

3. What units are typically used for the distance to a planet?

The distance to a planet is typically measured in astronomical units (AU) or kilometers (km). 1 AU is the average distance between the Earth and the Sun, which is approximately 149.6 million km.

4. Can we use this formula to calculate the diameter of any planet?

Yes, this formula can be used to calculate the diameter of any planet as long as we have accurate measurements of its angular size and distance from the observer.

5. How accurate is this method of calculating the diameter of a planet?

This method of calculating the diameter of a planet is considered to be very accurate, with a margin of error of only a few kilometers. However, small discrepancies may occur due to factors such as atmospheric distortion and variations in the planet's shape.

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