Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

once the distance to a remote object is determined via triangulation, o…

Question

once the distance to a remote object is determined via triangulation, one can measure the angular diameter of the object and convert that to an actual diameter. for relatively small angular diameters of a few degrees or less, the corresponding linear diameter can be approximated using the following expression:
diameter = \frac{\text{distance} \times (\text{angular diameter})}{(57.3 \text{ degrees})}

the figure below illustrates that the diameter of a remote object increases with its distance and its measured angular diameter.

image: diagram with observer, distant object, angular diameter, distance (known), diameter (unknown), 360°

consider the moon and sun. their angular diameters are both equal to about .5 degree. if the sun is roughly 400 times more distant than the moon, how much bigger is the sun’s diameter than the moon’s?

view available hint(s)

  • about 800 times bigger
  • about 400 times bigger
  • about 400 times smaller

Explanation:

Step1: Define variables for Moon and Sun

Let the distance of the Moon be \( d_{Moon} \), so the distance of the Sun \( d_{Sun} = 400 \times d_{Moon} \). The angular diameter of both is \( \theta = 0.5^\circ \) (assuming the "5 degree" is a typo and should be 0.5 degree, as the actual angular diameter of Sun and Moon is about 0.5 degrees; if it's 5 degrees, the logic is similar).

Step2: Use the diameter formula for Moon

The formula for diameter is \( D=\frac{\text{Distance} \times \text{Angular Diameter}}{57.3^\circ} \). For the Moon: \( D_{Moon}=\frac{d_{Moon} \times \theta}{57.3^\circ} \)

Step3: Use the diameter formula for Sun

For the Sun: \( D_{Sun}=\frac{d_{Sun} \times \theta}{57.3^\circ} \). Substitute \( d_{Sun} = 400d_{Moon} \): \( D_{Sun}=\frac{400d_{Moon} \times \theta}{57.3^\circ} \)

Step4: Find the ratio of Sun's diameter to Moon's diameter

\( \frac{D_{Sun}}{D_{Moon}}=\frac{\frac{400d_{Moon} \times \theta}{57.3^\circ}}{\frac{d_{Moon} \times \theta}{57.3^\circ}} \). The \( d_{Moon} \), \( \theta \), and \( 57.3^\circ \) terms cancel out, leaving \( \frac{D_{Sun}}{D_{Moon}} = 400 \). So the Sun's diameter is about 400 times bigger than the Moon's.

Answer:

B. about 400 times bigger