Answered: In 2014, the Mars rover named Curiosity…

Applications and Investigations in Earth Science (9th Edition)

9th Edition

ISBN:9780134746241

Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa

Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa

Chapter1: The Study Of Minerals

Section: Chapter Questions

Problem 1LR

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Question

In 2014, the Mars rover named Curiosity observed the Martian moon Phobos eclipsing most of the
Sun as shown in the figure below. Look up the “average” Sun-Mars distance in units of AU and use this value throughout this problem.
(a) Using the fact that the Sun subtends an angle of 0.5 degrees from the Earth, determine the
angular size of the Sun from Mars. (i.e., use ratios of Earth and Mars properties, instead of just
looking up the actual size of the Sun)
(b) Find the dimensions of Phobos and its average distance from Mars. Use this to check your answer about the
Sun’s angular size, by roughly determining the angular size of the Phobos and seeing that this

is somewhat consistent with the eclipse picture below. Ignore details of less than a factor of 2.

This is an ungraded review problem on angular size ratios.

The image shows a sequence of three photographs illustrating different stages of a solar eclipse.

- **Left Image**: The sun is partially obscured, with a crescent of light visible on the left side.
- **Center Image**: The eclipse has progressed to a more symmetrical appearance, resembling a ring of light, known as an annular eclipse.
- **Right Image**: The sun is again partially obscured, with the crescent of light now visible on the right side.

Each stage demonstrates the dynamic motion of the moon as it moves across the face of the sun, changing the amount and shape of sunlight visible from Earth.

Expert Solution

Step 1: Solution of a

Calculate the angular size of the Sun from Mars

To determine the angular size of the Sun from Mars, we can use ratios of Earth and Mars properties. We are given that the Sun subtends an angle of 0.5 degrees from Earth.

The average distance from the Sun to Mars can be found by looking up the “average” Sun-Mars distance in units of AU. Let’s assume this value is 1.5 AU.

Using the fact that angular size is inversely proportional to the distance, we can set up the following proportion:

Earth-Sun distance / Earth-Sun angular size = Mars-Sun distance / Mars-Sun angular size

Plugging in the values:

1 AU / 0.5 degrees = 1.5 AU / Mars-Sun angular size.

Plugging in the values:

1 AU / 0.5 degrees = 1.5 AU / Mars-Sun angular size

Rearranging the equation to solve for Mars-Sun angular size:

Mars-Sun angular size = (1.5 AU / 0.5 degrees) * 1 AU = 3 degrees

So, the angular size of the Sun from Mars is 3 degrees.