Simple Calculation of Astronomical Unit Problem

[unscientific]

Homework Statement

The period of Mars is 1.881 years. When Mars is opposite the Sun in the sky, its position with respect to the background of fixed stars is measured at sunset and again at sunrise. The effective baseline between the two positions is 11,700 km and the change in position on the sky is 30.8''. Find 1 AU.

Homework Equations

The Attempt at a Solution

The solution provided is shown below.

To convert 30.8'' arcseconds to radians, shouldn't it be $\frac{30.8}{60\times 60} \times \frac{\pi}{180}$?

[Edit: It appears there was a mistake in the solutions, the final answer turns out to be right.]

 

[SteamKing]

Yes, there are 3600 arc-seconds in one degree of arc.

However, your solution above omits the factor '180' in your calculations, even though the final answer is numerically correct.

You have only calculated the earth-Mars distance, D. The problem asks for one to calculate the value of 1 AU (astronomical unit), which is the distance from the Earth to the sun. This value is quite larger than what you have assumed is your answer.

 

[unscientific]

Yes, there are 3600 arc-seconds in one degree of arc.

However, your solution above omits the factor '180' in your calculations, even though the final answer is numerically correct.

You have only calculated the earth-Mars distance, D. The problem asks for one to calculate the value of 1 AU (astronomical unit), which is the distance from the Earth to the sun. This value is quite larger than what you have assumed is your answer.

Yes, I forgot to include the second part of solution which uses Kepler's 3rd Law to find the earth-sun distance, given we have found mars-sun distance.