Chapter 11.5, Problem 59E

(a)

To determine

To Find: The absolute magnitude of for object M35 in the constellation Gemini has an apparent magnitude of 5.3 and a parallax of about 0.018.

Answer to Problem 59E

The absolute magnitude is $1.58$ .

Explanation of Solution

Given:

The objects more than 30 parsecs, or 576 trillion miles from Earth, the formula relating the parallax $p$ , the absolute magnitude $M$ , and the apparent magnitude $m$ , is $M-m+5+5\log p$ .

The one parsec is about 3.26 light years or 19.2 trillion miles

Calculation:

Consider the formula is $M=m+5+5\log p$ .

Then, the magnitude is,

  $\begin{array}{c}M=\left(5.3\right)+5+5\log \left(0.018\right)\\ =\left(5.3\right)+5-8.7236\\ =1.58\end{array}$

(b)

To determine

To Find: The parallax to four decimal places

Answer to Problem 59E

The absolute magnitude is $0.0219$ .

Explanation of Solution

Given:

The stars with apparent magnitudes greater than 5 can only be seen with a telescope. If a star has an apparent magnitude of 8.6 and an absolute magnitude of 5.3,

Calculation:

Consider the formula is $M=m+5+5\log p$ .

Then, the parallax formula is,

  $\begin{array}{l}5\log p=m-\left(m+5\right)\\ P={10}^{\frac{\left(M-m-5\right)}{5}}\end{array}$

Then, the parallax is,

  $\begin{array}{c}P={10}^{\frac{\left(5.3-8.6-5\right)}{5}}\\ =0.0219\end{array}$