Chapter 5, Problem 60P

To determine

The period of Sun’s orbital motion.

Answer to Problem 60P

Solution:

  $\text{1}{\text{.85×10}}^{\text{8}}\text{years}$

Explanation of Solution

Given:

Mass of Galaxy = ${\text{4×10}}^{\text{41}}\text{kg}$

Mean distance from the center of Galaxy = 30000 light-years

Formula Used:

The period of Sun’s orbital motion about the center of the galaxy can be determined by using Kepler’s third law,

  $T^2=\left(\frac{4{\pi }^2}{G{m}_g}\right)r^3$

Where,

G -gravitational constant

$r$ -mean distance

$m_g$ -Mass of galaxy

Calculation:

Convert the distance light years to meters:

  $\begin{array}{l}\text{r}\text{=}{\text{3×10}}^{\text{4}}\text{light}\text{years}\end{array}$

  $\begin{array}{c}\text{=}{\text{3×10}}^{\text{4}}\text{lightyears}\left(\frac{\text{9}{\text{.46×10}}^{\text{15}}\text{m}}{\text{1}\text{light}\text{year}}\right)\end{array}$

  $\begin{array}{c}\text{ =}\text{2}{\text{.84×10}}^{\text{20}}\text{m}\end{array}$

Substitute the known values and solve:

  ${\text{T}}^{\text{2}}\text{=}\left(\frac{{\text{4π}}^{\text{2}}}{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left({\text{4×10}}^{\text{41}}\text{kg}\right)}\right){\left(\text{2}{\text{.84×10}}^{\text{20}}\text{m}\right)}^3$

   $\text{T=}\sqrt{\left(\frac{{\text{4π}}^{\text{2}}}{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left({\text{4×10}}^{\text{41}}\text{kg}\right)}\right){\left(\text{2}{\text{.84×10}}^{\text{20}}\text{m}\right)}^3}$

   $\text{=5}\text{.82}\times {\text{10}}^{15}\mathrm{s}$ .

Converting the term period, second to years, we get

  $\begin{array}{l}\text{T=}\left(\text{5}{\text{.82×10}}^{\text{15}}\text{s}\right)\left(\frac{\text{1}\text{year}}{\text{365}\text{days}}\right)\left(\frac{\text{1day}}{\text{24}\text{hours}}\right)\left(\frac{\text{1}\text{hour}}{\text{60}\text{}\text{minutes}}\right)\left(\frac{\text{1}\text{minute}}{\text{60}\text{s}}\right)\end{array}$

   $\begin{array}{c}\text{=1}{\text{.85×10}}^{\text{8}}\text{years}\end{array}$

Conclusion:

Thus, the period of Sun’s orbital motion $\text{=1}{\text{.85×10}}^{\text{8}}\text{years}$ .