Chapter 33, Problem 39GP

Solution

The gravitational force with which each galaxy attracts each other.

  $F=0.166\times {10}^{40}$ Newton

Given:

The nearest large galaxy is about $2\times {10}^6$ light year away from a person’s galaxy.

Both galaxies have mass of $3\times {10}^{41}$ kg.

Formula used:

  $F=G\frac{m_1{m}_2}{r^2}$

Here, m₁ and m₂ are masses.

r is the distance between the two masses.

G is gravitational constant

Calculation:

1 light year = $3\times {10}^8\times 365\times 24\times 60\times 60=9.5\times {10}^{15}\text{m}$

So,

  $2\times {10}^6\text{ly}$ light is equal to $9.5\times {10}^{15}\times 2\times {10}^6=19.0\times {10}^{21}\text{m}$

G is equal to $6.67\times {10}^{-11}$ Newtons kg-²m²

From $F=G\frac{m_1{m}_2}{r^2}$

  $F=6.67\times {10}^{-11}\frac{3\times {10}^{41}\times 3\times {10}^{41}}{{(19.0\times {10}^{21})}^2}$

  $\begin{array}{c}F=\frac{60.03\times {10}^{82}}{361\times {10}^{42}}\\ =0.166\times {10}^{40}\text{N}\end{array}$