Question

Find the acceleration of our galaxy, due to the nearest comparably sized galaxy. The approximate mass of each galaxy is $8\times 10^{11}$ solar masses and they are separated by $2$ million light-years. Each galaxy has a diameter of 100,000 light years.
(Assume 1 light year $=10^{16}m$, Gravitational constant $G=$ $10^{-10}\left(\frac{N{m}^2}{k{g}^2}\right)$ and mass of sun $=2.0\times 10^{30}Kg$ )

• A. $4\times 10^{-13}m/s^2$
• B. $2\times 10^{-13}m/s^2$
• C. $5\times 10^{-15}m/s^2$
• D. $5\times 10^{-13}m/s^2$

Answer 1

Answer: (A)

Here, masses of each galaxy are
$m_1=m_2=8\times 10^{11}$solar mass
$=8\times 10^{11}\times 2\times 10^{30}kg$
$=16\times 10^{41}kg$
Distance between galaxies is
$d=2\times 10^6$ light year
$=2\times 10^6\times 10^{16}m=2\times 10^{22}m$
Force of gravitational pull on each of galaxy is
$F=\frac{G{m}_1{m}_2}{d^2}=\frac{10^{-10}\times {\left(16\times 10^{41}\right)}^2}{{\left(2\times 10^{22}\right)}^2}$
Acceleration of galaxy is
$a=\frac{F}{m}=\frac{10^{-10}\times {\left(16\times 10^{41}\right)}^2}{{\left(2\times 10^{22}\right)}^2\times 16\times 10^{41}}$
$=\frac{10^{-10}\times 16\times 10^{41}}{{\left(2\times 10^{22}\right)}^2}=4\times 10^{-13}m{s}^{-2}$