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 $G =$ $10^{-10}\left(\frac{ Nm ^{2}}{ kg ^{2}}\right)$ 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}} $
$=\frac{10^{-10} \times 16 \times 10^{41}}{\left(2 \times 10^{22}\right)^{2}}=4 \times 10^{-13} ms ^{-2}$