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What will be the acceleration due to gravity on the surface of the moon, if its radius is

1/4th the radius of the earth and its mass is 1/80 times the mass of the earth.


Info is quite incorrect, as instead of 80, it is 96.
Let radius of moon= r_m
radius of earth= r_e
mass of moon= m_m
mass of earth= m_e
Let acc. due to gravity on earth= g_e
acc. due to gravity on moon= g_m

Given, r_e=4r_m and  m_e=96m_m
We know, 
g_e= \frac{Gm_e}{r_e^2} ............(i)
g_m= \frac{Gm_m}{r_m^2} ...........(ii)
Where G is the Universal gravitational constant

Divide (i) by (ii)
 \frac{g_e}{g_m} = \frac{Gm_e}{r_e^2} * \frac{r_m^2}{Gm_m}

 \frac{g_e}{g_m} = \frac{m_e}{r_e^2} * \frac{r_m^2}{m_m}
Substitute the values-:
 \frac{g_e}{g_m} = \frac{96m_m}{(4r_m)^2} * \frac{r_m^2}{m_m}
\frac{g_e}{g_m} = 6
g_m= \frac{g_e}{6}
We know, g_e=9.8m/s^2
So, g_m= \frac{9.8}{6} = 1.633 m/s^2
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