Given :
Los Angeles and Las Vegas.
Mass of Mars , [tex]M=6.42\times 10^{23}\ kg[/tex].
Radius of Mars ,
[tex]R=3397\ km\\\\R = 3.397\times 10^6\ m[/tex]
To Find :
The acceleration of gravity on the surface of Mars.
Solution :
Acceleration due to gravity is given by :
[tex]g_m = \dfrac{GM_m}{R_m^2}\\\\g_m =\dfrac{(6.67\times 10 ^{-11}\ m^3\ kg^{-1}\ s^{-2})(6.42\times 10^{23}\ kg) }{(3.397\times 10^6\ m)^2}\\\\g_m = 3.71\ m/s^2[/tex]
Therefore, the acceleration of gravity on the surface of Mars is 3.71 m/s².
Hence, this is the required solution.
