To develop this problem we will use the gravitational equations posed in the Newtonian theory and for which gravity is described as,
[tex]a = \frac{GM_{mars}}{R^2_{mars}}[/tex]
Where,
G = Gravitational Universal constant
[tex]M_{mars}[/tex]= Mass of mars
[tex]R_{mars}[/tex]= Radius of mars
Using the relation given we have,
[tex]a = \frac{G*(0.107*M_e)}{(0.53R_e)^2}[/tex]
[tex]a = \frac{GM_E}{R^2_E}*\frac{0.107}{0.53^2}[/tex]
The first term is known to all as gravity on earth, which is equivalent to [tex]9.8m / s ^ 2[/tex]
[tex]a = 9.8*0.381[/tex]
[tex]a = 3.73m/s^2[/tex]
Therefore the gravitational acceleration at the surface of Mars is [tex]3.73m/s^2[/tex]