Question

if you dropped a ball while standing on the surface of mars at what rate would it accelerate toward the ground

Answer 1

First of all you need to have in mind the following data:
Mass of Mars: 6.43 x 10^23 kg
Radius of Mars: 3.40 x 10^6 m
Formulas: F = G(m1)(m2)/(r^2), m1 = F(r^2)/G(m2), m2 = F(r^2)/G(m1), F = ma G = 6.67 x 10^-11
 We can say that the first and second objects with mass can be called as following:  
m1=  centre of Mars  
m2 = will be the ball.
The distance between them = the radious of mars. 

Answer 2

Answer:

3.71 m/s^2

Explanation:

The acceleration of gravity on the surface of a planet can be found by using the following formula:

$g=\frac{GM}{r^2}$

where

$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=6.39\cdot 10^{23} kg$ is the mass of the planet (in this case, Mars)

$r=3.39\cdot 10^6 m$ is the radius of the planet (in this case, Mars)

Substituting the numbers into the formula, we find

$g=\frac{(6.67\cdot 10^{-11})(6.39\cdot 10^{23} kg)}{(3.39\cdot 10^6 m)^2}=3.71 m/s^2$