Answer:
The acceleration due to gravity on the surface of Z-34 is equal to [tex]3g[/tex] .
Explanation:
Here we know that acceleration due to gravity on a planet of mass M and radius R is given by
[tex]g = \frac{GM}{R^{2}}[/tex]
Given that mass of earth is M and radius is R .
Mass of Z-34 = [tex]\frac{M}{3}[/tex] ,
Radius of Z-34 = [tex]\frac{R}{3}[/tex]
G = gravitational constant .
[tex]g_{z}[/tex] = acceleration due to gravity on Z-34 ,
[tex]\frac{g}{g_{z}} = \frac{M}{\frac{M}{3} } \times (\frac{\frac{R}{3} }{R})^{2} = \frac{1}{3} \\ \\g_{z} = 3g[/tex]
