Answer: [tex]5.70 m/s^{2}[/tex]
Explanation:
This question is incomplete, please remember to post the whole question :)
However, the table related to this question is attached, where it is shown the mass and radius of each planet expressed in terms of Earth's mass [tex]m_{E}=5.972(10)^{24} kg[/tex] and Earth's radius [tex]r_{E}=6371000 m[/tex].
For example, for planet Rams, the mass is [tex]m_{R}=9.3 m_{E}[/tex] and the radius is [tex]r_{R}=4 r_{E}[/tex].
Now, for Earth the acceleration due gravity is [tex]g_{E}=9.8 m/s^{2}[/tex] and according to Newton's Universal Law of Gravitation we can find the gravitational forcer exerted by Earth [tex]F_{E}[/tex] on an object placed on its surface with mass [tex]m[/tex]:
[tex]F_{E}=mg_{E}=G\frac{m_{E} m}{r_{E}^{2}}[/tex] (1)
Where [tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant
Simplifying:
[tex]g_{E}=G\frac{m_{E}}{r_{E}^{2}}[/tex] (2)
Doing the same with planet Rams:
[tex]F_{R}=mg_{R}=G\frac{m_{R} m}{r_{R}^{2}}[/tex] (3)
Simplifying:
[tex]g_{R}=G\frac{m_{R}}{r_{R}^{2}}[/tex] (4)
Remembering the relation between the Earth's mass and radius with Ram's mass and radius:
[tex]g_{R}=G\frac{9.3 m_{E}}{(4r_{E})^{2}}[/tex] (5)
Solving:
[tex]g_{R}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{9.3 (5.972(10)^{24} kg)}{(4(6371000 m))^{2}}[/tex] (6)
Finally:
[tex]g_{R}=5.70 m/s^{2}[/tex]
