Answer:
Part a)
a = 1.62 m/s/s
Part b)
a = 3.70 m/s/s
Explanation:
Part A)
Acceleration due to gravity on the surface of moon is given as
[tex]a = \frac{GM}{R^2}[/tex]
here we know that
[tex]M = 7.35 \times 10^{22} kg[/tex]
[tex]R = 1.74 \times 10^6 m[/tex]
now we have
[tex]a_g = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}[/tex]
[tex]a_g = 1.62 m/s^2[/tex]
Part B)
Acceleration due to gravity on surface of Mercury is given as
[tex]a = \frac{GM}{R^2}[/tex]
here we know that
[tex]M = 3.30 \times 10^{23} kg[/tex]
[tex]R = 2.44 \times 10^6 m[/tex]
now we have
[tex]a_g = \frac{(6.67 \times 10^{-11})(3.30 \times 10^{23})}{(2.44 \times 10^6)^2}[/tex]
[tex]a_g = 3.70 m/s^2[/tex]