Answer:
Part a)
[tex]g = 1 \times 10^{-5} m/s^2[/tex]
Part b)
[tex]v = 0.011 m/s[/tex]
Explanation:
Part a)
As we know that the acceleration due to gravity is given as
[tex]g = \frac{GM}{R^2}[/tex]
[tex]g = \frac{G(\frac{4}{3}\pi R^3 \rho)}{R^2}[/tex]
[tex]g = \frac{4}{3}\pi \rho G R[/tex]
now we know that
[tex]\rho = \frac{M}{\frac{4}{3}\pi R^3}[/tex]
[tex]\rho = \frac{5.98 \times 10^{24}}{\frac{4}{3}\pi (6.37 \times 10^6)^3}[/tex]
[tex]\rho = 5523.2 kg/m^3[/tex]
now we have
[tex]g = \frac{4}{3}\pi (5523.2) (6.67 \times 10^{-11})(6.5)[/tex]
[tex]g = 1 \times 10^{-5} m/s^2[/tex]
Part b)
As we know that the escape speed is given as
[tex]v = \sqrt{\frac{2GM}{R}}[/tex]
[tex]v = \sqrt{2gR}[/tex]
[tex]v = \sqrt{2(1\times 10^{-5})(6.5)}[/tex]
[tex]v = 0.011 m/s[/tex]