Answer
a) $4.38\times10^7m/s^2$
b) $2.8\times10^9N$
c) $9.4\times10^3m/s$
Work Step by Step
a) Acceleration due to gravity at any location at or above the surface of a star is given by $g_{star}=\frac{GM_{star}}{r^2}$
$$g_{star}=\frac{GM_{Sun}}{r^2_{Moon}}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(1.99\times10^{30}kg)}{(1.74\times10^6m)^2}=4.38\times10^7m/s^2$$
b) $W=mg_{star}=(65kg)(4.38\times10^7m/s^2)=2.8\times10^9N$
c) Initial velocity $v_0=0$,
$$v^2=v^2_0+2a(x-x_0)$$
$$=\sqrt{2a(x-x_0)}=\sqrt{2(4.38\times10^7m/s^2)(1m)}=9.4\times10^3m/s$$
