Answer
a) $F=16N$
(Note that we assumed a mass of 70kg; other assumptions are possible.)
b) $2.14\times 10^8m$
Work Step by Step
(a) The required force can be determined as
$F=\frac{4GmM_a}{r^3}$
We plug in the known values to obtain:
$F=4(6.67\times 10^{-11}N.m^2/Kg^2)\frac{(70Kg)(10^6\times 2.00\times 10^{30}Kg)(1.8m)}{(10^6mi\times 1609m/mi)^3}$
$F=16N$
(Note that we assumed a mass of 70kg; other assumptions are possible.)
(b) We can find the required distance as follows:
$r=(\frac{4GmM_a}{F})^{\frac{1}{3}}$
$\implies r=(\frac{4GmM_a}{10mg})^{\frac{1}{3}}=(\frac{4GM_a}{10g})^{\frac{1}{3}}$
We plug in the known values to obtain:
$r=[\frac{4(6.67\times 10^{-11})(2.00\times 10^{36})(1.8)}{10(9.81)}]^{\frac{1}{3}}=2.14\times 10^8m$