First we need to calculate the mass of the tiny sphere, which is given by
[tex]m=dV[/tex]
where
d is the density
V is the volume
Given the radius of the sphere, r, its volume is
[tex]V= \frac{4}{3} \pi r^3= \frac{4}{3} \pi (1.0 \cdot 10^{-5}m)^3 =4.2 \cdot 10^{-15} m^3[/tex]
So now we can calculate the mass of the sphere
[tex]m=dV=(1.0 \cdot 10^{18} kg/m^3)(4.2 \cdot 10^{-15} m^3)=4187 kg[/tex]
And the weight of the sphere at the Earth surface, where the gravitational acceleration is [tex]g=9.81 m/s^2[/tex], would be
[tex]W=mg=(4187 kg)(9.81 m/s^2)=41074 kg[/tex]
