Answer
The ratio of the sphere's gravitational force on the dust particle and the earth's gravitational force on the dust particle is $1.61\times 10^{-7}$.
Work Step by Step
Let $M_s$ be mass of the lead sphere.
Let $R_s$ be the distance from the dust to the sphere's center.
Let $M_d$ be the mass of the dust particle.
We can write an expression for the force of gravity $F_s$ of the sphere on the dust particle.
$F_s = \frac{G~M_s~M_d}{R_s^2}$
Let $M_e$ be the mass of the earth. Let $R_e$ be the earth's radius. We can write an expression for the force of gravity $F_e$ of the earth on the dust particle.
$F_e = \frac{G~M_e~M_d}{R_e^2}$
We can find the ratio of $\frac{F_s}{F_e}$.
$\frac{F_s}{F_e} = \frac{(\frac{G~M_s~M_d}{R_s^2})}{(\frac{G~M_e~M_d}{R_e^2})}$
$\frac{F_s}{F_e} = \frac{M_s~R_e^2}{M_e~R_s^2}$
$\frac{F_s}{F_e} = \frac{(5900~kg)(6.38\times 10^6~m)^2}{(5.98\times 10^{24}~kg)(0.50~m)^2}$
$\frac{F_s}{F_e} = 1.61\times 10^{-7}$
The ratio of the sphere's gravitational force on the dust particle and the earth's gravitational force on the dust particle is $1.61\times 10^{-7}$.
