Answer
2.4
Work Step by Step
We know that
$r=\sqrt\frac{GM_E}{g^\prime}$
Also, $G=6.67\times10^{-11} \frac{Nm^2}{ Kg^2}$, $M_E=6\times10^{24}Kg$ and $g^\prime=1.67\frac{m}{s^2}$
Putting these values in the equation, we get
$r=\sqrt\frac{6.67\times10^{-11}\times6\times10^{24}}{1.67}$
$r=1.5\times 10^7 m$
Now we find the ration $\frac{r}{r_e}$ where $r_e$ is the radius of the Earth ($6.4\times 10^6$):
$\frac{r}{r_e}=\frac{1.5\times 10^7}{6.4\times10^6}=2.4$
Therefore, the object must be 2.4 Earth radii from the center of Earth if it is to weigh the same as it does on the Moon.
