Answer:
Decreased by a factor of 4.5
Explanation:
"We have Newton formula for attraction force between 2 objects with mass and a distance between them:
[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]
where [tex]G =6.67408 × 10^{-11} m^3/kgs^2[/tex] is the gravitational constant on Earth. [tex]M_1, M_2[/tex] are the masses of the object and Earth itself. and R distance between, or the Earth radius.
So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:
[tex]\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}[/tex]
[tex]\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}[/tex]
[tex]\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}[/tex]
[tex]\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2[/tex]
Since [tex]M_2 = 2m_2[/tex] and [tex]r = R/3[/tex]
[tex]\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5[/tex]
So gravity would have been decreased by a factor of 4.5
