Answer
The percentage increase in the moon's gravitational pull is 6.87%
Work Step by Step
We can find the distance to the moon when the moon is over the other side of the Earth:
$d_o = (3.82\times 10^8~m)+(6.37\times 10^6~m)$
$d_o = 3.8837\times 10^8~m$
We can find the distance to the moon when the moon is directly over our heads:
$d_h = (3.82\times 10^8~m)-(6.37\times 10^6~m)$
$d_h = 3.7563\times 10^8~m$
Let a person's mass be $M_p$
Let $M_m$ be the mass of the moon
We can find the moon's gravitational pull on the person when the moon is over the other side of the Earth:
$F_o = \frac{G~M_m~M_p}{d_o^2}$
$F_o = \frac{G~M_m~M_p}{(3.8837\times 10^8)^2}$
$F_o = \frac{G~M_m~M_p}{15.08\times 10^{16}}$
We can find the moon's gravitational pull on the person when the moon is directly over our heads:
$F_h = \frac{G~M_m~M_p}{d_h^2}$
$F_h = \frac{G~M_m~M_p}{(3.7563\times 10^8)^2}$
$F_h = \frac{G~M_m~M_p}{14.11\times 10^{16}}$
We can find the percentage increase in the moon's gravitational pull when it is directly overhead:
$\frac{F_h}{F_o} = \frac{\frac{G~M_m~M_p}{14.11\times 10^{16}}}{\frac{G~M_m~M_p}{15.08\times 10^{16}}} = \frac{15.08\times 10^{16}}{14.11\times 10^{16}} = 1.0687$
We can see that the percentage increase is 6.87%
The percentage increase in the moon's gravitational pull is 6.87%
