MooneffectSome people believe that the Moon controls their activities. If the Moon moves from being directly on the opposite side of Earth from you to being directly overhead, by what percent does (a) the Moon’s gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? Assume that the Earth–Moon (center-to-center) distance is$\mathbf{3}\mathbf{.}\mathbf{82}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{m}$and Earth’s radius is$\mathbf{6}\mathbf{.}\mathbf{37}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}$.

The Earth-Moon (center-to-center) distance, ${\mathrm{r}}_{\mathrm{me}}=3.82\times {10}^8\mathrm{m}$

The radius of the Earth,${\mathrm{r}}_{\mathrm{e}}=6.37\times {10}^6\mathrm{m}$.

i)$M=Massofmoon$

ii)$M_e=Massofearth$

According to Newton’s law of gravitation, the force between the two objects is directly proportional to the masses of the bodies in the system and inversely proportional to the square of the distance between them.

Formula:

Gravitational force,

$F=\frac{GMm}{r^2}$ (i)

Using equation (i), Force between you and Moon at the initial position can be given as:

$F_1=\frac{GMm}{{\left(r_{me}+r_e\right)}^2}$

Using equation (i), Force between you and Moon at the final position can be given as:

$F_2=\frac{GMm}{{\left(r_{me}-r_e\right)}^2}$

Ratio of two forces:

$$\begin{gathered} \frac{F_2}{F_1}=\frac{GMm}{{\left(r_{me}-r_e\right)}^2}\times \frac{{\left(r_{me}-r_e\right)}^2}{GMm} \\ =\frac{{\left(r_{me}+r_e\right)}^2}{{\left(r_{me}-r_e\right)}^2} \\ =\frac{{\left[\left(3.82\times {10}^{8}m\right)+\left(6.37\times {10}^{6}m\right)\right]}^2}{{\left[\left(3.82\times {10}^{8}m\right)-\left(6.37\times {10}^{6}m\right)\right]}^2} \\ =1.069 \end{gathered}$$

$$\begin{gathered} \%I\mathrm{ncrease}\mathrm{in}\mathrm{force}=\left(\frac{F_2-F_1}{F_1}\times 100\right) \\ =\left(\frac{F_2}{F_1}-1\right)\times 100 \\ =\left(1.069-1\right)\times 100 \\ =6.9\% \end{gathered}$$

There is an increase in 6.9% in the net gravitational force.

Now, calculate the decrease in the weight using the decrease in the gravitational acceleration or gravitational force.

$$\begin{gathered} \left(\begin{array}{c}\%\mathrm{Decrease}\\ \mathrm{in}\mathrm{weight}\end{array}\right)=\frac{{\mathrm{W}}_2-{\mathrm{W}}_1}{\mathrm{Actiual}\mathrm{weight}} \\ =\frac{{\displaystyle \frac{\mathrm{GMm}}{{\left({\mathrm{r}}_{\mathrm{me}}-{\mathrm{r}}_{\mathrm{e}}\right)}^2}}-{\displaystyle \frac{\mathrm{GMm}}{{\left({\mathrm{r}}_{\mathrm{me}}-{\mathrm{r}}_{\mathrm{e}}\right)}^2}}}{{\displaystyle \frac{\mathrm{GMm}}{{\mathrm{r}}_{\mathrm{e}}^2}}}\times 100 \\ =\frac{4{\mathrm{r}}_{\mathrm{me}}\times {\mathrm{r}}_{\mathrm{e}}^3}{{\left({\mathrm{r}}_{\mathrm{me}}-{\mathrm{r}}_{\mathrm{e}}\right)}^2\times {\left({\mathrm{r}}_{\mathrm{me}}-{\mathrm{r}}_{\mathrm{e}}\right)}^2}\times \left(\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{e}}}\right)\times 100 \\ =\frac{4\left(3.82\times {10}^8\mathrm{m}\times {\left(6.37\times {10}^6\mathrm{m}\right)}^3\right)}{{\left(3.82\times {10}^8\mathrm{m}-6.37\times {10}^6\mathrm{m}\right)}^2\times {\left(3.82\times {10}^8\mathrm{m}+6.37\times {10}^6\mathrm{m}\right)}^2}\left(\frac{7.36\times {10}^{22}\mathrm{kg}}{5.98\times {10}^{24}\mathrm{kg}}\right)\times 100 \\ =2.28\times {10}^2\% \\ \approx 2.3\times {10}^2\% \end{gathered}$$

Therefore, there is a decrease of $2.3\times {10}^2\%$in weight.