Respuesta :
Solution :
Distance between the Earth and the moon, D = [tex]$3.84 \times 10^8 \ m$[/tex]
Radius of the earth, r = [tex]$6.37 \times 10^6 \ m$[/tex]
Gravitational force on the near side of the moon, [tex]$F_N = \frac{GM_m m}{(D-r)^2}$[/tex] ......(1)
Gravitational force on the far side of the moon, [tex]$F_F = \frac{GM_m m}{(D+r)^2}$[/tex] ..............(2)
where, [tex]$M_m$[/tex] = mass of moon
m = mass of the near side and far side respectively (both are taken as equal)
Therefore,
[tex]$\frac{F_N-F_F}{F_N}\times 100 = \frac{\frac{GM_m m}{D-r}^2-\frac{GM_m m}{D+r}^2}{\frac{GM_m m}{D-r}^2} \times 100 $[/tex]
[tex]$=1 - \frac{(D-r)^2}{(D+r)^2} \times 100$[/tex]
[tex]$=1- \left(\frac{1-\frac{r}{D}}{1+\frac{r}{D}}\right)^2 \times 100$[/tex]
[tex]$= 1- \left(\frac{0.98}{1.02}\right)^2 \times 100$[/tex]
= 7%
Hence, correct option is (b) About 7% greater
