131. With gB = 9.8128 m/s2 and gM = 9.7999 m/s2, we apply Eq. 4-26: R M − RB =
FG H
IJ K
v02 sin 2θ 0 v02 sin 2θ 0 v02 sin 2θ 0 g B − = −1 gM gB gB gM
which becomes R M − RB = RB
FG 9.8128 − 1IJ H 9.7999 K
and yields (upon substituting RB = 8.09 m) RM – RB = 0.01 m = 1 cm.