Answer
(a) To the moon.
(b) 1.62 N.
(c) 2.7 mN.
(d) 1.617 N toward the moon.
Work Step by Step
(a) The rock will fall to the moon.
Since the moon is much closer to the rock than the earth, and the gravitational force is inversely proportional to the square of the distance between the two bodies.
(b) Knowing that the gravitational field strength of the moon $g_{m} $= 1.62 N/kg, we can get the gravitational force exerted by the moon on the 1 kg rock.
(b)
$W_{m} $= m.$ g_{m} $= 1 x 1.62 = 1.62 N
$W_{m}$ = 1.62 N, toward the moon
(c)Knowing that the distance between the earth and the moon is r= 3.845 x $10^{8}$m and the mass of the earth is $M_{e}$ = 5.974 x $10^{24}$ kg. we can get the gravitational force exerted by the earth on the I kg rock using the Newton's law of universal gravitaion
W = G$ \frac{m_{1} m_{2}}{r^{2}}$
where G= 6.674 x $10^{-11} $N.$m^{2}$/.$Kg^{2}$
W = G$ \frac{m_{1} m_{2}}{r^{2}}$
=(6.674 x$ 10^{-11})$ $\frac{ (5.974 \times10^{24}) }{(3.845\times 10^{8}) \times (3.845\times 10^{8}) }$$
= 2.7 \times $$10^{-3} $ N, toward the earth
(d)Taking the direction toward the moon to be the positive direction
From (b) and (c) we can get the net force.
$F_{n}$=$W_{m} $ -$W_{e} $.
= 1.62 - (2.7$ \times $ $10^{-3}$)
= 1.6173 N, toward the moon
