g(t) = g0 + g1 sinωt
Now, the observed gravity of Earth would be a mean value,
g = g0 + g1 /2πω (cos 2πω-cos 0) = g0
Since gravity of Earth depends on gravity constant (and vice versa), we can conclude that,
G(t) = G+ (GMT/ R2T) g1 sinωt
The simple equation mg=mdv/dt slightly changes to
m a = m dv/dt = mg + mg1 sinωt
so,
v (t) = g t + g1/ω [cos ωt-1]
Obviously, frequency "ω" should be enormous, and its clear that when it tends to infinite, the last term vanishes.
It does not seem very likely, because as a result of all that,
S (t) = 1/2 gt2 -g1/ω2 sin ωt - (g1/ω) t
which looks a little weird… Again, the most interesting thing is that if ω is big enough, the last two terms would be very little and when ω tends to infinite, they simply vanish (into thin air, as it were); and of course, the last term, "- (g1/ω t)", that is so intriguing.
so,
v (t) = g t + g1/ω [cos ωt-1]
Obviously, frequency "ω" should be enormous, and its clear that when it tends to infinite, the last term vanishes.
It does not seem very likely, because as a result of all that,
S (t) = 1/2 gt2 -g1/ω2 sin ωt - (g1/ω) t
which looks a little weird… Again, the most interesting thing is that if ω is big enough, the last two terms would be very little and when ω tends to infinite, they simply vanish (into thin air, as it were); and of course, the last term, "- (g1/ω t)", that is so intriguing.
