Information given:
Mass of A = 5.00kg
Mass of B = 5.00kg
Mass of C = 5.00kg
We know G = 6.67 x 10^{-11} Nm^{2}kg^{-2}.
(a) Using F = GmM/r^{2},
F_{BA} = [6.67 x 10^{-11} Nm^{2}kg^{2} (5kg) (5kg)] / (0.300 m)^{2}
F_{BA} = 1.85 × 10^{−8} N
F_{BC} = [6.67 x 10^{-11} Nm^{2}kg^{2} (5kg) (5kg)] / (0.400 m)^{2}
F_{BC} =1.04 × 10^{−8} N
we find that the A pulls upward on B with 1.85 × 10^{-8} N, and C pulls to the righton B
with 1.04 × 10^{−8} N
Thus, the net force canbe found by using Pythagoras:
(F_{N})^{2} = (1.85 × 10^{−8} N)^{2} + (1.04 × 10^{−8} N)^{2}
(F_{N}) = (2.13 × 10^{−8} N
(b) Tan θ = (1.85 × 10^{−8} N) / (1.04 × 10^{−8} N)
θ = 60.7° to the line BC
(c) Information given:
Altitude of satellite = 160 km = 1.60 x 10^{5} m
We know G = 6.67 x 10^{-11} Nm^{2}kg^{-2}.
We know Mass of the Earth = 5.98 x 10^{24} kg
We know Radius of the Earth = 6.37 x 10^{6} m
radius of orbit r = Altitude of satellite + Radius of the Earth
r = 1.60 x 10^{5} m + 6.37 x 10^{6} m
r = 6.53 x 10^{6} m
The centripetal acceleration is given by F_{c} = mv^{2}/r and is provided by the gravitation force F = GmM/r where r is the radius of orbit and M is the mass of the earth and m is the mass of the satellite
So mv^{2}/r = GmM/r^{2}
v^{2} = GM/r
v^{2} = (6.67 x 10^{-11} Nm^{2}kg^{-2}) ( 5.98 x 10^{24} kg) / (6.53 x 10^{6} m)
v = 7.82 x 10^{3} ms^{-1}
(d) T = 2πr/v
T = 2π(6.53 x 10^{6} m) / (7.82 x 10^{3} ms^{-1})
T = 5.23 x 10^{s }s
T =
87.4 minutes |