Answer:
period of rotation is 9.9843 years
Explanation:
Given data
sides = 1.0×10^12 m
to find out
period of rotation
solution
first we use the gravititional formula i.e
F(g) = 2 F cos 30
here F = G × mass(sun)² / radius²
F = 6.67 × × (1.99× )² / (1× )²
so
F(g) = 2 × 6.67 × × (1.99× )² / (1× )² × (1.73/2)
F(g) = 4.57 × N
and
by the law of sines
r/sin30 = s/sin120
so r will be
r = 1.0×10^12 (0.5/.87)
r = 5.75 × m
we know that gravititional force = centripetal force
so here
centripetal force = mv² /r
centripetal force = 1.99× v² / 5.75 ×
so
4.57 × = mv² /r
4.57 × = 1.99× v² / 5.75 ×
v² = 4.57 × × 5.75 ×
and
v = 11480 m/sec
and we know period of rotation formula
t = d /v
t = 2 r /v
t = 2 × × 5.75 × [tex]10^{11} / 11480
t = 314 × [tex]10^{5} / 31536000 year
t = 9.9843 year
so period of rotation is 9.9843 years