Answer:
v = 2.75 10⁴ m / s
Explanation:
For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system
F = ma
where force is the force of gravity
F = [tex]G \frac{m M}{r^2}[/tex]
acceleration is centripetal
a = [tex]\frac{v^2}{r}[/tex]
we substitute
G m M / r² = m v² / r
[tex]\frac{GM}{r}[/tex] = v²
v = [tex]\sqrt{GM/r}[/tex]
indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system
r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m
let's calculate
v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11
v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8
v = 2.75 10⁴ m / s
