Respuesta :
Answer:
a) M = 5.46 10³¹ kg and b) T = 1.17 10²s
Explanation:
a) Let's use the law of universal gravitation and Newton's second law
F = G m M / r²
F = ma
F = m v² / r
G m M / r² = m v² / r
G M / r = v² (1)
M = r v² / G
Let's use the slowest planet, the velocity module is constant
v = d / t
v = 2π r / T
T = 7.6 years (365 days / year) (24 h / 1 day) (3600s / 1 h) = 2,397 10⁸ s
v = 45.7 km / s (1000m / 1km) = 45700 m / s
r = T v / 2π
r = 2,397 10⁸ 45700 /2π
r = 1,743 10¹² m
We calculate the mass of the Star
M = 1,743 10¹² 45700² / 6.67 10⁻¹¹
M = 5.46 10³¹ kg
b) Faster period of the planet
From equation1 we cleared the radius of the orbit
r₂² = G M / v₂²
v₂ = 56.7 km / s (1000m / 1km) = 56700 m / s
r₂² = 6.67 10⁻¹¹ 5.46 10³¹/56700²
r₂² = 1,133 10¹² m2
r₂ = 1.06 10⁶ m
We calculate the period
v = 2π r / T
T = 2π r / v
T = 2π 1.06 10⁶/56700
T = 1.17 10²s
