Respuesta :
Answer:
(a)[tex]M = 1.62x10^{31}Kg[/tex] (b)[tex]T = 1.06y[/tex]
Explanation:
(a) What is the mass of the star?
The Universal law of gravitation shows the interaction of gravity between two bodies:
[tex]F = G\frac{Mm}{r^{2}}[/tex] (1)
Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.
For this particular case M is the mass of the star and m is the mass of the planet. Since it is a circular motion the centripetal acceleration will be:
[tex]a = \frac{v^{2}}{r}[/tex] (2)
Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):
[tex]ma = G\frac{Mm}{r^{2}}[/tex]
By replacing (2) in equation (1) it is gotten:
[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex] (3)
Therefore, the mass of the star can be determine if M is isolated from equation (3):
[tex]M = \frac{rv^{2}}{G}[/tex] (4)
But r can be known from Kepler's third law, since it gave the semi-major axis:
[tex]a^{3} = T^{2}[/tex]
[tex]a = \sqrt[3]{(7.60)^{2}}[/tex]
[tex]a = \sqrt[3]{57.76}[/tex]
[tex]a = 3.86y[/tex]
However, a can be expressed in astronomical units:
[tex]3.86y. \frac{1AU}{1y}[/tex] ⇒ [tex]3.86AU[/tex]
One astronomical unit is defined as the distance between the Earth and the Sun ([tex]1.50x10^8 Km[/tex]):
[tex]3.86AU. \frac{1.50x10^8km}{1AU}[/tex] ⇒ [tex]5.79x10^{8}Km[/tex]
r and v will be expressed in meters before being replaced in equation (4):
[tex]r = 5.79x10^{8}Km . \frac{1000m}{1Km}[/tex] ⇒ [tex]5.79x10^{11}m[/tex]
[tex]v = 43.3Km . \frac{1000m}{1Km}[/tex] ⇒ [tex]43300m[/tex]
[tex]M = \frac{(5.79x10^{11}m)(43300m)^{2}}{6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2}}[/tex]
[tex]M = 1.62x10^{31}Kg[/tex]
So the mass of the star is [tex]1.62x10^{31}[/tex]Kg
(b) What is the orbital period of the faster planet, in years?
To find the period, the equation for orbital velocity can be used:
[tex]v = \frac{2\pi r}{T}[/tex] (5)
Notice that the distance of the faster planet from the Star (r) is needed, that can be found using equation (4) in terms of r and the mass of the star:
[tex]r = \frac{MG}{v^{2}}[/tex]
It is necessary to express the velocity of the faster planet in meters.
[tex]58.6Km . \frac{1000m}{1Km}[/tex] ⇒ [tex]58600m[/tex]
[tex]r = \frac{(1.62x10^{31}Kg)(6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})}{(58600m/s)^{2}}[/tex]
[tex]r = 3.14x10^{11}m[/tex]
Equation (5) can be rewritten in terms of T:
[tex]T = \frac{2\pi r}{v}[/tex]
[tex]T = \frac{2\pi(3.14x10^{11}m)}{58600m/s}[/tex]
[tex]T = 33667579 s[/tex]
There are 31536000 seconds in 1 year:
[tex]T = 33667579 s. \frac{1y}{31536000s}[/tex] ⇒ [tex]1.06y[/tex]
[tex]T = 1.06y[/tex]
So the period of the faster planet is 1.06 years
