Use the expression -
P²k = a³---(i)
Where P is time in years, a is in AU, k is the function of the Sun's mass and set to unity.
Further, we're looking for
the mass of the new star we have slightly modified Kepler's Third Law. After
all, the inverse of a constant is also a constant. The reason we do it is
because we want an answer in the form of K:1, where K is the mass of the HD 10180
and the mass of the Sun is 1, so k needs to start out on the other side.
Convert 600 days in years = 600/365 = 1.64 years
Put this in the expression mentioned at (i) -
1.64²k = a³
You gave us a = 1.4 AU
1.64²k = 1.4³ ; isolating k
k = 1.4³/1.64²
k = 2.744/2.702 = 1.015
So, the requisite ratio = 1.015 : 1
The ratio of the star's mass to the sun's mass is about 1.02 : 1
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Centripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
[tex]\texttt{ }[/tex]
Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
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Given:
mass of the sun = M_sun = 1.99 × 10³⁰ kg
radius of the orbit = R = 1.4 AU = 2.094 × 10¹¹ m
Orbital Period of planet = T = 600 days = 600 × 24 × 3600 = 5.184 × 10⁷ seconds
Asked:
mass of the star = M = ?
Solution:
Firstly , we will use this following formula to find the mass of the star:
[tex]F = ma[/tex]
[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]
[tex]G M = \omega^2 R^3[/tex]
[tex]\frac{GM}{R^3} = \omega^2[/tex]
[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]
[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]
[tex]M = \frac{4 \pi^2 R^3}{GT^2}[/tex]
[tex]M = \frac{4 \pi^2 (2.094 \times 10^{11})^3}{6.67 \times 10^{-11} \times (5.184 \times 10^7)^2}[/tex]
[tex]\boxed {M \approx 2.0 \times 10^{30} \texttt{ kg} }[/tex]
[tex]\texttt{ }[/tex]
Next , we will calculate the ratio of the star's mass to the sun's mass as follows:
[tex]M : M_{sun} = (2.0 \times 10^{30}) : (1.99 \times 10^{30})[/tex]
[tex]\boxed{M : M_{sun} \approx 1.02 : 1}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Circular Motion
