The mass of the star hd 68988 is about 2.3 × 10³⁰ kg.
The mass of the star hd 68988 is about 1.16 times of our sun's mass.
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Further explanation
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 )
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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 = 10.5 × 10⁶ km = 1.05 × 10¹⁰ m
Orbital Period of planet = T = 6.3 days = 6.3 × 24 × 3600 = 544320 seconds
Unknown:
mass of the star = M = ?
Solution:
Firstly , we will use this following formula to find the orbital period:
[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 (1.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 544320^2}[/tex]
[tex]\boxed {M \approx 2.3 \times 10^{30} \texttt{ kg} }[/tex]
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[tex]M : M_{sun} = 2.3 \times 10^{30} : 1.99 \times 10^{30}[/tex]
[tex]M : M_{sun} \approx 1.16[/tex]
[tex]\boxed {M \approx 1.16 \times M_{sun}}[/tex]
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Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
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Answer details
Grade: High School
Subject: Physics
Chapter: Circular Motion
