Answer :
The mass, in Msun, of a star observed to have a planet with an orbital radius of 4.00AU and a period of 8.00 years is calculated using Kepler's Third Law, which relates the period, the semimajor axis, and the masses of two orbiting bodies.
The formula is given below:
[tex][tex][tex][tex] \frac{a^3}{P^2} = \frac{G(m_1+m_2)}{4\pi^2} [/tex][/tex][/tex][/tex]where [tex]m_1[/tex] and [tex]m_2[/tex] are the masses of the two objects in solar masses, [tex]a[/tex] is the semimajor axis of the planet's orbit in astronomical units (AU), [tex]P[/tex] is the planet's orbital period in years, [tex]G[/tex] is the gravitational constant in units of [(AU^3)/(Msun yr^2)] (Note: Msun represents the mass of the Sun)
The given value is [tex]a = 4.00AU[/tex] and [tex]P = 8.00 years[/tex]. The gravitational constant is [tex]G = 4\pi^2[/tex].We want to calculate the mass of the star, which is denoted by [tex]m_1[/tex]. Let [tex]m_2 = m_p[/tex] be the mass of the planet.
Plugging in the values, we get:[tex] \frac{(4.00\text{AU})^3}{(8.00\text{yr})^2} = \frac{(4\pi^2) \cdot (m_1 + m_p)}{4\pi^2} [/tex]Simplifying, we get:[tex] m_1 + m_p = \frac{(4.00\text{AU})^3}{(8.00\text{yr})^2} [/tex]Subtracting [tex]m_p[/tex] from both sides gives:[tex] m_1 = \frac{(4.00\text{AU})^3}{(8.00\text{yr})^2} - m_p [/tex]We can now substitute [tex]m_p = 1[/tex], since the mass of the planet is given in units of 1 Jupiter mass.
Thus:[tex] m_1 = \frac{(4.00\text{AU})^3}{(8.00\text{yr})^2} - 1 = \boxed{1.63 \text{ Msun}} [/tex]
Therefore, the mass, in Msun, of a star observed to have a planet with an orbital radius of 4.00AU and a period of 8.00 years is approximately 1.63 Msun.
Learn more about orbital radius here.
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