2022-01-07

On October 15, 2001, a planet was discovered orbiting around the star HD 68988. Its orbital distance was measured to be 10.5 million kilometers from the center of the star, and its orbital period was estimated at 6.3 days. What is the mass of HD 68988? Express your answer in kilograms and in terms of our sun’s mass.

alexandrebaud43

The planet has an orbital radius $r=10.5×{10}^{9}m$ and the orbital period is $T=6.3days$. The period T of the planet is given in days, so the period in seconds is
$T=\left(6.3days\right)\left(\frac{86400s}{1day}\right)=54.43×{10}^{4}s$
Equation (13.12) states the relationship between the period T and the distance r in the form
$T=\frac{2\pi {r}^{\left(\frac{3}{2}\right)}}{\sqrt{GM}}$ (1)
Where M is the mass of the star and G is the general constant of gravity and equals $6.673×{10}^{-11}N\cdot \frac{{m}^{2}}{k}{g}^{2}$. We solve equation (1) for M to be in the form
$M=\frac{4{\pi }^{2}{r}^{3}}{{T}^{2}G}$ (2)
Now, we plug the values for r, T and G into equation (2) to get M
${M}_{s}=\frac{4{\pi }^{2}{r}^{3}}{{T}^{2}G}$
$=\frac{4{\pi }^{2}{\left(10.5×{10}^{9}m\right)}^{3}}{{\left(54.43×{10}^{4}s\right)}^{2}\left(6.67×{10}^{-11}N\cdot \frac{{m}^{2}}{k}{g}^{2}\right)}$
$=2.3×{10}^{30}kg$
The mass of the Sun is ${M}_{Sun}=1.99×{10}^{30}kg$, so in terms of the mass of the Sun, the mass of the star is
$M=\left(2.3×{10}^{30}kg\right)\left(\frac{{M}_{Sun}}{1.99×{10}^{30}kg}\right)=1.2{M}_{Sun}$

temzej9

Sun's mass is equivalent to $1.99×{10}^{30}kg$. The calculated mass of star HD68988 is $2.3×{10}^{30}kg$. The mass of the star is,
$M=\left(2.3×{10}^{30}kg\right)\cdot \left(\frac{1M}{1.99}×{10}^{30}kg\right)=1.16M$
Therefore, the mass of HD68988 is 1.16M.

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