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Planets beyond the solar system.On October 15, 2001, a planet was discovered orbiting around the star HD68988. Its orbital distance was measured to be10.5million kilometers from the center of the star, and its orbital period was estimated at 6.3 days.
A.What is the mass of HD68988? Express your answer in kilograms.
B.What is the mass of HD68988? Express your answer in terms of our sun's mass.

Respuesta :

Answer:

Part a)

[tex]M = 2.31 \times 10^{30} kg[/tex]

Part b)

[tex]M = 1.15 M_s[/tex]

Explanation:

Part a)

As we know that the period of revolution of the planet is given by

[tex]T = 2\pi\sqrt{\frac{r^3}{GM}}[/tex]

now we know that

[tex]r = 10.5 \times 10^6 km = 1.05 \times 10^{10} m[/tex]

T = 6.3 days

[tex]T = 5.44 \times 10^5 s[/tex]

now we have

[tex]5.44 \times 10^5 = 2\pi\sqrt{\frac{(1.05\times 10^{10})^3}{(6.67 \times 10^{-11})M}}[/tex]

[tex]M = 2.31 \times 10^{30} kg[/tex]

Part b)

As we know that mass of sun is given as

[tex]M_s = 2.0 \times 10^{30} kg[/tex]

now we have

[tex]\frac{M}{M_s} = \frac{2.31 \times 10^{30}}{2 \times 10^{30}}[/tex]

[tex]M = 1.15 M_s[/tex]

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