Answer:
Mass of HD 68988 is ≅ 1.2 times the mass of sun.
Explanation:
Given :
Orbital distance [tex]r = 10.5 \times 10^{9}[/tex] m
Time period [tex]T = 6.3[/tex] days
Now we have to convert time period in seconds
[tex]T = 6.3 \times 24 \times 60\times 60 = 544320[/tex] sec
From the kepler's third law,
 [tex]T = \frac{2\pi r^{\frac{3}{2} } }{\sqrt{Gm} }[/tex]
 [tex]m = \frac{4\pi^{2} r^{3} }{T^{2}G }[/tex]
Where [tex]G = 6.67 \times 10^{-11}[/tex]
  [tex]m = \frac{4\pi^{2} (10.5 \times 10^{9} )^{3} }{(544320)^{2} 6.67 \times 10^{-11} }[/tex]
  [tex]m[/tex] ≅ [tex]2.32 \times 10^{30}[/tex] Kg
Now we have to calculate mass of HD 68988 in terms of sun mass,
Mass of sun = [tex]1.99 \times 10^{30}[/tex] Kg
Therefore mass of HD 68988 is  [tex]\frac{2.32 \times 10^{30}}{1.99 \times 10^{30}}[/tex] ≅ 1.2 times the mass of sun.
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