Question

On October 15, 2001, a planet was discovered orbiting around the star HD68988. 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.

Answer 1

Mass of HD 68988 is ≅ 1.2 times the mass of sun.

Step-by-step explanation:

Given :

Orbital distance
`r = 10.5 * 10^(9)` m

Time period
`T = 6.3` days

Now we have to convert time period in seconds

`T = 6.3 * 24 * 60* 60 = 544320` sec

From the kepler's third law,

`T = \frac{2\pi r^{(3)/(2) } }{√(Gm) }`

`m = (4\pi^(2) r^(3) )/(T^(2)G )`

Where
`G = 6.67 * 10^(-11)`

`m = (4\pi^(2) (10.5 * 10^(9) )^(3) )/((544320)^(2) 6.67 * 10^(-11) )`

`m` ≅
`2.32 * 10^(30)` Kg

Now we have to calculate mass of HD 68988 in terms of sun mass,

Mass of sun =
`1.99 * 10^(30)` Kg

Therefore mass of HD 68988 is
`(2.32 * 10^(30))/(1.99 * 10^(30))` ≅ 1.2 times the mass of sun.