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

## 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.

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.

## Newton's modification on Kepler's Third Law:

Kepler's third law relates the orbital distance and the orbital period. The modification of newton's law relates the mass of the object to the orbital period and orbital distance. The modified Kepler's third law is formulated as

{eq}T^{2}=\displaystyle \frac{4\pi^{2} r^{3}}{GM} {/eq}

Where:

{eq}T {/eq} is the period,

{eq}r {/eq} is the orbital distance,

{eq}G {/eq} is the gravitational constant, and

{eq}M {/eq} is the mass.

(A) The mass of HD68988 is {eq}M=2.31\times10^{30} kg {/eq}

(B) The mass of HD68988 is {eq}1.16 {/eq} times the mass of the sun or {eq}1.16 {/eq} suns.

Solution:

(A)

(i) First convert the period from days to seconds

{eq}6.3 days \times \frac{24 hrs}{1 day} \times \frac{60 mins}{1 hr} \times \frac{60 s}{1 min}=5.4432\times10^{5} s {/eq}

(ii)Using the modified Kepler's law

{eq}T^{2}=\displaystyle \frac{4\pi^{2} r^{3}}{GM} {/eq}

finding the mass

{eq}M=\displaystyle \frac{4\pi^{2} r^{3}}{GT^{2}} {/eq}

{eq}M=\displaystyle \frac{4 \pi^{2} (1.05\times10^{10} m)^{3}}{(6.67\times10^{-11} \frac{m^{3}}{kg \cdot s^{2}})(5.4432\times10^{5} s)^{2}} {/eq}

{eq}M=2.31\times10^{30} kg {/eq}

(B) To find in terms of the suns' mass, we get the ratio

{eq}\frac{M_{HD68988}}{M_{sun}}=\frac{2.31\times10^{30} kg}{1.99\times10^{30} kg} {/eq}

{eq}=1.16 {/eq}