- #1
kuahji
- 390
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In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the term "hot Jupiter"). The orbit was just {1 \over 9} the distance of Mercury from our sun, and it takes the planet only 3.09 days to make one orbit (assumed to be circular). What is the mass of the star? Express your answer in kilograms.
Ok, so I using a formula found in the book T=2\pir^{3/2}/\sqrt{Gm}
Where r is the semi-major axis, G the gravitational constant, and m the mass of the star.
Solving for m
m=(2\pir^{3/2}/T)^2/G
From the initial problem I found T=4.23x10^4 (converting 3.09 days/rev to seconds).
The orbital radius of Mercury in back of the text is 5.79x10^10, dividing that by nine yields 6.43x10^9.
Plugging in all the quantities I get 8.77x10^31kg. However this is not the correct answer. Any ideas where I could be going wrong?
Ok, so I using a formula found in the book T=2\pir^{3/2}/\sqrt{Gm}
Where r is the semi-major axis, G the gravitational constant, and m the mass of the star.
Solving for m
m=(2\pir^{3/2}/T)^2/G
From the initial problem I found T=4.23x10^4 (converting 3.09 days/rev to seconds).
The orbital radius of Mercury in back of the text is 5.79x10^10, dividing that by nine yields 6.43x10^9.
Plugging in all the quantities I get 8.77x10^31kg. However this is not the correct answer. Any ideas where I could be going wrong?
