Question

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 19 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 (a)in kilograms and (b)as a multiple of our sun's mass.

Answer 1

ote, I'll be using notation AeB = A * 10^B.

This question requires some assumptions.
First, Jupiter-sized does not imply Jupiter mass.
Size does not matter, instead, it's the mass that matters, but let's assume that statement means it also has the mass of Jupiter, which is approximately 1.9e27 kg.

That planet orbits the star and the orbit radius is 1/9 that of Mercury. But Mercury's orbit is not circular, and has its semi-major axis (longest "radius" in ellipse) as 5.8e7 km, and eccentricity of 0.205
eccentricity e = (a-p)/(a+p), where a is the longest "radius" in the, and p is the shortest.
e = (a-p)/(a+p) = 1 - 2/((a/p)+1) = 0.205
2/((a/p)+1) = 1-0.205 = 0.795
(a/p)+1 = 2/0.795 = 2.516...
a/p = 2.516...-1 = 1.516...
We know a = 5.8*10^7 km, so p = a/1.516... =~ 2.3e7 km.
So much work, just for one assumption, and we will assume that the planet's orbit is 1/9 the mean of a and p = (1/9) * (4.5e7km) = 4.5e6 km = 4.5e9 m
[One can also assume the radius = 1/9 the semi-major axis, which sort of simplifies thing. It's all assumptions, since the question was not clear.]

Now, period of orbit (T) = 2*pi * sqrt( r^3/G(M1+M2) )
Where G is the Gravitional constant, M1 and M2 are the mass of the 2 bodies.
G = 6.67428e-11 N m^2 kg^-2
T = 2*pi * sqrt( r^3/G(M1+M2) ) = 3.09 days = 266976 s
sqrt(r^3/G(M1+M2)) =~ 42490
r^3 / G(M1+M2) =~ 1.8e9 s^2

We have r = 4.5e9 m from above.
G(M1+M2) = r^3 / 1.8e9 =~ 5.1e19
M1+M2 = 5.1e19 / G = 7.6e29 kg
Mass of that planet was less than 1%, and relatively negligible.
So mass of star = 7.6e29 kg.

(You may want to redo this with higher precision, but given the vague information such as 1/9 of an undefined "distance of Mercury", probably does not make sense to do higher precision).

Now, part B, mass of sun is about 2.0e30 kg, so as mass of star as multiple of sun's mass = 0.38.

Next, for speed.
Distance of one revolution = 2 * pi * r =~ 2.8e10 m
Time to make one revolution = 3.09 days = 266976 s
Speed =~ 1.05e5 m/s
Found this in math tutoring from some one else. Hope it helps. It is to the same question you asked.

Answer 2

wow, holy crap. I had no idea I had to account for the eccentricity