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 in kilograms?

(b) as a multiple of our sun's mass.

(c) How fast (in km/s) is this planet moving?

Answer 1

using notation AeB = A * 10^B.

mass of Jupiter, which is approximately 1.9e27 kg.

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.

the planet's orbit is 1/9 the mean of a and p = (1/9) * (4.5e7km) = 4.5e6 km = 4.5e9 m

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.

B)

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

C)

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