Question

Hot Jupiters. In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star $\mathbf{HD}\mathbf{179949}$ (hence the term “hot Jupiter”). The orbit was just$\frac{\mathbf{1}}{\mathbf{9}}$ the distance of Mercury from our sun, and it takes the planet only$\mathbf{3}\mathbf{.}\mathbf{09}$days to make one orbit (assumed to be circular). (a) What is the mass of the star? Express your answer in kilograms and as a multiple of our sun’s mass. (b) How fast (in km>s) is this planet moving?

Short answer

a)The mass of the star is $2.21\times {10}^{30}\text{\hspace{0.17em}kg}$, and${\mathrm{m}}_{\mathrm{star}}=1.11{\mathrm{m}}_{\mathrm{sun}}$

b) The planet is moving at a speed of $1.51\times {10}^5\text{\hspace{0.17em}m}/\text{s}$.

Step-by-step solution

Identification of the given data

• The orbit is$\frac{1}{9}$ the distance of the Mercury from the sun.
• The planet takes about$3.09$ making one orbit.

 Step 2: Significance of the Kepler’s third law in identifying the mass of the star and speed of the planet

This law illustrates that a planet’s orbital period’s square is equal to its orbit’s “semi-major axes”.

The product of the cube of the radius and the Kepler's constant, along with the time period, will provide the star's mass. Moreover, dividing the perimeter of the star by the time period gives the star's speed.

Determination of the mass and also the speed of the star

a) From Kepler’s third law, the orbital period of the star can be expressed as:

$\begin{array}{c}\mathrm{T}=\frac{2{\mathrm{\pi r}}^{3/2}}{\sqrt{{\mathrm{Gm}}_{\mathrm{star}}}}\\ {\mathrm{m}}_{\mathrm{star}}=\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^3}{{\mathrm{T}}^2\mathrm{G}}\end{array}$

Here, T is the planet's orbital period, which is$3.09\text{\hspace{0.17em}days=2.67}\times {10}^{5\text{}}\text{\hspace{0.17em}s}$r is the orbit radius of theearth $\frac{5.79\times {10}^{10}\text{\hspace{0.17em}m}}{9}=6.43\times {10}^9\text{m}$. The value of G, which is the gravitational constant is $6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$.

Substituting the values in the above equation, we get,

$\begin{array}{l}{m}_{star}=\frac{{\text{4π}}^{\text{2}}{\left({\text{6.43×10}}^{\text{9}}\text{\hspace{0.17em}m}\right)}^{\text{3}}}{{\left({\text{2.67×10}}^{\text{5}}\text{\hspace{0.17em}s}\right)}^{\text{2}}\left({\text{6.67×10}}^{\text{-11}}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)}\\ m_{star}=2.21\times {10}^{30}\text{\hspace{0.17em}kg}\end{array}$

Hence, the mass of the star is $2.21\times {10}^{30}\text{\hspace{0.17em}kg}$.

Apart from that, it has been observed that $\frac{{\mathrm{m}}_{\mathrm{star}}}{{\mathrm{m}}_{\mathrm{sun}}}=1.11$, hence,${\mathrm{m}}_{\mathrm{star}}=1.11{\mathrm{m}}_{\mathrm{sun}}$

b) The orbital velocity of the planet is expressed as:

$\mathrm{v}=\frac{2\mathrm{\pi r}}{\mathrm{T}}$

Here v is the orbital velocity, r is the orbital radius of the planet,$6.43\times {10}^9\text{\hspace{0.17em}m}$ and T is the earth's orbital period $2.67\times {10}^5\text{\hspace{0.17em}s}$.

Substituting the values in the above equation, we get-

$\begin{array}{c}\mathrm{v}=\frac{2\mathrm{\pi }(6.43\times {10}^9\text{\hspace{0.17em}m})}{2.67\times {10}^5\text{\hspace{0.17em}s}}\\ \mathrm{v}=1.51\times {10}^5\text{\hspace{0.17em}s}\end{array}$

Thus, the planet is moving at a speed of $1.51\times {10}^5\text{\hspace{0.17em}s}$.