Chapter 11, Problem 20P

(a)

To determine

To Calculate:The size of the semi major axis of the planet’s orbit.

Answer to Problem 20P

  $1.33\text{AU}$

Explanation of Solution

Given data:

  $M=1.05M_{Sun}$

Here, $M_{Sun}$ is the mass of the sun.

Orbital period: $T=1.50\text{yr}$

FormulaUsed:

According to Kepler’s third law of planetary motion, the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

  $T^2=\frac{4{\pi }^2}{GM}{r}^3$

Here, $T$ is the time period, $r$ is the length of the semi major axis of the orbit, $G$ is the gravitational constant, and $M$ is the mass of the star.

Calculation:

  $T^2=\frac{4{\pi }^2}{G\left(1.05M_{Sun}\right)}{r}^3$

Rearrange this equation for $r$

  $r={\left(\frac{1.05G{M}_{Sun}{T}^2}{4{\pi }^2}\right)}^{1/3}$

Convert the orbital period of the planet from years to seconds.

  $\begin{array}{c}T=\left(\text{1}\text{.50}\text{y}\right)\left(\frac{\text{365}\text{days}}{\text{1}\text{y}}\right)\left(\frac{\text{24}\text{hours}}{\text{1}\text{day}}\right)\left(\frac{\text{60}\text{min}}{\text{1}\text{hour}}\right)\left(\frac{\text{60}\text{s}}{\text{1}\text{min}}\right)\\ \text{=4}{\text{.37×10}}^{\text{7}}\text{s}\end{array}$

Calculate the size of the semi major axis of the planet’s orbit

  $r={\left(\frac{1.05G{M}_{Sun}{T}^2}{4{\pi }^2}\right)}^{1/3}$

Substitute $\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{ for G,}\text{2}{\text{.0×10}}^{\text{30}}{\text{kg for M}}_{\text{Sun}}\text{ and 4}{\text{.73×10}}^{\text{7}}\text{s for T}$

  $\begin{array}{c}r={\left(\frac{\text{1}\text{.05}\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{2}{\text{.0×10}}^{\text{30}}\text{kg}\right){\left(\text{4}{\text{.73×10}}^{\text{7}}\text{s }\right)}^{\text{2}}}{{\text{4π}}^{\text{2}}}\right)}^{\text{1/3}}\\ \text{=}{\left(\text{7}{\text{.93928×10}}^{\text{33}}{\text{m}}^{\text{3}}\right)}^{\text{1/3}}\\ \text{=}\left(\text{1}{\text{.995×10}}^{\text{11}}\text{m}\right)\left(\frac{\text{1}\text{AU}}{\text{1}{\text{.50×10}}^{\text{11}}\text{m}}\right)\\ \text{=1}\text{.33}\text{AU}\end{array}$

Conclusion:

The size of the semi major axis of the planter’s orbit is $1.33\text{AU}$ .

(b)

To determine

To Calculate:The mass of the planet as compared to the mass of jupiter.

Answer to Problem 20P

The mass of the planet is $24.6$ times the mass of the Jupiter that is $m=\left(24.6\right)M_J$ .

Explanation of Solution

Given data:

Mass of the planet = $m$

Mass of star= $M$

Formula Used:

Use conservation of momentum principle for the planet of mass $m$ and star of mass $M$ .

  $mv=MV$

Here, $v$ is the speed of the orbiting planet and $V$ is the speed of the star.

Calculation:

Since, $M=1.05M_{Sun}$ , rewrite the above equation and rearrange it for $m$ .

  $m=\frac{1.05M_{Sun}V}{v}\mathrm{........}\left(1\right)$

Calculate the orbital speed of the planet.

  $v=\frac{2\pi r}{T}$

Substitute $\text{1}{\text{.995×10}}^{\text{11}}\text{m for }r\text{ and 4}{\text{.73×10}}^{\text{7}}\text{s for }T$ .

  $\begin{array}{c}v=\frac{2\pi r}{T}\\ \text{=}\frac{\text{2π}\left(\text{1}{\text{.995×10}}^{\text{11}}\text{m}\right)}{\text{4}{\text{.73×10}}^{\text{7}}\text{s}}\\ \text{=2}{\text{.65×10}}^{\text{4}}\text{m/s}\end{array}$

Calculate the mass of the planet $m$ using equation $\left(1\right)$ .

  $m=\frac{1.05M_{Sun}V}{v}$

Substitute $\text{2}{\text{.0×10}}^{\text{30}}{\text{kg for M}}_{\text{Sun}}\text{,}\text{592}\text{m/s}$ for speed of the planet $V,\text{ and 2}\text{.65}\times {\text{10}}^4\text{m/s for }v$ .

  $\begin{array}{c}m=\frac{1.05M_{Sun}V}{v}\\ \text{=}\frac{\left(\text{1}\text{.05}\right)\left(\text{1}{\text{.99×10}}^{\text{30}}\text{kg}\right)\left(\text{592}\text{m/s}\right)}{\text{2}{\text{.65×10}}^{\text{4}}\text{m/s}}\\ \text{=466}{\text{.78×10}}^{\text{26}}\text{kg}\end{array}$

Compare the mass of the planet with the mass of the Jupiter $M_J$ of value $1.90\times {10}^{27}\text{kg}$ .

  $\begin{array}{c}\frac{m}{M_J}=\frac{466.78\times {10}^{26}kg}{1.90\times {10}^{27}\text{kg}}\\ =24.6\\ m=\left(24.6\right)M_J\end{array}$

Conclusion:

The mass of the planet is $24.6$ times the mass of the Jupiter that is $m=\left(24.6\right)M_J$ .