Question

A distant star system is discovered in which a planet with

twice the radius of the earth and rotating 3.0 times as fast as the

earth orbits a star with a total power output of $6.8\times {10}^{29}W$.

a. If the star’s radius is 6.0 times that of the sun, what is the

electromagnetic wave intensity at the surface? Astronomers

call this the surface flux. Astronomical data are provided

inside the back cover of the book.

b. Every planet-day (one rotation), the planet receives$9.4\times {10}^{22}J$.

of energy. What is the planet’s distance from its star? Give

your answer in astronomical units (AU), where 1 AU is the

distance of the earth from the sun.

Short answer

a. The wave intensity at the surface is $3.1\times {10}^{9}W/m^2$

b. The planet's distance from its star is,$19.4au$

Step-by-step solution

Part a Step 1: Given data

The radius of the star is 6 times of the radius of the sun that is,

$$\begin{gathered} R=6\times R_{sun} \\ R=\left(6\times 6.96\times {10}^{8}m\right) \end{gathered}$$

The power output of the star,$P=6.8\times {10}^{29}W$

Determination of intensity

Consider the energy radiation in spherical symmetry, the intensity of the EM wave at the surface is,

$$\begin{gathered} I=\frac{P}{4{\mathrm{\pi R}}^2} \\ \Rightarrow I=\frac{6.8\times {10}^{29}W}{4\mathrm{\pi }{\left(6\times 6.96\times {10}^{8}m\right)}^2} \\ \Rightarrow I=3.1\times {10}^{9}W/m^2 \end{gathered}$$

Part b: Step 1: Introduction

The energy from the star will be radiated into the space equally in all directions. The ratio of the power output by

the star to the power falling on the planet’s surface is equal to the ratio of the surface area of a sphere with radius

equal to the planet’s orbital radius to the cross-sectional area of the planet that can be considered as a disc.

The planet completes one full rotation 3 times faster than the earth. So, the time taken by the planet to complete a rotation is, $t=\frac{24}{3}h=8h$

Energy received by the planet is,role="math" localid="1649875552528" $U_{Planet}=9.4\times {10}^{22}J$

Determination of planet's distance from its star

According to the question,

$$\begin{gathered} \frac{P_{star}}{P_{planet}}=\frac{4{\mathrm{\pi R}}_{\mathrm{orbit}}^2}{{\mathrm{\pi R}}_{\mathrm{planet}}^2} \\ {R}_{orbit}=\sqrt{\frac{P_{star}{R}_{\mathrm{planet}}^2}{4P_{planet}}} \\ {R}_{orbit}=\frac{R_{planet}}{2}\sqrt{\frac{P_{star}}{P_{planet}}} \\ {R}_{orbit}=R_{earth}\sqrt{\frac{6.8\times {10}^{29}W}{\frac{U_{planet}}{t}}} \\ {R}_{orbit}=6.37\times {10}^{6}m\sqrt{\frac{6.8\times {10}^{29}W\times \left(8\times 3600\right)s}{9.4\times {10}^{22}J}} \\ {R}_{orbit}=2.9\times {10}^{12}m \\ {R}_{orbit}=\left(2.9\times {10}^{12}\right)\times \left(6.685\times {10}^{-12}\right)au \\ {R}_{orbit}=19.4au \end{gathered}$$