Detecting planets

As given, we can consider the sun (or star) to be a disk (circle), and assume its " ILLUMINATING REGION " to be the area of this disk . What this means is that we have a circular area, emitting light uniformly across its surface, and that the unobstructed area which we can see is directly responsible for our perception of its brightness. This is the technique described in finding planets. Now, in this particular situation, we're given a planet with volume 8 times that of the Earth. Right here is where the math kicks in! Given a sphere, E (earth) and another one P (random planet) and the ratio of their volumes (which in this case is 1:8), we can use the formula for their volumes to find the volume of P, as compared to the earth's volume. $(4/3 * 3.14 * (R_p)^3) /(4/3 \times 3.14 \times (R_e)^3)$ is equal to 8. Therefore, as constants get cancelled, $(R_p/R_e)^3 =8$ . On taking cube roots, we get that $R_p$ is 2 times $R_e$ . Hence, $R_p = 2 \times 6370$ km. Now, area of the planet's disk is $3.14(2 \times 6370)^2$ and that of the sun/star is $3.14 \times (700000)^2$ . Thus, the fraction of the star being blocked out by the planet is the ratio of their difference in areas to the area of the star. This "Blocking Out" is responsible for dimming in the intensity of starlight. The required fraction is: $** (A_{star} - A_{planet})/A_star **$ This value comes out to be equal to $1-{(4 \times 6370^2)/(700000^2)}$ which is .033124 This figure gives us the dimming in intensity of starlight due to a planet 8 times the volume of the earth.

Radius of the earth: 6370 km

Volume of the earth: 1.08 × 10¹² km³

The volume of the planet = 8 × Volume of the earth: 8.66 × 10¹² km³

Using the 'volume of a sphere' equation, the radius of the planet is:

∛((8.33 × 10¹²)⋅4/3⋅π) =12740km

Since we only see a disk, we need the area of the planet:

π⋅(12740)² =509904363.8 km³

And the area of the sun (as a disk) is:

π⋅(7 × 10⁵)² =1.539 × 10¹²

To find the percentage dimmed:

(Area of planet ÷ Area of Sun) × 100 =0.03313

First we are assuming that the distance between the star and the plant is much much less than the distance between earth and the star. Secondly, the percentage in which the starlight dims is equal to the ratio of the cross sectional area of the plant to the star times 100. We will show that the radius of the earth is $r_{e}$ , the radius of the plant is $r_{p}$ and the radius of the star is $r_{s}$ . Assuming that the earth, the plant and the star are all approximately spherical; the volume is equal to $\frac{4 \pi r^{3}}{3}$ where the volume of the plant is eight times that of the earth. Hence $8 \frac{4 \pi r_{e}^{3}}{3}=\frac{4 \pi r_{p}^{3}}{3}$ which simplifies to $r_{p}=2r_{e}$ .

Now the cross sectional area of the spheres are equal to $\pi r^{2}$ hence using our original statement of the percentage of the ratio of the planet to the star; $\frac{\pi (2r_{e})^{2}}{\pi (r_{s})^{2}}\times100$ . Therefore substituting the values of the radius of the earth and radius of the sun we get that the amount that the starlight will dim is equal to $0.033\%$ .

The volume of Earth is $\frac{4}{3} \times \pi \times 6370^{3}$ cubic km. By multiplying this by 8 and then dividing by $[\frac{4}{3} \times \pi]$ you get the radius of the planet cubed. So now take the cube root and you get the radius of the planet which is 12740 km.

Since we are assuming that the light is a disk because of us viewing it from far away, we just need to take the ratio of the planet's area / star's area and multiply by 100 to get percent of dimness. $[\pi \times (12740)^{2}] / [\pi \times (7 \times 10^{5})^{2}] \times 100 = 0.033124 \%$