Question

Imagine we find a star that has the same radius as the Sun, but its temperature is 9.6 times greater. That star should be _____ times as luminous as the Sun.

Suppose we find a planet with an orbital period of 200 days around a star with the same mass as the Sun, but only 75 percent as luminous. What is the planet’s semi-major axis, in AU?

Answer 1

We use the equation below to solve for the star’s luminosity, relative to the sun’s, where L = luminosity and T = surface temperature, when the radius is the same. Here, surface temperature equals 9.6 solar

Ans.

So, Luminocity is 8493.46 times!

From Doppler's Mathematical Insight the semi-major axis (a) is equal to

$\sqrt[3]{\frac{GM}{4\pi r^{2}}P^{2}}$

Here, P is time period. 200/365 = 0.5479 times the P of earth. Putting this in above equation we get,

Semi-major axis of this planet is 0.6696 times the earth.