**Question:** Find the number of Watts per square meter received when Earth is at 1 AU from the Sun. How many AUs do you need to be from the Sun for the brightness to be 2 times bigger?

**Answer:** You can find the example calculation on page 524 of the textbook, and it shows the Sun's apparent brightness is 1300 watts per square meter. For the brightness to be 2 times bigger, first you have to realize that you have to be closer to the Sun, so the answer has to less than 1 AU. Apparent brightness of light follows an inverse square law (brightness decreases as the square of the distance, see the luminosity-distance formula), so to be 2 times brighter, the distance has to be square root of 2 times closer or be at 0.707 AU.

**Question:** Mercury's orbit is quite elliptical, and it's about 1.5 times farther from the Sun at aphelion than at perihelion. Use the module to determine the amount of light
received at perihelion, and divide it by the amount of light at aphelion.

**Answer:** First of all, perihelion is the closest point to the Sun while the aphelion is the furthest point. So, the ratio being asked is larger than 1. Again, the light follows the inverse square law, so the amount of light at the near distance is (distance ratio)^{2} times larger. That is, (1.5)^{2}=2.25 times larger.

**Question:** It is estimated that liquid water is stable on a planet when the amount of light received is between about 1000 and 2000 watts per square meter. What is distance (in AU) where the light is at 1000 Watts per square meter? ___ AU

where the light is at 2000 Watts per square meter? ___ AU

**Answer:** In question 1, you learned that the apparent brightness of the Sun is 1300 watts per square meter. For it to drop to 1000 w/m^{2},
you have to move away from the Sun, so you know the answer is larger than 1 AU.
Again, the light obeys inverse square law. To decrease the brightness by a factor of 1300/1000=1.3, you have to increase the distance by square root of 1.3 which is 1.14. So, light is 1000 Watts per square meter at 1.14 AU.

Similarly, light is 2000 Watts per square meter or 2000/1300=1.54 times brighter when you are square root of 1.54 or 1.24 times closer to the Sun. So,
the light is at 2000 Watts per square meter when you are 1/1.24 = 0.81 AU away from the Sun.

**Question:** Star A has a radius 3 times the Sun's radius. If its temperature is the same as the Sun's, then the luminosity of Star A is __ times larger than that of the Sun.

**Answer:** The luminosity-diameter relation found on page 534 of your textbook (also given as Stefan-Boltzmann law in the lecture note), you know that luminosity of a star is proportional to the square of the radius and to the 4th power of temperature, i.e.

**Question:**Star B has the same size as the Sun. If it's temperature is 3 times higher than the Sun's, then the luminosity of Star B is __ times larger than that of the Sun.

**Answer:**This is the same question as Question 4, but this time temperature is changed. Luminosity depends on temperature to the 4th power, so increasing T by a factor of 3 means luminosity which increases as (temperature ratio)^{4} should go up as 3^{4}=3x3x3x3=9x9=81 times larger than the Sun.

**Question:**If a black hole has a mass 10 times the Sun's mass, what will its radius be in kilometers?

**Answer:**The black hole radius is 3 x (mass in solar unit) = 3 x 10 = 30 km as you learned in the textbook or in the lecture note.

**Question:** Suppose a spaceship is moving at 0.9c. How many times slower does the spaceship's clock appear to be running than a "stationary" clock? (Your answer should be a number greater than 1.)

**Answer:** From page 444 of your textbook (also in the lecture note),
you know that (time in moving frame) = (time in rest frame) x (square root of (1-(v/c)^{2}). Here, v/c=0.9 and (square root of (1-(v/c)^{2})
=(square root of (1-0.9x0.9))=0.44. In other words, in one minute of your time, the clock in the spaceship appears to tick only by 0.44 minute. So, the clock inside the spaceship is running 1/0.44=2.3 times slower.