HOMEWORK 2 SOLUTIONS COSMOLOGY PHYSICS 20B

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1 HOMEWORK S COSMOOGY PHYSICS 0B (5pts) 1. If it takes Venus 5 days to orbit the Sun, roughly how far is Venus from the Sun? et us define the period of Venus orbit around the Sun as P V the semi-major axis distance from the Sun as A V. and Venus Since the eccentricity of Venus orbit is very close to zero (e 0.007) the semi-major axis distance is approximately the distance between the Sun and Venus. We can then use the following relation to solve for the distance between Venus and the Sun. P V P E = A3 V A 3 E where P E is the period of the Earth s orbit around the Sun and A E is the Earth s semi-major axis distance from the Sun. Converting 5 days to years and plugging in the values we get the following: (0.616 yr) (1 yr) = A3 V (1 AU) 3 () (0.616 yr) A V = 3 (1 yr) AU 0.74 AU (3) The distance between the Sun and Venus is approximately 0.74 AU. (1) (5pts). Rigel is the brightest star in the Orion constellation. Its spectrum peaks at a wavelength of approximately 77 Å. What is the temperature at the surface of Rigel? We can use the following relation to solve for the temperature of Rigel given its mas wavelength. λ MAX = 600 nm 5000 K T First we must convert 77 Åto nm. Note: 1 Å= m (4) m 109 nm 1 m = 7.7 nm (5) 1

2 Rearranging equation 4 we can solve for the temperature of Rigel. T = 5000 k 600 nm λ MAX = 5000 K 600 nm 7.7 nm K (6) The temperature at the surface of the star Rigel is K. (5pts) 3. If the Earth orbited the Sun at a distance of 8 AU, how much fainter would the Sun appear to be in the daytime sky? For this question we use the apparent brightness equation. b = 4πR (7) where b is the apparent brightness, is the luminosity, and R is the distance from the source. We want to compare the apparent brightness of the Sun when the Earth is 8 AU away and when its 1 AU away. When the Earth is 1 AU away we have the following: b 1 = 1 4π(1 AU) (8) where b 1 and 1 is the apparent brightness and luminosity of the Sun at 1 AU away. And for the Earth at 8 AU from the Sun we have b 8 = 8 4π(8 AU) (9) where b 8 and 8 is the apparent brightness and luminosity of the Sun at 8 AU away. Since the source aka the Sun is the same in both set ups the luminosity is the same. 1 = 8 = We can then rearrange equation 8 and 9 to be in terms of b and R. 4π = b 8(8 AU) = b 1 (1 AU) (10) b 8 = 1 64 b 1 = b 1 (11) If the Earth was 8 AU away it would be 64 times fainter than if it was at 1 AU or 1.6 % the brightness at 1 AU.

3 (10pts) 4. How much weaker ( or stronger ) is the force of gravity on the surface of Pluto versus that on the surface of the Earth? [ The mass and radius of the Earth and Pluto are M E = kg, R E = 6400 km, M P = kg, R P = 100 km. ] In this problem we want to solve for the force of gravity at the surface of the Earth and Pluto. We first start with the force due to gravity equation F = Gm 1m d (1) where G is the gravitational constant, m 1 and m are the masses of objects 1 and, respectively, and d is the distance between the objects. To solve for the force of gravity at the surface we need to place an object at the surface and calculate the force on it for both planets. et s assume we place a person on the surface of the Earth with mass, m. Therefore F will be the force of gravity on the person at the surface. Solving for the force of gravity on a person at the surface of the Earth we get F E = GmM E R E (13) Doing the same process for a person on the surface of Pluto we get F P = GmM P R P (14) What we have done here is solved for the force of a person on the surface of the Earth and Pluto due to the individual Planet s gravitational pull. We then now equate the two equations using the fact that Gm is the same in both equations. Solving for F P Gm = F P R P M P = F E R E M E (15) F P = R E M P M E RP F E = (6400 km)( kg ) ( kg)(100 km) F E 1 16 F E 0.06 F E (16) The force on the surface of Pluto is 6.% or 16 times weaker than the force on the surface of the Earth. 3

4 (5pts) 5. You are 10 meters away from a candle and 100 meters away from a 150 Watt light bulb, but both the candle and the bulb look to be the same brightness to you. How uminous is the candle ( in Watts )? Again we use the apparent brightness equation b = 4πR (17) For the candle and the light bulb we have the following equations c b c = 4πRc (18) b b b = 4πRb (19) where b c, b b, R c, R b, c and b is the apparent brightness, distance and luminosity of the candle and light bulb, respectively. We are given the values of R c, R b and b are given values in the problem question as well as the understanding that b c = b b. We can then solve for the luminosity of the candle, c. c 4πR c = b 4πR b c = b Rb Rc (150 W atts)(10 m) = (100 m) = 1.5 W atts (0) The uminosity of the candle is 1.5 Watts. (5pts) 6. Neptune is approximately km from the Sun and takes roughly 165 years to complete one trip around the Sun. How quickly is Neptune moving along its orbit? We want to solve for Neptune s speed. ike from HW 1 the speed is T otal Distance Speed = (1) T otal T ime et s confine ourselves to one orbit around the Sun with radius R N. In one orbit, the total distance traveled is given by the circumference. T otal Distance T raveled = πr N = π ( km ) = km () 4

5 Converting kilometers to meters and years to seconds we can solve for the speed. Speed = T otal Distance T otal T ime = m s = 544 m s or 19, 59km hr (3) Neptune s speed as it orbits the Sun is 544 m s km or 19,59 hr. 5