SCCC 115
Homework 4 - Due 11 October 2005
Prof. Christina Lacey
R&D: 6-6, 6-11     P: 1-10, 5-4, 6-7, 6-8

R&D: 6-6:

Which are the terrestrial planets/ Why are they given that name?

The terrestrial planets are Mercury, Venus, Earth, and Mars. They are called terrestrial because all of these planets are similar to Earth in that they are all primarily rocky planets and all the planets share the same basic composition.

R&D: 6-11

Comets generally vaporize upon striking Earth's atmosphere. How, then, do we know their composition?

The ices that make up a comet become gaseous when the comet nears the Sun. Astronomers use spectrometers to analyze the EM emission from the atoms and molecules that make up the gases in the comets. Astronomers also study the reflected light from the dust contained in the comets. How the light is reflected tells astronomers about the size and composition of the dust.

1-10
Given the angular size of the Venus is 55 '' when the planet is 45,000,000 km from Earth, calculate Venus' diameter (in km).

                    diameter = distance $\times \frac {\mbox{angular diameter}}{57.3^\circ}$

$$D_{Venus} = d \frac{\alpha}{57.3^\circ} = 4.5 \times 10^7 km \frac{55 ''\; \frac{1'}{60 ''}\; \frac{1^\circ}{60 '}}{57.3^\circ}$$

$$D_{Venus} = 1.2 \times 10^4  km =12,000 km$$ (1)

P 5-4

A telescope (space-based) can achieve an angular resolution of 0.05'' for red light ($\lambda$= 500 nm). What is the angular resolution for (a) IR light ($\lambda$= 3.5 $\mu$m) (b) UV light ($\lambda$= 140 nm)?

(a) From the information provided by the numbers from the red light, we can find the diameter of the mirror.

$$\alpha = \mbox{angular resolution ($^{\prime\prime}$)} = 0.25 \frac{\lambda (\mu m)}{D_{mirror} (m)}$$

$$\Rightarrow D_{mirror} (m)= 0.25 \; \frac{\lambda (\mu m)}{\alpha ('')}$$

$$D_{mirror} (m) = 0.25 \; \frac{500 nm\; \frac{1\times 10^{-9}  m}{1 nm} \; \frac{1 \mu m}{1\times 10^{-6} m}}{0.05 ''} = 3.5 m$$

$$\mbox {So:}\quad \alpha ('') = 0.25 \; \frac{3.5  \mu m}{3.5 m}= 0.25 '' \quad (IR)$$

(b)

$$\alpha ('') = 0.25 \; \frac{140  nm \; \frac{1\times 10^{-9}... ...frac{1 \mu m}{1\times 10^{-6} m}}{3.5 m} = 0.01 '' \quad (UV)$$ (2)

P 6-7

A short period comet is conventionally defined as a comet having an orbital period of less than 200 years. What is the maximum possible aphelion distance for a short period comet with a perihelion of 0.5 AU? Where does this place the comet relative to the outer planets?

$$P= 200 yr ;~~~~ p^2 (yrs)=\frac{a^3 (AU)}{M(M_\odot)}$$ (3)

$$perihelion= 0.5 A.U. = a(1-e); ~~~~~ aphelion= a(1+e)$$ (4)

Need to find a, the semi-major axis, then solve for e, eccentricity to finally calculate aphelion. Comet orbits Sun, so M=1$M_\odot$.

$$p^2=a^3 ~~~\rightarrow a=p^{2/3}= (200 yrs)^{2/3}= 34.2~ A.U.$$ (5)

$$perihelion= 0.5 A.U. = a(1-e)~~~~ \rightarrow e= 1-\frac{perihelion}{a}= 1- \frac{0.5 A.U.}{34.2 A.U.}= .985$$ (6)

$$aphelion=a(1+e)= 34.2~ A.U. (1+.985)= 67.9~ A.U.$$ (7)

This comet orbit is beyond Pluto's orbit.

P 6-8

How many time has Mariner 10 orbited the Sun?

Mariner 10 was launched in Nov. 1973, so it has been 32 years. The orbital period of Mariner 10 is 176 days.

number of orbits= $32 \times \frac {365 days}{176 days}$= 66.4 times.