SCCC 115
Homework 1 - Due 11 Sept. 2003
Prof. Christina Lacey

Question 1: Q: How many miles and m are in a light year?

A: Distance traveled is: $d=v\times t$, where v= velocity and t is the time traveled. For light, $v$=speed of light=$c$ and $t=$ 1 year.

$$d = vt=ct$$

$$= 3\times 10^8 m/s \left ( {1 yr} \frac{365 days}{1 yr} \frac{24 hr}{1 day} \frac{3600 s}{1 hr}\right ) =9.5 \times 10^{15} m$$

$$1 \mbox{l.y.} = 9.5 \times 10^{15} \, m \frac{1\, km}{1 \times 10^3\, m}= 9.5 \times 10^{12} \, km$$

$$1 \mbox{l.y.} = 9.5 \times 10^{15}\, m \frac{1 mile}{1.609\, km \frac{1\, km}{1 \times 10^3\, m}}=5.9\times 10^{12} miles$$

P 1-2
a) Write the following in scientific notation:

$$1000=1 \times 10^3 \quad 0.000001=1 \times 10^{-6} \quad 1001=1.001 \times 10^3 \quad$$

$$1,000,000,000,000,000= 1 \times 10^{15} \quad 123,000 = 1.23 \times 10^{5} \quad$$

$$0.000456 = 4.56 \times 10^{-4}$$ (1)

b) Write the following in ``normal'' notation:

$$3.16 \times 10^7= 31,600,000 \quad\quad 2.998\times 10^5=299,800$$

$$6.67 \times 10^{-11}=0.0000000000667 \quad\quad 2\times 10^0 = 2$$

c: Calculate:

$$(2\times 10^3)+10^{-2} = 2,000.01 = 2.00001\times 10^3$$

$$(1.99 \times 10^{30})/(5.98\times10^{24})=3.32 \times 10^5= 332,000$$

$$(3.15 \times10^7)\times (2.998 \times10^5)= 9.47 \times10^{12}$$ (2)

P 1-6

Through how many degrees, arcminutes, or arcseconds does the moon move in:

a) 1 hour of time
The moon takes 29.5 days to revolve around the Earth once:

$$\frac{\alpha^\circ}{1 hr}=\frac{360^\circ}{29.5  days}$$

$$\alpha^\circ= \frac {360^\circ}{29.5  days} \frac{1  day}{24 hr}= 0.5^\circ /hr$$ (3)

b) 1 minute of time

$$\alpha = \frac {.5^\circ}{1 hr} \frac{1  hr}{60 min}= 0.8.3\times 10^{-3\;\circ}/min$$

$$=\frac{0.008^\circ}{min} \frac{60^\prime}{1^\circ} =0.5^\prime/min$$

c) 1 second of time

$$\alpha = \frac {0.5^\prime}{min} \frac{1  min}{60 sec}  \frac{60^{\prime\prime}}{1^\prime} =0.5^{\prime\prime}/sec$$ (4)

The moon moves its own diameter in an hour.

P 1-9
At what distance is an object if its parallax, as measured from either end of a 1000 km baseline is:
a) parallax=1$^\circ$

$$distance = baseline\times \frac{360^\circ/2 \pi}{parallax}$$

$$= 1000 km \times \frac{360^\circ/2 \pi}{1^\circ} = 1000 km \frac{57.3^\circ}{1^\circ}=5.7\times10^4 km$$

b) parallax=1$^\prime$

$$d = b \times \frac{360^\circ/2 \pi}{parallax}$$

$$= 1000 km \times \frac{360^\circ/2 \pi}{1^\prime\frac{1^\circ}{60^\prime}} = 1000 km 57.3\times 60=3.4\times10^6 km$$ (5)

c) parallax=1 $^{\prime\prime}$

$$d = b \times \frac{360^\circ/2 \pi}{parallax}$$

$$= 1000 km \times \frac{360^\circ/2 \pi} {1^{\prime\prime}\frac{1^\p... ...e\prime}}\frac{1^\circ}{60^\prime}} = 1000 km 57.3\times 3600=2.1\times10^8 km$$ (6)

P 1-11

Calculate the parallax, using Earth's diameter as a baseline, of the Sun's nearest neighbor, Proxima Centauri, which lies 4.3 l.y. from Earth.

$$baseline = b= Earth's\; diameter$$

$$= 2\times R_{Earth} = 2\times(6400  km)=1.3\times 10^4 km$$

$$\alpha = \frac{b (57.3^\circ)}{d}=\frac{1.3\times 10^4 km(57.3^\circ)}{4.3  ly \frac{9.5\times10^{12} km}{1 ly}}$$

$$= 1.8 \times 10^{-8} ^\circ$$

$$\alpha = 1.8 \times 10^{-8} ^\circ  \frac{60^\prime}{1^\circ} \frac{60^{\prime\prime}}{1^\prime} = 6.6 \times 10^{-5} ^{\prime\prime}$$ (7)

P 1-13

The moon lies roughly 384,000 km from Earth and the Sun lies about 150,000,000 km away. If both have the same angular size as seen from Earth, how many times larger is the Sun than the moon?
Let $D$=diameter and $d=$distance:

$$D=d\frac{\alpha}{57.3^\circ}$$ (8)

So:

$$D_{sun} = d_{sun}\frac{\alpha_{sun}}{57.3^\circ}$$ (9)

$$D_{moon} = d_{moon}\frac{\alpha_{moon}}{57.3^\circ}$$ (10)

We can divide Eq 13 by Eq 14:

$$\frac{D_{sun}}{D_{moon}} = \frac{d_{sun}\frac{\alpha_{sun}} {57.3^\circ}}{d_{moon}\frac{\alpha_{moon}}{57.3^\circ}}$$

Remember that $\alpha_{sun}=\alpha_{moon}$:

$$\frac{D_{sun}}{D_{moon}} = \frac{d_{sun}}{d_{moon}} = \frac{150,000,000 km}{384,000 km}=395$$ (11)

Thus, the Sun is 395 times larger than the moon in diameter.

P 2-9
Jupiter's moon Callisto orbits Jupiter at a distance of 1.88 million km ( $=1.88\times10^6$ km). Callisto's orbital period about Jupiter is 16.7 days (=P). (Assume Callisto's mass is negligible compared to Jupiter). What is the mass of Jupiter?

From Eq. 2.7 in textbook:

$$P(yrs)^2 = \frac{a (AU)^3}{M_{total} (M_\odot)}$$

$$M_{total} (M_\odot) = \frac{a  (AU)^3}{P (yrs)^2}$$ (12)

One can assume that Callisto's orbit is roughly circular, so:

$$a = 1.88\times10^6 km \; \frac{1 AU}{1.5\times10^8  km}=1.3\times10^{-2} AU$$

$$p = 16.7  days \frac{1 yr}{365 days}=4.6 \times 10^{-2} yr$$

$$M_{total} = M_{Callisto}+M_{Jupiter} \simeq M_{Jupiter}$$ (13)

$$M_{total} (M_\odot) = \frac{a  (AU)^3}{P (yrs)^2}$$

$$= \frac{(1.3 \times 10^{-2})^3}{(4.6 \times 10^2)^2} =\frac{2.2 \times 10^{-6}}{2.1 \times 10^{-3}}$$

$$= 1.0 \times 10^{-3} M_\odot=1.0 \times 10^{-3} M_\odot \frac{2\times 10^{30} kg}{1 M_\odot}$$

$$= 2\times 10^{27} \mbox {kg}$$ (14)

P 2-15
The Moon's mass is $M_{moon}=7.4\times 10^{22}$ kg and the radius of the Moon is $R_{moon}=1700$ km. What is the speed of a spacecraft moving in a circular orbit just above the lunar surface? What is the escape speed?

$$G = 6.67\times 10^{-11} \mbox{N m$^2$/s$^2$}$$

$$\mbox{ 1 N (Newton)= kg m/s$^2$}$$

$$v_{circ} = \sqrt{\frac{GM}{R}}=\sqrt{\frac{GM_{moon}}{R_{moon}}}$$

$$= \sqrt{\frac{6.67\times 10^{-11} \mbox{N m$^2$/kg$^2$}\;\; 7.4\times 10^{22} kg}{1700 km \frac{1000  m}{1 km}}}$$

$$= \sqrt{\frac{6.67\times 10^{-11} \frac{ kg m}{s^2}\frac{m^2}{ kg^2 }\;\; 7.4\times 10^{22}  kg}{1.7\times10^6 m}}$$

$$= \sqrt{2.9\times 10^6 \frac{m^2}{s^2}}$$

$$v_{circ} = 1.7\times 10^3 \frac {m}{s}$$

$$v_{escape} = \sqrt{\frac{2GM}{R}}=\sqrt{2} v_{circ}$$

$$= 2.4\times 10^3 \frac{m}{s}$$ (15)