SCCC 115
Homework 5 - Due 20 Oct. 2005
Prof. Christina Lacey
R&D: 7-7, 7-19, 8-16, 8-19     P: 6-12,6-15, 7-2, 9-6, 8-7

R&D: 7-7:

What clue does the differentiation of the Earth provide to our planet's history?

Earth is observed to be differentiated with higher density material toward the core. Thus at one time Earth must have been molten in order for the higher density material to sink toward the core.

R&D: 7-19:

What if Earth had no moon. Do you think we would know anything about tidal forces?

Yes, we would still feel the tidal effects from the Sun.

For example:

$$F_{tidal}=\frac {GMmR}{r^3}$$ (1)

in the case of the Moon exerting tidal forces on the Earth, M is the mass of the Moon, m is the mass of the body (for example the mass of a person on the surface of the Earth where one is calculating tidal effects, R is the radius of the Earth, and r is the distance between the centers of mass of the two bodies, M and m.

So the tidal forces of the moon and the Sun on the center of the Earth are proportional to (the test m and R are the same for both situations):

$$F_{moon} \propto \frac {M_{moon}}{r^3} = \frac {7.35 \times 10^{22} kg}{(3.84 \times 10^5 km )^3}= 1.3 \times 10^6  kg/m^3$$

$$F_{Sun} \propto \frac {M_\odot}{r^3}= \frac {2\times 10^{30} kg}{(1.5 \times 10^8 km)^3}= 6.0 \times 10^5  kg/m^3$$

$$\frac{F_{moon}}{F_{Sun}} = {1.3 \times 10^6  kg/m^3}{6.0 \times 10^5  kg/m^3 }= 2$$

So the Moon's tidal force on the Earth is twice as strong as the Sun's tidal force.

R&D: 8-16:

Because the Moon always keeps one face toward the Earth, an observer on the moon's near side would see Earth appear almost stationary in the lunar sky. Still, the Earth would change its appearance as the moon orbited Earth. How would Earth's appearance change?

Earth's appearance changes due to its 24 hour rotation. Also, the Earth would appear to move North or South in the lunar sky over the several days as the orbital inclination of the Moon to Earth changes. The moon's orbit is eccentric, so at times the Earth is closer and therefore bigger in the sky.

R&D: 8-19:

Explain why Mercury is never seen overhead at midnight in the Earth's sky.

Mercury's orbit is interior to Earth's orbit, so it must appear close to the Sun in the sky. The farthest Mercury can be from the Sun in the sky is 28$^\circ$ (maximum elongation). Because of this we only see Mercury just after sunset or just before sunrise.

P 6-10

The asteroid Icarus has a perihelion distance of 0.2 AU and an eccentricity of 0.7. What is the aphelion distance from the Sun?

Remember that:

$$perihelion = a(1-e)=0.2 AU$$

$$aphelion = a(1+e)=?$$ (2)

Solve the perihelion equation for $a$, the semi-major axis:

$$a(1-e) = 0.2 AU$$

$$a = \frac{0.2 AU}{1-e}$$ (3)

Substitute this result for a into the equation for aphelion, then put in e=0.7:

$$aphelion = \frac{0.2 AU}{1-e}(1+e)=\frac{0.2 AU}{1-0.7}(1+0.7) = \frac{0.2 AU}{0.3}{1.7}= 1.1 AU$$ (4)

P 6-12
How long would it take for a radio signal to complete the round trip between Earth and Saturn? Assume Saturn is at its closest point to Earth. How far would a spacecraft orbiting the planet in a circular orbit with a radius of 100,000 km about Saturn travel in that time?

Saturn is closest to Earth when Earth is at aphelion and Saturn is at perihelion. From Table 6.1, Earth: a= 1.0 AU, e = 0.017; Saturn: a= 9.5 AU and e= 0.054. Both Earth and Saturn have very low eccentricities, thus one can assume circular orbits for both planets.

So the closest distance between Earth and Saturn is:

9.5 AU - 1.0 AU = 8.5 AU $\frac{1.5 \times 10^8 km}{1 AU}$= 1.3 $\times 10^9$ km

So a round trip to Saturn is 2 $\times (1.3\times 10^9$ km), because signal travels to Saturn, then returns. The speed of the radio signal is c, the speed of light.

time = $\frac{\mbox{distance}}{\mbox{velocity}}= \frac{ 2 \times (1.3\times 10^9  km)}{3 \times 10^5 km}= 8.7 \times 10^3 s= 144  min=2.4  hrs$

Now for the spacecraft (assume satellite's mass is negligible to Saturn's mass.

distance traveled= d= velocity $\times$ time =vt

$$v = \sqrt {\frac{GM}{r}}$$

$$\mbox{so} \quad \quad d = vt=\sqrt {\frac{GM_{Saturn}}{r}}\; t=\sqrt {\frac{6.67\times 10^{... ....7 \times 10 ^{26} kg} {100,000 kg \frac{100 m}{1 km}}}\; (8.9 \times 10^3 s)$$

$$d = \sqrt{3.8 \times 10^8  \frac{kg \frac{m}{s^2} m}{kg}} (8.7 \times 10^3 s)$$

$$= 1.9 \times 10^4 \frac{m}{s}(8.7 \times 10^3 s)=1.7 \times 10^8 m = 1.7 \times 10^5 km$$ (5)

So Mission Control cannot maneuver spacecraft in real time because it would take over two hours for a command to be sent and returned to Earth, far to slow for maneuvering a high-speed spacecraft.

P 6-15

Consider a planet growing by the accretion of material from the solar nebula. As the planet grows, its density remains roughly constant. Does the force of gravity at the surface of the planet increase, decrease, or stay the same? Specifically, what would happen to the surface gravity and escape speed as the radius of the planet doubled? Give reasons for your answer.

If density remains constant, then :

$$M={\rho}{4/3 \pi r^3}  \rightarrow  M\propto {\rho}{r^3}$$ (6)

So if density is constant and the radius doubles, then mass increases by a factor of $2^3 = 8$.

So the escape speed becomes when you substitute in M $\propto (2r)^3= 8 r^3$:

$$v=\sqrt{\frac{2GM}{r}}\propto \sqrt{\frac{\rho r^3}{r}} \propto \sqrt{\frac {8 r^3}{2r}}= \sqrt{4 \times r^2} = 2r$$ (7)

So the escape speed increases.

$$a=\frac{v^2}{r} \propto \frac{(2r)^2}{2r} =2r$$ (8)

So again, the acceleration due to gravity, which is also known as the surface gravity, increases as r increases.

P 7-2
What would earth's surface gravity and escape speed be if the entire planet had a density equal to that of the crust, say, $\rho=$ 3,000 kg/m$^3$?

$$a = \frac{v^2}{r} \quad \quad \mbox{and} \quad v=\sqrt{\frac{GM}{r}}$$

$$a = \frac{v^2}{r}= \frac{\frac{GM}{r}}{r} = \frac{G M_\oplus}{r^2_\oplus}$$

$$M_\oplus = \rho V = \mbox{density $\times$ Volume}= \rho \frac{4}{3} \pi r^3$$

$$M_\oplus = 3,000 \frac{kg}{m^3} \frac{4}{3} \pi (6.4 \times 10^3 m )^3= 3.3 \times 10^{24} \frac{kg}{m^3}m^3=3.3 \times 10^{24} kg$$

$$a = \frac{G M_\oplus}{r^2_\oplus} = \frac{6.67 \times 10^{-11}  \frac... ...ac{kg  m^2}{s^2}\frac {m^2}{kg^2} 3.3 \times 10^{24} kg}{(6.4 \times 10^6 m)^2}$$

$$\mbox {surface gravity} = a= 5.4 \frac{m}{s^2}$$ (9)

The escape speed is:

$$v_{escape} = \sqrt\frac{2GM_\oplus}{r_\oplus}= \sqrt{\frac{2\times 6.67 \times 10^{-11}  \frac{Nm^2}{kg^2} \; 3.3 \times 10^{24} kg}{6.4 \times 10^6 m}}$$

$$= \sqrt{\frac{2\times 6.67 \times 10^{-11}  \frac{\frac{kg m}{s^2} m^2}{kg^2} \; 3.3 \times 10^{24} kg}{6.4 \times 10^6 m}}$$

$$= \sqrt{6.9 \times 10^7 \frac{m^2}{s^2}}= 8.3 \times 10^3  m/s= 8.3  km/s$$

The surface gravity and escape velocity are less than normal, because the density and therefore the mass are less.

P 8-6
The angular resolution of the Hubble telescope is $\alpha=0.05^{\prime \prime}$. What is the smallest feature (smallest diameter) one can see on the moon and Mercury with the Hubble.

distance to moon= d = 380,000 km
distance from Earth to Mercury= d = 1 AU- .47 AU = .53 AU = 8.0 $\times 10^7$ km = 380,000 km

Diameter=$\alpha$ distance $/57.3^\circ$

Moon:

$$D = \alpha  d/57.3^\circ = 0.05 '' \frac {1^\circ}{3600 ''} 3.8 \times 10^8  m /57.3^\circ = 93 m$$ (10)

Mercury:

$$D = \alpha  d/57.3^\circ = 0.05 '' \frac {1^\circ}{3600 ''} 8 \times 10^{10}  m /57.3^\circ = 19,000 m = 19 km$$ (11)

P 8-7
What was the orbital period of the Apollo 11 command module, orbiting 10 km above the lunar surface.

distance between the center of the moon and the Apollo 11 = $R_{moon}+ 10$ km
distance= r=1738 km+10 km = 1748 km

$$v_{Apollo} = \sqrt{\frac{6.67 \times 10^{-11}  \frac{kg m}{s^2} \frac{m^2}{kg^2}7.35 \times 10^{22} kg }{1748 km \frac {1000 m}{1 km}}}$$

$$= \sqrt{2.8 \times 10^6 \frac{m^2}{s^2}}=1.7\times 10^3 m/s =1.7  km/s$$ (12)

Period of the the Apollo spacecraft about the Moon:

$$P=\frac{\mbox{circumference}}{\mbox{velocity}} = \frac{2 \pi... ...\frac{2 \pi (1748 km)}{1.7  km/s}=6.5 \times 10^3  s= 1.8  hrs$$ (13)

P 9-6

See fig. 9.4.