Solar System Homework #1 Answers


Solar System Homework #1 Answers

1. a) Look up the mass of the Moon and the radius of the Moon in the book, and compute the density of the Moon, in gm/cm³.

Part 1: Data

Looking on page 178, we see that the mass of the Moon is 7.35 X 10²² kilograms, and the radius of the Moon is 1738 kilometers.

Part 2: Equation

We must use the density formula: density = mass / volume

Part 3: Unit Conversion

Nothing is in the right units! We must convert the mass into grams:

M_Moon = 7.35X 10²² kg X 1000 gm / kg = 7.35 X 10²⁵ grams

We must convert the radius into centimeters:

R_Moon = 1738 km X (1000m / km) X (100cm / m) = 1.738 X 10⁸ cm

Part 4: Computation

So V = (4/3)(pi)R³ = (4/3)(3.14)(1.738 X 10⁸ cm)³

V = 2.20 X 10²⁵ cm³

And Density = M / V = 7.35 X 10²⁵ gm / 2.20 X 10²⁵ cm³

density = 3.34 gm/cm³

Part 5: The Answer

The density of the Moon in 3.34 grams per cubic centimeter, in good agreement with the value in the book.

b) Based on that average density, would you say the moon is made mostly of ice (density = 0.9 gm/cm³), rock (density = 3.0 gm/cm³), or iron (density = 9.0 gm/cm³)?

The average density of the Moon is closest to the value for the density of rock. Thus, it is possible for us to come to the conclusion that the Moon is made mostly of rock, with little water or iron.

2. In late August of 2003, Mars was at its closest point to Earth in about 60,000 years! This was due to the coincidence that Mars and Earth lined up with the Sun when the Earth was close to its farthest point from the Sun (called aphelion) and Mars was close to its closest approach to the Sun (called aphelion).

a) When Mars is at perihelion and Earth is at aphelion, how far apart are the two planets, in kilometers? How many miles is that, if there are approximately 1.6 kilometers in a mile?

ANSWER: Mars is at a distance of 206,600,000 km from the Sun at perihelion (Chapter 10, p. 251). At aphelion, Earth is at a distance of 152,100,000 km from the Sun. This is a difference of 54,500,000 km, or 34,000,000 miles.

b) What is the angular diameter of Mars when the Earth and Mars are separated by this distance, in seconds of arc? This is about how large Mars appeared to be in August 2003.

Part 1: Data

Actual diameter of Mars = 3394 km x 2 = 6788 kjm

Distance at closest approach = 54,500,000 km

Part 2: Equation

We use the formula for Angular Diameter:

Angular Diameter = 206265 (Actual Diameter / Distance)

Part 3: Unit Conversion

Both Actual Size and Distance are in kilometers. No unit conversions are needed.

Part 4: Computation

Angular Diameter = 206265 (6788 km / 54,500,000 km)

Angular Diameter = 25.7 seconds of arc

Part 5: The Answer

So Mars has an angular diameter of 25.7 seconds of arc when it it at its closest to Earth

c) Mars is about 200 million kilometers from Earth on average. What is the average angular diameter of Mars? How many times smaller is this than the August 2003 figure?

Part 1: Data

Actual diameter of Mars = 3394 km x 2 = 6788 kjm

Average distance = 200,000,000 km

Part 2: Equation

We again use the formula for Angular Diameter:

Angular Diameter = 206265 (Actual Diameter / Distance)

Part 3: Unit Conversion

Both Actual Size and Distance are in kilometers. No unit conversions are needed.

Part 4: Computation

Angular Diameter = 206265 (6788 km / 200,000,000 km)

Angular Diameter = 7 seconds of arc

Part 5: The Answer

So Mars has an angular diameter of only 7 seconds of arc at its average distance from Earth. This is 3.67 times smaller than its angular size at closest approach.

3. a) Jupiter is about 5 times farther away from the Sun than the Earth is, and has about 300 times the mass of the Earth. Compare the gravitational force between Jupiter and the Sun to the gravitational force between the Earth and the Sun.

Part 1: Data

Masses: M_Jup = 300 M_Ear

distances from the Sun: R_{Sun Jup} = 5 R_{Sun
Ear}

Part 2: Equation

We want to compare the gravitational force exerted by the Sun on Jupiter to the gravitational force exerted by the Sun on Earth. Since when we compare we divide, we want to compute

F_{Sun Jup} / F_{Sun Ear}

First, we write out the relevant formulas

F_{Sun Jup} = G M_Sun M_Jup / (R_{Sun Jup})²

F_{Sun Ear} = G M_Sun M_Ear / (R_{Sun Ear})²

Now we perform the divisions. See the gravity handout for details:

F_{Sun Jup} / F_{Sun Ear} = G M_Sun M_Jup (R_{Sun
Ear})² / G M_Sun M_Ear (R_{Sun Jup})²

Note how, happily, both G and the mass of the Sun drop out! We are left with

F_{Sun Jup} / F_{Sun Ear} = M_Jup (R_{Sun Ear})² / M_Ear (R_{Sun Jup})²

Part 3: Unit Conversion

No unit conversions are needed!

Part 4: Computation

We substitute from the data above. M_Jup becomes 300 M_Ear and R_{Sun Jup} becomes 5 R_{Sun Ear}

F_{Sun Jup} / F_{Sun Ear} = (300 M_Ear) (R_{Sun Ear})² / M_Ear (5 R_{Sun Ear})²

Do the squaring first:

F_{Sun Jup} / F_{Sun Ear} = (300) (M_Ear) (R_{Sun Ear})² / M_Ear (25) (R_{Sun Ear})²

Note that the 5 is squared to 25! The mass of the Earth and the distance between the Earth and Sun both drop out and we are left with just numbers:

F_{Sun Jup} / F_{Sun Ear} = (300) / (25) = 12

We get rid of the fraction by multiplying both sides by F_{Sun Ear}:

F_{Sun Jup} = 12 F_{Sun Ear}

Part 5: The Answer

The gravitational force between the Sun and Jupiter is 12 times greater than the gravitational force between the Sun and Earth. Earth may be closer to the Sun than Jupiter, but Jupiter's mass more than makes up for it's severe distance.

b) Perform the same computation for Saturn which is 10 times farther away from the Sun than Earth, and 100 times more massive than the Earth.

Part 1: Data

Masses: M_Sat = 100 M_Ear

distances from the Sun: R_{Sun Sat} = 10 R_{Sun Ear}

Part 2: Equation

We want to compare the gravitational force exerted by the Sun on Saturn to the gravitational force exerted by the Sun on Earth. Since when we compare we divide, we want to compute

F_{Sun Sat} / F_{Sun Ear}

First, we write out the relevant formulas

F_{Sun Sat} = G M_Sun M_Sat / (R_{Sun Sat})²

F_{Sun Ear} = G M_Sun M_Ear / (R_{Sun Ear})²

Now we perform the divisions. See the gravity handout for details:

F_{Sun Sat} / F_{Sun Ear} = G M_Sun M_Sat (R_{Sun
Ear})² / G M_Sun M_Ear (R_{Sun Sat})²

Note how, happily, both G and the mass of the Sun drop out! We are left with

F_{Sun Sat} / F_{Sun Ear} = M_Sat (R_{Sun Ear})² / M_Ear (R_{Sun
Sat})²

Part 3: Unit Conversion

No unit conversions are needed!

Part 4: Computation

We substitute from the data above. M_Sat becomes 100 M_Ear and R_{Sun Sat} becomes 10 R_{Sun Ear}

F_{Sun Sat} / F_{Sun Ear} = (100 M_Ear) (R_{Sun Ear})² / M_Ear (10 R_{Sun Ear})²

Do the squaring first:

F_{Sun Sat} / F_{Sun Ear} = (100) (M_Ear) (R_{Sun Ear})² / M_Ear (100) (R_{Sun Ear})²

Note that the 10 is squared to 100! The mass of the Earth and the distance between the Earth and Sun both drop out and we are left with just numbers:

F_{Sun Sat} / F_{Sun Ear} = (100) / (100) = 1

We get rid of the fraction by multiplying both sides by F_{Sun Ear}:

F_{Sun Sat} = 1 X F_{Sun
Ear}

Part 5: The Answer

Amazingly, The gravitational force between the Sun and Saturn is the same as than the gravitational force between the Sun and Earth. Earth may be closer to the Sun than Saturn, but Saturn's mass exactly makes up for it's severe distance.

Updated 8/16/99

By James E. Heath