Lecture 2 Solutions: Math Survival Kit
Practice Problem Solutions
Student note: These are model solutions written to show every step, units, and checks. I do not expect your work to be this detailed; shorter solutions are fine if your setup, units, and reasoning are correct.
HW note: Homework uses only a subset of these problems; the full set is included for future study and self checking.
Scientific Notation & Prefixes
Problem 1: Express in Scientific Notation
(a) 4,500,000,000 years (age of the solar system)
Restatement: Write \(4{,}500{,}000{,}000\) in scientific notation.
Given: \(4{,}500{,}000{,}000\) years
Find: Scientific notation form
Equation: \(a \times 10^n\) with \(1 \le a < 10\)
Solution: \[4{,}500{,}000{,}000 = 4.5 \times 10^9 \text{ years}\]
Unit check: Units remain years.
Sanity check: Moving the decimal 9 places to the left gives \(10^9\), which matches the scale.
Answer: \(4.5 \times 10^9\) years.
(b) 0.00000000656 m (wavelength of H-alpha)
Restatement: Write \(0.00000000656\) m in scientific notation and convert to nm.
Given: \(0.00000000656\) m
Find: Scientific notation in meters and in nm
Equation: \(1 \text{ nm} = 10^{-9} \text{ m}\)
Solution: \[0.00000000656 \text{ m} = 6.56 \times 10^{-9} \text{ m}\]
Convert to nm: \[6.56 \times 10^{-9} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}} = 6.56 \text{ nm}\]
Unit check: m converts to nm correctly.
Sanity check: The number is extremely small, so a nanometer-scale value is reasonable.
Answer: \(6.56 \times 10^{-9}\) m = 6.56 nm.
Note: The H-alpha line is 656 nm, which corresponds to \(6.56 \times 10^{-7}\) m. The given number here is \(100\) times smaller.
Problem 2: Calculate
(a) \((2 \times 10^{30})(3 \times 10^8)\)
Restatement: Multiply two numbers in scientific notation.
Given: \(2 \times 10^{30}\) and \(3 \times 10^8\)
Find: Product in proper scientific notation
Equation: \((a \times 10^m)(b \times 10^n) = (ab) \times 10^{m+n}\)
Solution: \[2 \times 3 = 6\] \[10^{30} \times 10^8 = 10^{38}\] \[\Rightarrow 6 \times 10^{38}\]
Unit check: Not applicable (pure numbers).
Sanity check: Multiplying two large powers of ten should give a larger power of ten.
Answer: \(6 \times 10^{38}\).
(b) \(\frac{9 \times 10^{15}}{3 \times 10^7}\)
Restatement: Divide two numbers in scientific notation.
Given: \(9 \times 10^{15}\) and \(3 \times 10^7\)
Find: Quotient in proper scientific notation
Equation: \((a \times 10^m)/(b \times 10^n) = (a/b) \times 10^{m-n}\)
Solution: \[\frac{9}{3} = 3\] \[\frac{10^{15}}{10^7} = 10^{15-7} = 10^8\] \[\Rightarrow 3 \times 10^8\]
Unit check: Not applicable (pure numbers).
Sanity check: Dividing by a smaller power of ten should reduce the exponent.
Answer: \(3 \times 10^8\).
(c) \((4 \times 10^3)^2\)
Restatement: Square a number in scientific notation.
Given: \(4 \times 10^3\)
Find: \((4 \times 10^3)^2\) in proper scientific notation
Equation: \((a \times 10^n)^2 = a^2 \times 10^{2n}\)
Solution: \[4^2 = 16\] \[(10^3)^2 = 10^6\] \[16 \times 10^6 = 1.6 \times 10^7\]
Unit check: Not applicable (pure numbers).
Sanity check: Squaring should make the number larger; \(10^6\) is larger than \(10^3\).
Answer: \(1.6 \times 10^7\).
Problem 3: SI Prefixes
A radio telescope observes at 1.4 GHz. Express in Hz.
Restatement: Convert 1.4 GHz to Hz.
Given: \(1.4\) GHz
Find: Frequency in Hz
Equation: \(1 \text{ GHz} = 10^9 \text{ Hz}\)
Solution: \[1.4 \text{ GHz} = 1.4 \times 10^9 \text{ Hz}\]
Unit check: GHz converts to Hz correctly.
Sanity check: GHz is a large radio frequency; \(10^9\) Hz is the right scale.
Answer: \(1.4 \times 10^9\) Hz.
Unit Conversions
Problem 4: Speed of Light
Convert \(c = 3 \times 10^8\) m/s to km/hr.
Restatement: Convert a speed from m/s to km/hr.
Given: \(c = 3 \times 10^8\) m/s
Find: \(c\) in km/hr
Equation: Use factor-label conversion
Solution: \[3 \times 10^8 \frac{\text{m}}{\text{s}} \times \frac{1 \text{ km}}{1000 \text{ m}} \times \frac{3600 \text{ s}}{1 \text{ hr}}\] \[= 3 \times 10^8 \times \frac{3600}{1000} \frac{\text{km}}{\text{hr}} = 3 \times 10^8 \times 3.6 \frac{\text{km}}{\text{hr}}\] \[= 10.8 \times 10^8 \frac{\text{km}}{\text{hr}} = 1.08 \times 10^9 \frac{\text{km}}{\text{hr}}\]
Unit check: m cancels m and s cancels s, leaving km/hr.
Sanity check: The value should be about a billion km/hr; this matches.
Answer: \(1.08 \times 10^9\) km/hr.
Problem 5: Wavelength Conversion
Red light has wavelength 700 nm. Express in:
(a) meters
Restatement: Convert 700 nm to meters.
Given: \(700\) nm
Find: Wavelength in meters
Equation: \(1 \text{ nm} = 10^{-9} \text{ m}\)
Solution: \[700 \text{ nm} \times \frac{10^{-9} \text{ m}}{1 \text{ nm}} = 700 \times 10^{-9} \text{ m} = 7.0 \times 10^{-7} \text{ m}\]
Unit check: nm cancels, leaving meters.
Sanity check: Visible light is around \(10^{-7}\) m, so this is correct.
Answer: \(7.0 \times 10^{-7}\) m.
(b) micrometers
Restatement: Convert 700 nm to micrometers.
Given: \(700\) nm
Find: Wavelength in micrometers
Equation: \(1 \mu\text{m} = 10^{-6} \text{ m} = 1000 \text{ nm}\)
Solution: \[700 \text{ nm} \times \frac{1 \mu\text{m}}{1000 \text{ nm}} = 0.70 \mu\text{m}\]
Unit check: nm cancels, leaving micrometers.
Sanity check: \(0.7 \mu\text{m}\) is in the red part of visible light, so this matches.
Answer: \(0.70 \mu\text{m}\).
Problem 6: Distance to Proxima Centauri
Proxima Centauri is 4.2 light-years away.
(a) Convert to parsecs
Restatement: Convert 4.2 ly to pc.
Given: \(4.2\) ly
Find: Distance in pc
Equation: \(1 \text{ pc} = 3.26 \text{ ly}\)
Solution: \[4.2 \text{ ly} \times \frac{1 \text{ pc}}{3.26 \text{ ly}} = \frac{4.2}{3.26} \text{ pc} \approx 1.29 \text{ pc}\]
Unit check: ly cancels, leaving pc.
Sanity check: A few light-years is about one parsec, so 1.29 pc is reasonable.
Answer: \(\approx 1.29\) pc.
(b) Convert to meters
Restatement: Convert 4.2 ly to meters.
Given: \(4.2\) ly
Find: Distance in meters
Equation: \(1 \text{ ly} = 9.46 \times 10^{15} \text{ m}\)
Solution: \[4.2 \text{ ly} \times \frac{9.46 \times 10^{15} \text{ m}}{1 \text{ ly}} = 4.2 \times 9.46 \times 10^{15} \text{ m}\] \[= 39.7 \times 10^{15} \text{ m} = 3.97 \times 10^{16} \text{ m}\]
Unit check: ly cancels, leaving meters.
Sanity check: Stellar distances are around \(10^{16}\) m, so this fits.
Answer: \(\approx 4.0 \times 10^{16}\) m.
Ratio Method
Problem 7: Telescope Comparison
30-meter telescope vs. Hubble (2.4 m). How much more light?
Restatement: Compare light-collecting areas using diameter ratio.
Given: - \(D_1 = 30\) m - \(D_2 = 2.4\) m
Find: \(A_1/A_2\)
Equation: \(A \propto D^2\)
Solution: \[\frac{A_1}{A_2} = \left(\frac{D_1}{D_2}\right)^2 = \left(\frac{30}{2.4}\right)^2 = (12.5)^2 = 156.25\]
Unit check: Ratio is dimensionless.
Sanity check: A much larger diameter should collect far more light; a factor of about \(10^2\) makes sense.
Answer: About \(1.56 \times 10^2\), or 156 times more light.
Problem 8: Planet Volumes
Jupiter’s radius is 11 times Earth’s. How many Earths fit inside?
Restatement: Use volume scaling to compare sizes.
Given: \(R_J/R_E = 11\)
Find: \(V_J/V_E\)
Equation: \(V \propto R^3\)
Solution: \[\frac{V_J}{V_E} = \left(\frac{R_J}{R_E}\right)^3 = (11)^3 = 1331\]
Unit check: Ratio is dimensionless.
Sanity check: Volume grows quickly with radius; 1331 is reasonable.
Answer: About 1330 Earths.
Problem 9: Inverse-Square Brightness
Star X appears 16 times fainter than identical Star Y. How much farther is Star X?
Restatement: Use inverse-square scaling to relate brightness and distance.
Given: \(B_X = B_Y / 16\)
Find: \(d_X/d_Y\)
Equation: \(B \propto 1/d^2\) so \(B_X/B_Y = (d_Y/d_X)^2\)
Solution: \[\frac{1}{16} = \left(\frac{d_Y}{d_X}\right)^2\] \[\frac{d_Y}{d_X} = \frac{1}{4}\] \[\frac{d_X}{d_Y} = 4\]
Unit check: Ratio is dimensionless.
Sanity check: Farther should be dimmer; 4 times farther gives 16 times fainter.
Answer: Star X is 4 times farther away.
Rate Problems
Problem 10: Light Travel Time to Neptune
Neptune is 30 AU from the Sun. How long for light to travel?
Restatement: Compute light travel time for 30 AU.
Given: - \(d = 30\) AU - \(1 \text{ AU} = 1.50 \times 10^{11}\) m - \(c = 3 \times 10^8\) m/s
Find: \(t\) in hours
Equation: \(t = d/c\)
Solution: \[d = 30 \times 1.50 \times 10^{11} \text{ m} = 4.5 \times 10^{12} \text{ m}\] \[t = \frac{4.5 \times 10^{12} \text{ m}}{3 \times 10^8 \text{ m/s}} = 1.5 \times 10^4 \text{ s}\]
Convert to hours: \[1.5 \times 10^4 \text{ s} \times \frac{1 \text{ hr}}{3600 \text{ s}} \approx 4.2 \text{ hr}\]
Unit check: m divided by (m/s) gives seconds, then converts to hours.
Sanity check: Light takes about 8.3 minutes per AU; \(30 \times 8.3 \text{ min} \approx 250 \text{ min} \approx 4.2\) hr.
Answer: About 4.2 hours.
Problem 11: Earth’s Orbital Distance per Day
Earth moves at 30 km/s. How far in one day?
Restatement: Compute distance traveled in one day and express in AU.
Given: - \(v = 30\) km/s - \(t = 1\) day - \(1 \text{ day} = 86{,}400\) s - \(1 \text{ AU} = 1.50 \times 10^8\) km
Find: \(d\) in AU
Equation: \(d = v t\)
Solution: \[d = 30 \frac{\text{km}}{\text{s}} \times 86{,}400 \text{ s} = 2.592 \times 10^6 \text{ km}\] \[d = 2.592 \times 10^6 \text{ km} \times \frac{1 \text{ AU}}{1.50 \times 10^8 \text{ km}} = 1.728 \times 10^{-2} \text{ AU}\]
Unit check: km cancels, leaving AU.
Sanity check: \(2\pi\) AU per year is about \(6.3\) AU; \(6.3/365 \approx 0.017\) AU/day, which matches.
Answer: \(\approx 0.017\) AU per day.
Problem 12: Signal Delay to Mars
Mars is 1.5 AU from Earth. How long for a radio signal?
Restatement: Compute the light-travel time for 1.5 AU.
Given: - \(d = 1.5\) AU - \(1 \text{ AU} = 1.50 \times 10^{11}\) m - \(c = 3 \times 10^8\) m/s
Find: \(t\) in minutes
Equation: \(t = d/c\)
Solution: \[d = 1.5 \times 1.50 \times 10^{11} \text{ m} = 2.25 \times 10^{11} \text{ m}\] \[t = \frac{2.25 \times 10^{11} \text{ m}}{3 \times 10^8 \text{ m/s}} = 7.5 \times 10^2 \text{ s}\] \[t = 7.5 \times 10^2 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}} = 12.5 \text{ min}\]
Unit check: m divided by (m/s) gives seconds, then converts to minutes.
Sanity check: 1 AU is about 8.3 minutes, so 1.5 AU should be about 12.5 minutes.
Answer: About 12.5 minutes one way.
Synthesis
Problem 13: Betelgeuse Supernova
Restatement: A supernova occurred at Betelgeuse in 1300 CE, 700 light-years away. When do we see it, and what is the lookback time?
Key elements a full answer should include: - Light travel time equals distance in light-years - The observed date is 700 years after the event - Lookback time is 700 years
Sample response: “If the explosion occurred in 1300 CE and the star is 700 light-years away, the light arrives 700 years later. We would see it around 2000 CE. The lookback time is 700 years, because that is how long the light traveled.”
Grading guidance: Full credit requires the correct date and the correct definition of lookback time; partial credit if one part is missing.
Problem 14: Dimensional Check
Restatement: Determine whether \(v = d \cdot t\) is dimensionally correct for velocity.
Key insight: Velocity has dimensions \([L][T^{-1}]\); multiplying distance and time gives \([L][T]\).
Answer: The equation is not dimensionally correct. Velocity should be \(v = d/t\), not \(v = d \cdot t\).
Common misconception: Treating any combination of distance and time as a velocity without checking units.
Problem 15: Why Ratios?
Restatement: Explain why the ratio of the Sun’s mass to Earth’s is more useful than the difference.
Key insight: Ratios reveal scale; differences between huge numbers are not informative.
Answer: The difference is basically the Sun’s mass again, which tells you little. The ratio tells you the Sun is about \(3.3 \times 10^5\) times more massive, which is meaningful and scale-independent.
Common misconception: Thinking subtraction is the natural way to compare sizes at cosmic scales.
Problem 16: Sanity Check
Restatement: A student gets \(10^9\) m for a nearby star distance. Explain why this is wrong and give the right order of magnitude.
Key insight: Stellar distances are around \(10^{16}\) m; \(10^9\) m is a solar-scale number.
Answer: \(10^9\) m is \(10^6\) km, which is about \(1.4\) times the Sun’s radius. That is far too small for any stellar distance. Nearby stars are around a few light-years away, or roughly \(10^{16}\) m.
Common misconception: Confusing solar system scales with interstellar scales or dropping powers of ten in conversions.