Homework 2 Solutions

Gravity & Orbits II + Light as Information

Solutions to Homework 2.

Author

Dr. Anna Rosen

Published

February 11, 2026

Note

Student note: These are model solutions written to show reasoning, units, and checks. Your solutions can be shorter if your setup and logic are correct.

Grade memo note: Use these to write your reflection (what you got right, what broke, and what you will do differently next time).

Part A — Gravity & Orbits

Problem 1 — Energy Signs and Orbital Fate

Restatement: Interpret \(E < 0\), \(E = 0\), and \(E > 0\) for orbital energy.

Key insight: Zero potential energy is defined at infinity, so bound orbits have negative total energy.

Answer:

(a) Why is gravitational potential energy negative?

We define \(U = 0\) at \(r = \infty\) (infinite separation). To bring two masses together from infinity, gravity does positive work—meaning the system releases energy. Therefore \(U\) is lower (more negative) at finite separation:

\[U = -\frac{GMm}{r} < 0\]

Physically: negative \(U\) means you’d have to add energy to separate the masses to infinity.

(b) Meaning of total energy \(E = K + U\):

Energy	Orbit Type	Fate
\(E < 0\)	Elliptical (bound)	Object returns; trapped in orbit
\(E = 0\)	Parabolic (marginally unbound)	Object barely escapes, reaches infinity with \(v = 0\)
\(E > 0\)	Hyperbolic (unbound)	Object escapes with excess velocity

(c) Correcting the student’s statement:

The student said: “It has positive energy, so it’s gaining energy as it approaches.”

This is wrong. Total mechanical energy \(E = K + U\) is conserved—it doesn’t change as the comet approaches.

What happens is:

As the comet gets closer, \(U\) becomes more negative (deeper in the potential well)
Kinetic energy \(K\) increases by the same amount (it speeds up)
Total \(E\) stays constant and positive

“Positive energy” means the comet’s total energy is above the escape threshold, not that it’s gaining energy.

Common misconception: Confusing the sign of energy with energy gain or loss.

Problem 2 — Orbital Velocity Practice

Given:

\(G = 6.67 \times 10^{-8}\) cm³ g⁻¹ s⁻²
\(M_\odot = 2 \times 10^{33}\) g
\(r = 1\) AU \(= 1.5 \times 10^{13}\) cm
Mars orbital radius: \(r_{\text{Mars}} = 1.52\) AU

Find: Orbital velocities for Earth and Mars.

Equation:

\[v_{\text{orb}} = \sqrt{\frac{GM}{r}}\]

Let’s unpack each piece:

\(G\): gravitational constant (sets strength of gravity)
\(M\): central mass (Sun)
\(r\): orbital radius
\(v_{\text{orb}}\): circular orbital speed

What this equation says: Faster orbits require larger central mass or smaller orbital radius. Speed scales as \(M^{1/2} r^{-1/2}\).

Steps:

(a) Calculate Earth’s orbital velocity:

First compute \(GM/r\):

\[\frac{GM_\odot}{r} = \frac{(6.67 \times 10^{-8}\,\text{cm}^3\,\text{g}^{-1}\,\text{s}^{-2})(2 \times 10^{33}\,\text{g})}{1.5 \times 10^{13}\,\text{cm}}\]

Separate the numerical and unit parts:

\[= \frac{6.67 \times 2}{1.5} \times 10^{-8+33-13} \times \frac{\text{cm}^3 \cdot \cancel{\text{g}^{-1}} \cdot \text{s}^{-2} \cdot \cancel{\text{g}}}{\text{cm}}\]

\[= \frac{13.34}{1.5} \times 10^{12}\,\text{cm}^2\,\text{s}^{-2}\]

\[\approx 8.9 \times 10^{12}\,\text{cm}^2\,\text{s}^{-2}\]

Now take the square root:

\[v_{\text{orb}} = \sqrt{8.9 \times 10^{12}\,\text{cm}^2\,\text{s}^{-2}} \approx 3.0 \times 10^6\,\text{cm/s}\]

Convert to km/s using the identity trick:

\[v_{\text{orb}} = 3.0 \times 10^6\,\text{cm/s} \times \frac{1\,\text{m}}{10^2\,\text{cm}} \times \frac{1\,\text{km}}{10^3\,\text{m}} = 3.0 \times 10^{6-2-3}\,\text{km/s} = 30\,\text{km/s}\]

Answer (a): \(\boxed{v_{\oplus} \approx 3.0 \times 10^6\,\text{cm/s} = 30\,\text{km/s}}\)

(b) Unit check:

\[\left[\frac{GM}{r}\right] = \frac{[L^3 M^{-1} T^{-2}][M]}{[L]} = \frac{[L^3 T^{-2}]}{[L]} = [L^2 T^{-2}]\]

\[[v] = \sqrt{[L^2 T^{-2}]} = [L T^{-1}] \quad \checkmark\]

(c) Mars’s orbital velocity using ratios:

Since \(v \propto r^{-1/2}\) (at fixed \(M\)):

\[\frac{v_{\text{Mars}}}{v_\oplus} = \sqrt{\frac{r_\oplus}{r_{\text{Mars}}}} = \sqrt{\frac{1}{1.52}} = \sqrt{0.658} \approx 0.81\]

\[v_{\text{Mars}} = 0.81 \times 30\,\text{km/s} \approx 24\,\text{km/s}\]

Answer (c): \(\boxed{v_{\text{Mars}} \approx 24\,\text{km/s}}\)

(d) Sanity check:

Mars orbits farther from the Sun, so gravity is weaker, requiring a slower orbital speed. Our answer (\(24 < 30\) km/s) confirms this. \(\checkmark\)

Problem 3 — Escape Velocity Comparison

Given:

Earth: \(M_\oplus = 6 \times 10^{27}\) g, \(R_\oplus = 6.4 \times 10^8\) cm
Moon: \(M_{\text{Moon}} = 7.3 \times 10^{25}\) g, \(R_{\text{Moon}} = 1.7 \times 10^8\) cm
\(G = 6.67 \times 10^{-8}\) cm³ g⁻¹ s⁻²

Find: Escape velocities for Earth and Moon.

Equation:

\[v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\]

Let’s unpack each piece:

Factor of 2: comes from requiring total energy \(E = 0\) (marginally unbound)
\(G\): gravitational constant
\(M\): mass of the body
\(R\): radius of the body (launch point)
\(v_{\text{esc}}\): minimum speed to escape to infinity

What this equation says: Escape speed rises if mass grows (deeper well) or radius shrinks (closer to the center of mass).

Steps:

(a) Earth’s escape velocity:

\[\frac{2GM_\oplus}{R_\oplus} = \frac{2 \times (6.67 \times 10^{-8}) \times (6 \times 10^{27})}{6.4 \times 10^{8}}\]

\[= \frac{2 \times 6.67 \times 6}{6.4} \times 10^{-8+27-8}\,\text{cm}^2\,\text{s}^{-2}\]

\[= \frac{80.04}{6.4} \times 10^{11}\,\text{cm}^2\,\text{s}^{-2}\]

\[\approx 1.25 \times 10^{12}\,\text{cm}^2\,\text{s}^{-2}\]

\[v_{\text{esc},\oplus} = \sqrt{1.25 \times 10^{12}\,\text{cm}^2\,\text{s}^{-2}} \approx 1.12 \times 10^6\,\text{cm/s}\]

Convert to km/s using the identity trick:

\[v_{\text{esc},\oplus} = 1.12 \times 10^6\,\text{cm/s} \times \frac{1\,\text{m}}{10^2\,\text{cm}} \times \frac{1\,\text{km}}{10^3\,\text{m}}\]

\[= 1.12 \times 10^{6-2-3}\,\text{km/s} = 1.12 \times 10^1\,\text{km/s} = 11.2\,\text{km/s}\]

Answer (a): \(\boxed{v_{\text{esc},\oplus} \approx 11.2\,\text{km/s}}\)

(b) Moon’s escape velocity:

\[\frac{2GM_{\text{Moon}}}{R_{\text{Moon}}} = \frac{2 \times (6.67 \times 10^{-8}) \times (7.3 \times 10^{25})}{1.7 \times 10^{8}}\]

\[= \frac{2 \times 6.67 \times 7.3}{1.7} \times 10^{-8+25-8}\,\text{cm}^2\,\text{s}^{-2}\]

\[= \frac{97.4}{1.7} \times 10^{9}\,\text{cm}^2\,\text{s}^{-2}\]

\[\approx 5.7 \times 10^{10}\,\text{cm}^2\,\text{s}^{-2}\]

\[v_{\text{esc,Moon}} = \sqrt{5.7 \times 10^{10}\,\text{cm}^2\,\text{s}^{-2}} \approx 2.4 \times 10^5\,\text{cm/s}\]

Convert to km/s using the identity trick:

\[v_{\text{esc,Moon}} = 2.4 \times 10^5\,\text{cm/s} \times \frac{1\,\text{m}}{10^2\,\text{cm}} \times \frac{1\,\text{km}}{10^3\,\text{m}}\]

\[= 2.4 \times 10^{5-2-3}\,\text{km/s} = 2.4 \times 10^0\,\text{km/s} = 2.4\,\text{km/s}\]

Answer (b): \(\boxed{v_{\text{esc,Moon}} \approx 2.4\,\text{km/s}}\)

(c) Ratio of escape velocities:

\[\frac{v_{\text{esc},\oplus}}{v_{\text{esc,Moon}}} = \frac{11.2}{2.4} \approx 4.7\]

Answer (c): \(\boxed{\text{Earth's escape velocity is } \sim 4.7 \times \text{ larger than the Moon's}}\)

(d) Interpretation for Apollo astronauts:

The Moon’s escape velocity is about 5× smaller than Earth’s. Since rocket fuel requirements scale steeply with required velocity change (\(\Delta v\)), the lunar module could be much smaller and lighter than the Saturn V needed to escape Earth. This made the return trip from the lunar surface much easier.

Unit check: \([2GM/R] = [L^2 T^{-2}]\), so \(v_{\text{esc}}\) has units \([L T^{-1}]\) \(\checkmark\)

Sanity check: The Moon is smaller and less massive, so it should have lower escape velocity. \(2.4 < 11.2\) km/s. \(\checkmark\)

Problem 4 — Kepler III with Newton

Given: Newton’s generalized Kepler’s Third Law:

\[P^2 = \frac{4\pi^2}{G(M+m)}a^3\]

Find: Simplified form, unit verification, binary star comparison.

Equation unpacked:

\(P\): orbital period
\(a\): semi-major axis
\(M + m\): total mass of the system
The factor \(4\pi^2/G\) sets the relationship between geometry and mass

Steps:

(a) Simplified form for \(m \ll M_\odot\):

When the orbiting body is much less massive than the central body:

\[M + m \approx M_\odot\]

\[P^2 \approx \frac{4\pi^2}{GM_\odot}a^3\]

Answer (a): \(\boxed{P^2 \approx \dfrac{4\pi^2}{GM_\odot}a^3}\)

(b) Verify for Earth (\(a = 1\) AU, \(P = 1\) yr):

Using CGS units:

\(a = 1.5 \times 10^{13}\) cm
\(G = 6.67 \times 10^{-8}\) cm³ g⁻¹ s⁻²
\(M_\odot = 2 \times 10^{33}\) g

\[P^2 = \frac{4\pi^2 a^3}{GM_\odot} = \frac{4 \times 9.87 \times (1.5 \times 10^{13})^3}{(6.67 \times 10^{-8})(2 \times 10^{33})}\]

Numerator: \(4 \times 9.87 \times 3.375 \times 10^{39} \approx 1.33 \times 10^{41}\)

Denominator: \(6.67 \times 2 \times 10^{25} \approx 1.33 \times 10^{26}\)

\[P^2 \approx \frac{1.33 \times 10^{41}}{1.33 \times 10^{26}} = 10^{15}\,\text{s}^2\]

\[P \approx \sqrt{10^{15}} \approx 3.16 \times 10^7\,\text{s}\]

Convert to years:

\[P = 3.16 \times 10^7\,\text{s} \times \frac{1\,\text{yr}}{3.16 \times 10^7\,\text{s}} = 1\,\text{yr} \quad \checkmark\]

Answer (b): \(\boxed{P \approx 1\,\text{yr}}\)

(c) Unit check:

\[\left[\frac{a^3}{G(M+m)}\right] = \frac{[L^3]}{[M^{-1}L^3T^{-2}][M]} = \frac{[L^3]}{[L^3T^{-2}]} = [T^2] \quad \checkmark\]

Both sides have dimensions of time squared.

Answer (c): \(\boxed{[P^2] = [\text{s}^2]\ \text{on both sides}}\)

(d) Binary star comparison (two equal \(1\,M_\odot\) stars):

For the Sun-Earth system: \(M + m \approx M_\odot\)

For the binary: \(M + m = 2M_\odot\)

The “Kepler constant” is \(P^2/a^3 = 4\pi^2/[G(M+m)]\)

\[\frac{(P^2/a^3)_{\text{binary}}}{(P^2/a^3)_{\odot}} = \frac{M_\odot}{2M_\odot} = \frac{1}{2}\]

Answer (d): \(\boxed{\left(\frac{P^2}{a^3}\right)_{\text{binary}} = \frac{1}{2}\left(\frac{P^2}{a^3}\right)_{\odot}}\)

At the same orbital separation, the binary has a shorter period because the total mass is larger.

Sanity check: Doubling total mass should strengthen gravity and shorten the period at fixed separation, so a smaller \(P^2/a^3\) constant is physically consistent. \(\checkmark\)

Problem 5 — Complete Workflow: Weighing Jupiter

Given:

Io’s orbital period: \(P = 1.77\) days
Io’s semi-major axis: \(a = 4.22 \times 10^{10}\) cm

Find: Jupiter’s mass using Newton’s Kepler III.

Equation:

Start from Newton’s two-body form:

\[P^2 = \frac{4\pi^2 a^3}{G\left(M_J + m_{\text{Io}}\right)}\]

Assuming \(m_{\text{Io}} \ll M_J\), solve for \(M_J\):

\[M_J \approx \frac{4\pi^2 a^3}{GP^2}\]

Steps:

(a) Use Newton’s Kepler III to solve for Jupiter’s mass.

Convert period to seconds:

\[P = 1.77\,\cancel{\text{days}} \times \frac{24\,\cancel{\text{hr}}}{1\,\cancel{\text{day}}} \times \frac{3600\,\text{s}}{1\,\cancel{\text{hr}}}\]

\[= 1.77 \times 24 \times 3600\,\text{s}\]

\[= 1.77 \times 86400\,\text{s} \approx 1.53 \times 10^5\,\text{s}\]

Compute the components:

\[a^3 = (4.22 \times 10^{10})^3 = 75.2 \times 10^{30} = 7.52 \times 10^{31}\,\text{cm}^3\]

\[P^2 = (1.53 \times 10^5)^2 = 2.34 \times 10^{10}\,\text{s}^2\]

\[GP^2 = (6.67 \times 10^{-8})(2.34 \times 10^{10}) = 1.56 \times 10^{3}\,\text{cm}^3\,\text{g}^{-1}\]

\[4\pi^2 a^3 = 4 \times 9.87 \times 7.52 \times 10^{31} = 2.97 \times 10^{33}\,\text{cm}^3\]

\[M_J = \frac{4\pi^2 a^3}{GP^2} = \frac{2.97 \times 10^{33}}{1.56 \times 10^{3}} \approx 1.9 \times 10^{30}\,\text{g}\]

(b) Unit check: verify \(\dfrac{a^3}{GP^2}\) has units of mass.

\[\left[\frac{a^3}{GP^2}\right] = \frac{[L^3]}{[M^{-1}L^3T^{-2}][T^2]} = \frac{[L^3]}{[M^{-1}L^3]} = [M] \quad \checkmark\]

(c) Express your answer in grams and in solar masses.

In grams:

\[M_J \approx 1.9 \times 10^{30}\,\text{g}\]

In solar masses:

\[\frac{M_J}{M_\odot} = \frac{1.9 \times 10^{30}}{2 \times 10^{33}} = 0.95 \times 10^{-3} \approx 10^{-3}\]

Answer: \[\boxed{M_J \approx 1.9 \times 10^{30}\,\text{g} \approx 10^{-3}\,M_\odot}\]

(d) Observable quantities needed:

Io’s orbital period (measured from repeated observations)
Io’s orbital size (measured from angular separation and Jupiter’s distance)

Neither quantity requires knowing Jupiter’s mass—we measure the orbit and infer the mass.

Sanity check: Jupiter is about 1/1000 of a solar mass, which matches the known value. \(\checkmark\)

Problem 6 — The \(\sqrt{2}\) Factor

Restatement: Derive and interpret \(v_{\text{esc}} = \sqrt{2}\,v_{\text{orb}}\).

Key elements: Compare formulas and connect to energy.

Answer:

(a) Algebraic derivation:

\[v_{\text{orb}} = \sqrt{\frac{GM}{r}}\]

\[v_{\text{esc}} = \sqrt{\frac{2GM}{r}} = \sqrt{2} \cdot \sqrt{\frac{GM}{r}} = \sqrt{2}\,v_{\text{orb}}\]

Result: \(\boxed{v_{\text{esc}} = \sqrt{2}\,v_{\text{orb}}}\)

(b) Speed increase needed for escape:

A spacecraft in circular orbit at speed \(v_{\text{orb}}\) must increase to \(\sqrt{2}\,v_{\text{orb}}\):

\[\frac{v_{\text{esc}}}{v_{\text{orb}}} = \sqrt{2} \approx 1.41\]

The spacecraft must increase its speed by a factor of \(\sqrt{2}\) (about 41% faster).

(c) Energy interpretation:

For a circular orbit:

Kinetic energy: \(K = \frac{1}{2}mv_{\text{orb}}^2 = \frac{GMm}{2r}\)
Potential energy: \(U = -\frac{GMm}{r}\)
Total energy: \(E = K + U = -\frac{GMm}{2r}\) (bound, negative)

To escape (\(E = 0\)), we need to add energy equal to \(|E| = \frac{GMm}{2r}\), which equals the orbital kinetic energy \(K\).

So escaping requires doubling the kinetic energy. Since \(K \propto v^2\):

\[\frac{K_{\text{esc}}}{K_{\text{orb}}} = 2 \Rightarrow \frac{v_{\text{esc}}^2}{v_{\text{orb}}^2} = 2 \Rightarrow v_{\text{esc}} = \sqrt{2}\,v_{\text{orb}}\]

Physical insight: The binding energy of a circular orbit equals its kinetic energy. To unbind, you must add exactly that much energy again.

Problem 7 — Virial Theorem Capstone

Given:

Star cluster kinetic energy: \(K = 1.0 \times 10^{50}\) erg
System in virial equilibrium: \(2K + U = 0\)

Find: Potential energy, total energy, and effect of energy injection.

Equation:

The virial theorem for a gravitationally bound system in equilibrium:

\[2K + U = 0\]

This connects random motions (kinetic energy of stars) to the gravitational binding (potential energy).

Steps:

(a) Find gravitational potential energy:

From the virial theorem:

\[U = -2K = -2 \times (1.0 \times 10^{50}\,\text{erg}) = -2.0 \times 10^{50}\,\text{erg}\]

Answer (a): \(\boxed{U = -2.0 \times 10^{50}\,\text{erg}}\)

(b) Total energy and bound state:

\[E = K + U = (1.0 \times 10^{50}) + (-2.0 \times 10^{50}) = -1.0 \times 10^{50}\,\text{erg}\]

Since \(E < 0\), the cluster is gravitationally bound.

Answer (b): \(\boxed{E = -1.0 \times 10^{50}\,\text{erg};\text{ cluster is bound}}\)

(c) After tidal interaction adds \(\Delta K = 0.6 \times 10^{50}\) erg:

New kinetic energy:

\[K' = K + \Delta K = (1.0 + 0.6) \times 10^{50} = 1.6 \times 10^{50}\,\text{erg}\]

If the cluster size hasn’t changed yet (immediate aftermath), potential energy remains:

\[U' \approx U = -2.0 \times 10^{50}\,\text{erg}\]

New total energy:

\[E' = K' + U' = (1.6 - 2.0) \times 10^{50} = -0.4 \times 10^{50}\,\text{erg}\]

Since \(E' < 0\), the cluster is still bound, but less tightly.

Answer (c): \(\boxed{E' = -0.4 \times 10^{50}\,\text{erg};\text{ cluster remains bound but will expand}}\)

Physical follow-up: The system is no longer in virial equilibrium (\(2K' \neq -U'\)). It will expand until stars slow down and it re-virializes at a larger radius. If the injected energy had been \(\geq 1.0 \times 10^{50}\) erg, the cluster would become unbound (\(E \geq 0\)) and disperse.

Unit check: \(K\), \(U\), and \(E\) are all energies, so every term carries units of erg and can be added directly. \(\checkmark\)

Sanity check: For virial equilibrium, \(E = K + U = -K\). Starting at \(K = 1.0 \times 10^{50}\) erg gives \(E = -1.0 \times 10^{50}\) erg, and adding kinetic energy makes \(E\) less negative, exactly as we found. \(\checkmark\)

Part B — Light as Information

Problem 8 — Kirchhoff’s Laws in Action

Restatement: Identify the source for each spectrum type.

Key insight: Absorption lines require a cooler gas in front of a hot continuum; emission lines require a hot, low-density gas.

Answer:

(a) Smooth rainbow with dark lines (absorption spectrum):

Source: Cool gas in front of a hot source (the hot source supplies the continuum).

Reasoning: The hot object emits all wavelengths (continuous blackbody). The cooler foreground gas absorbs photons at specific wavelengths (its atomic transitions), creating dark gaps in the rainbow. This is how stellar atmospheres work: the hot interior produces the continuum, the cooler outer layers absorb at characteristic wavelengths.

(b) Smooth rainbow with no lines (continuous spectrum):

Source: A hot dense object with no intervening cool gas.

Reasoning: A hot, dense material (solid, liquid, or dense gas/plasma) produces a continuous spectrum via blackbody radiation. Without a cooler gas in the line of sight, no absorption occurs.

(c) Dark background with bright lines (emission spectrum):

Source: A hot thin gas (low density) with no continuum source behind it.

Reasoning: A low-density gas produces light only at discrete wavelengths corresponding to its atomic transitions. Without a background continuum, only these bright lines are visible against the dark sky. Examples: nebulae, neon signs, laboratory discharge tubes.

Common misconception: Thinking absorption lines mean the source is “missing” energy rather than energy being selectively absorbed by a cooler intervening gas.

Problem 9 — Why Skies Are Red at Sunset

Restatement: Explain sunset colors using Rayleigh scattering.

Key insight: Scattering probability scales as \(\lambda^{-4}\), so short wavelengths scatter much more efficiently.

Answer:

Rayleigh scattering intensity:

\[I_{\text{scattered}} \propto \lambda^{-4}\]

This means blue light (\(\lambda \sim 450\) nm) scatters about \((700/450)^4 \approx 6\) times more than red light (\(\lambda \sim 700\) nm).

At noon (short path through atmosphere):

Sunlight passes through relatively little atmosphere
Some blue scatters away, but most direct sunlight remains nearly white (or slightly yellow)
The scattered blue light makes the sky blue

At sunset (long path through atmosphere):

Sunlight travels through a much longer atmospheric path (grazing angle)
Blue light is progressively scattered out along the entire path
By the time light reaches your eye directly, most blue and even green has been removed
The remaining direct sunlight is enriched in red/orange wavelengths

Why \(\lambda^{-4}\) makes this strong:

The fourth power is steep. Doubling the wavelength reduces scattering by \(2^4 = 16\) times. This dramatic wavelength dependence means even modest increases in path length cause significant color shifts—blue disappears long before red.

Summary:

Condition	Atmospheric Path	Direct Sun Color
Noon	Short	White/yellowish
Sunset	Very long	Red/orange

Problem 10 — Photon Energy from Wavelength

Given:

Wavelength: \(\lambda = 600\) nm (orange light)
Speed of light: \(c = 3.00 \times 10^{10}\) cm/s
Planck’s constant: \(h = 6.626 \times 10^{-27}\) erg·s

Find: Frequency and photon energy.

Equations:

\[\nu = \frac{c}{\lambda}, \qquad E = h\nu = \frac{hc}{\lambda}\]

Steps:

(a) Convert wavelength to cm (identity trick):

\[\lambda = 600\,\text{nm} \times \frac{10^{-9}\,\text{m}}{1\,\text{nm}} \times \frac{10^2\,\text{cm}}{1\,\text{m}}\]

\[= 600 \times 10^{-9+2}\,\text{cm} = 600 \times 10^{-7}\,\text{cm} = 6.0 \times 10^{-5}\,\text{cm}\]

Answer (a): \(\boxed{\lambda = 6.0 \times 10^{-5}\,\text{cm} = 600\,\text{nm}}\)

(b) Calculate frequency:

\[\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{10}\,\text{cm/s}}{6.0 \times 10^{-5}\,\text{cm}}\]

\[= \frac{3.00}{6.0} \times 10^{10-(-5)}\,\text{s}^{-1} = 0.5 \times 10^{15}\,\text{s}^{-1} = 5.0 \times 10^{14}\,\text{Hz}\]

Answer (b): \(\boxed{\nu = 5.0 \times 10^{14}\,\text{Hz}}\)

(c) Calculate photon energy:

\[E = h\nu = (6.626 \times 10^{-27}\,\text{erg}\cdot\text{s})(5.0 \times 10^{14}\,\text{s}^{-1})\]

\[= 6.626 \times 5.0 \times 10^{-27+14}\,\text{erg} = 33.1 \times 10^{-13}\,\text{erg}\]

\[= 3.3 \times 10^{-12}\,\text{erg}\]

Answer (c): \(\boxed{E \approx 3.3 \times 10^{-12}\,\text{erg}}\)

Unit check:

\([\nu] = [c/\lambda] = [LT^{-1}]/[L] = [T^{-1}]\) (frequency) \(\checkmark\)
\([E] = [h\nu] = [ML^2T^{-1}][T^{-1}] = [ML^2T^{-2}]\) (energy) \(\checkmark\)

Sanity check:

Visible light photons have energies of a few eV. Converting:

\[1\,\text{eV} = 1.6 \times 10^{-12}\,\text{erg}\]

\[E = \frac{3.3 \times 10^{-12}\,\text{erg}}{1.6 \times 10^{-12}\,\text{erg/eV}} \approx 2.1\,\text{eV}\]

This is correct for orange/red visible light (visible photons range from about 1.8–3.1 eV). \(\checkmark\)