| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_{\text{esc,E}} = \sqrt{\frac{2 G M_E}{R_E}}\] | To leave Earth, a spacecraft must at least reach the Earth escape velocity, given by the standard gravitational formula with mass \(M_E\) and radius \(R_E\). |
| 2 | \[E_{\text{esc,E}} = \frac{1}{2} v_{\text{esc,E}}^{2} = \frac{G M_E}{R_E}\] | Escape energy per unit mass from Earth equals the gravitational potential energy that must be overcome, \(G M_E / R_E\). |
| 3 | \[v_{\text{esc,M}} = \sqrt{\frac{2 G M_M}{R_M}}\] | Analogous expression for escaping the Moon’s weaker gravitational field with mass \(M_M\) and radius \(R_M\). |
| 4 | \[E_{\text{esc,M}} = \frac{1}{2} v_{\text{esc,M}}^{2} = \frac{G M_M}{R_M}\] | Escape energy per unit mass from the Moon. |
| 5 | \[\frac{E_{\text{esc,E}}}{E_{\text{esc,M}}} = \frac{M_E / R_E}{M_M / R_M} \approx 22\] | Substituting numerical values \(M_E = 5.97\times10^{24}\,\text{kg},\, R_E = 6.37\times10^{6}\,\text{m},\, M_M = 7.35\times10^{22}\,\text{kg},\, R_M = 1.74\times10^{6}\,\text{m}\) shows Earth’s gravity well is about \(22\) times deeper than the Moon’s. |
| 6 | \[\Delta v_{\text{out}} \gg \Delta v_{\text{return}}\] | The outbound leg (Earth \(\to\) Moon) needs a much larger total \(\Delta v\) to climb out of Earth’s gravity well than the return leg needs to leave the Moon’s weaker gravity. |
| 7 | \[F_{\text{atm}} \text{ assists on return}\] | On the way back, Earth’s atmosphere provides free braking (aerodynamic drag and heating), so less propellant is required for deceleration; launch from Earth, however, must fight this same atmosphere, costing extra fuel. |
| 8 | \[m_{\text{fuel,out}} > m_{\text{fuel,ret}}\] | Because the larger \(\Delta v\) and atmospheric losses occur on departure from Earth, the mass of propellant needed going to the Moon is substantially greater than that needed to come home. |
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A mass \( m_1 \) traveling with an initial velocity \( v \) has an elastic collision with a mass \( m_2 \) that is initially at rest.
A \(90 \, \text{kg}\) individual is cycling up a hill inclined at \(30^\circ\) on a \(12 \, \text{kg}\) bicycle. The hill is quite steep, and the coefficient of static friction is \(0.85\). The cyclist ascends \(12 \, \text{m}\) up the hill and then pauses at the summit. They then start descending from rest and travel \(9 \, \text{m}\) before firmly applying the brakes, causing the wheels to lock.
A car accelerates uniformly from rest to \( 29.4 \) m/s in \( 6.93 \) s along a level stretch of road. Ignoring friction, determine the average power in both watts and horsepower (\( 1 \text{ horsepower} = 745.7 \text{ Watts} \)) required to accelerate the car if:
When a falling meteoroid is at a distance above the Earth’s surface of \( 3.00 \) times the Earth’s radius, what is its acceleration due to the Earth’s gravitation?
A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius \( R \) and is vertical. In terms of \( R \) and constants, find the minimum height \( h \) above the bottom of the loop the block must start from so it makes it around the loop.
A bullet of mass \(0.0500 \, \text{kg}\) traveling at \(50.0 \, \text{m/s}\) is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is \(0.300 \, \text{kg}\) and it is initially at rest. The collision is completely inelastic and after impact the bullet + wooden block move together until the center of mass of the system rises a vertical distance \(h\) above its initial position.
The box in the diagram is sliding to the right across a horizontal table, under the influence of the forces shown. Which force(s) is doing negative work on the box?
The gravitational force that the Moon exerts on Earth is often cited as the source of the tides we witness. However, the gravitational force the Sun exerts on Earth is over \(100\) times greater than the force the Moon exerts on Earth.
Why is the force from the Moon credited for the tides, and not the force from the Sun?
A kickball is rolled by the pitcher at a speed of 10 m/s and it is kicked by another student. The kickball deforms a little during the kick, and then rebounds with a velocity of 15 m/s as its shape restores to a perfect sphere. Select all that must be true about the kickball and the kicking foot system.
If a small motor does 520 J of work to move a toy car 260 meters in a time of 37 seconds.
\(\text{More fuel is required because a spacecraft must overcome Earth’s far deeper gravitational potential and atmospheric drag, whereas leaving the Moon needs far less }\Delta v\text{ and Earth’s atmosphere assists braking on return.}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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