| Derivation or Formula | Reasoning |
|---|---|
| \[g’ = g\left(\dfrac{R_E}{R_E + h}\right)^2\] | Gravitational acceleration \(g’\) at altitude \(h\) follows the inverse-square law; \(R_E\) is Earth’s radius and \(g\) is the surface value \(9.8\,\text{m/s}^2\). |
| \[R_E = 6.37\times10^6\,\text{m}, \; h \approx 4.0\times10^5\,\text{m}\] | Typical low-Earth-orbit (LEO) values: International Space Station altitude. |
| \[g’ \approx 9.8\left(\dfrac{6.37}{6.37+0.40}\right)^2 \approx 9.8(0.941)^2 \approx 8.7\,\text{m/s}^2\] | Numerical evaluation shows only a \(\sim\!11\%\) reduction from surface gravity; the gravitational force is still substantial. |
| \[W’ = m g’\] | True weight (gravitational force) on the astronaut of mass \(m\) remains non-zero. |
| \[\Sigma F_r = m a_r = m g’\] | In circular orbit the required centripetal force equals the gravitational pull, producing continuous free fall toward Earth. |
| \[N = 0\] | Apparent weight is the normal force \(N\) from the spacecraft on the astronaut; in free fall this contact force vanishes, so \(N=0\). |
| \[\text{Apparent weightlessness}\] | Because \(N = 0\), astronauts feel “weightless” even though the gravitational force \(m g’\) is nearly as large as on Earth. |
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The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
A cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
A conical pendulum is formed by attaching a ball of mass \( m \) to a string of length \( L \), then allowing the ball to move in a horizontal circle of radius \( R \).
The heaviest train ever pulled by a single engine was over \( 2 \, \text{km} \) long. A force of \( 1.13 \times 10^5 \, \text{N} \) is needed to get the train to start moving. If the coefficient of static friction is \( 0.741 \) and the coefficient of kinetic friction is \( .592 \), what is the train’s mass?
Find the net gravitational force on a \(2.0 \, \text{kg}\) sphere midway between a \(4.0 \, \text{kg}\) sphere and a \(7.0 \, \text{kg}\) sphere that are \(1.2 \, \text{m}\) apart.
An object moves at constant speed in a circular path of radius \( r \) at a rate of \( 1 \) revolution per second. What is its acceleration in terms of \(r\)?
Two satellites are in circular orbits around Earth. Satellite A has speed \(v_A\). Satellite B has an orbital radius nine times that of satellite A. What is the speed of satellite B?
What is a man’s apparent weight at the equator if his weight is \(500 \, \text{N}\)? The Earth’s radius is \(6.37 \times 10^{6} \, \text{m}\).
A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?
A \(1.00 \, \text{kg}\) mass is attached to a \(0.800 \, \text{m}\) long string and spun in a vertical circle. The mass completes \(2.00\) revolutions in \(1.00 \, \text{s}\).
\(\text{No—gravity still acts ( }g’\!\approx\!8.7\,\text{m/s}^2\text{); astronauts are in free fall so their apparent weight }N=0\)
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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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