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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]g = \frac{G M}{R^2}[/katex] | The formula to calculate the gravitational field at the surface of a planet where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet. |
| 2 | [katex]g_{\text{Earth}} = \frac{G M_{\text{Earth}}}{R_{\text{Earth}}^2}[/katex] | Apply the gravitational field formula to Earth. Given that the gravitational field on Earth (standard gravity) is approximately [katex]9.8 \, \text{m/s}^2[/katex] and [katex]R_{\text{Earth}}[/katex] is approximately 6400 km or [katex]6.4 \times 10^6[/katex] m. |
| 3 | [katex]M_{\text{new planet}} = \frac{1}{2} M_{\text{Earth}}[/katex] | The mass of the new planet is defined to be half that of Earth. |
| 4 | [katex]g_{\text{new planet}} = \frac{G (\frac{1}{2} M_{\text{Earth}})}{R_{\text{new planet}}^2}[/katex] | Substitute the mass of the new planet into the gravitational field formula. |
| 5 | [katex]20 = \frac{G (\frac{1}{2} M_{\text{Earth}})}{R_{\text{new planet}}^2}[/katex] | Given that [katex]g_{\text{new planet}} = 20 \, \text{N/kg}[/katex], we set this equal to the derived expression for gravitational field on the new planet. |
| 6 | [katex]g_{\text{Earth}} \cdot R_{\text{Earth}}^2 = 9.8 \cdot (6.4 \times 10^6)^2[/katex] | Substitute the given values for Earth’s gravitational field and radius to calculate this product for further use. |
| 7 | [katex]\frac{1}{2} \cdot 9.8 \cdot (6.4 \times 10^6)^2 = 20 R_{\text{new planet}}^2[/katex] | Equating the expressions from Earth and the new planet to find the radius of the new planet. |
| 8 | [katex]R_{\text{new planet}} = \sqrt{\frac{1}{2} \cdot \frac{9.8 \cdot (6.4 \times 10^6)^2}{20}}[/katex] | Solving for [katex]R_{\text{new planet}}[/katex] using the simplified expression. |
| 9 | [katex]R_{\text{new planet}} \approx 3200 \, \text{km}[/katex] | The radius of the new planet is equal to half that of Earth because of the proportional division of mass and the square of the radius in calculating gravitational field strength. |
Just ask: "Help me solve this problem."
Two identical satellites are placed in orbit of two different planets. Satellite A orbits Mars, and Satellite B orbits Jupiter. The orbital speeds of each satellite are the same. Which satellite has a greater orbital radius?
A 1.5 kg object is located at a distance of 1.7 x106 m from the center of a larger object whose mass is 7.4 x 1022 kg.
A communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be [katex]6 \times 10^{24} \text{ kg}[/katex] the the radius of Earth to be [katex]6.4 \times 10^6 \text{ m}[/katex]. What is the satellite’s velocity?
A 3300-m-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The earth’s radius is 6400 km and the earth’s mass is 5.97 x 1024.
The International Space Station has a mass of 4.2 x105 kg and orbits Earth at a distance of 4.0 x102 km above the surface. Earth has a radius of 6.37 x106 m, and mass of 5.97 x1024 kg. Calculate the following:
3200 km
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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] |
General Metric Conversion Chart
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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