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Assuming circular motion and that earths period of orbit around the sun is one year we can make a general equation relating Period and radius of orbit.
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] \frac{mv^2}{r} = \frac{GMm}{r^2} [/katex] | Equating centripetal force to gravitational force for a stable orbit. |
| 2 | [katex] v = \frac{2\pi r}{T} [/katex] | Expressing orbital velocity in terms of the orbit’s circumference and period. |
| 3 | [katex] \left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r^2} [/katex] | Substituting the expression for [katex] v [/katex] into the force equation. |
| 4 | [katex] \frac{4\pi^2 r^3}{T^2} = GM [/katex] | Rearranging the equation to isolate [katex] GM [/katex]. |
| 5 | [katex] \frac{R_{\text{Earth}}^3}{T_{\text{Earth}}^2} = \frac{R_{\text{Saturn}}^3}{T_{\text{Saturn}}^2} [/katex] | Recognizing that [katex] GM [/katex] is constant, and applying the formula to both Earth and Saturn. |
| 6 | [katex] T_{\text{Saturn}} = T_{\text{Earth}} \times \left( \frac{R_{\text{Saturn}}}{R_{\text{Earth}}} \right)^{\frac{3}{2}} [/katex] | Solving for [katex] T_{\text{Saturn}} [/katex] in terms of known quantities. |
| 7 | [katex] T_{\text{Saturn}} = 365 , \text{days} \times (9)^{\frac{3}{2}} [/katex] | Substituting [katex] T_{\text{Earth}} = 365 , \text{days} [/katex] and [katex] R_{\text{Saturn}} = 9 , \text{AU} [/katex]. |
| 8 | [katex] T_{\text{Saturn}} \approx 27 , \text{years} [/katex] | Calculating the period of Saturn’s orbit. |
This table illustrates the steps taken to determine the orbital period of Saturn, showing how the increase in the radius of Saturn’s orbit by a factor of 9 results in an increase in the orbital period squared, leading to a period of approximately 27 years.
Just ask: "Help me solve this problem."
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 race car travels in a circular track of radius \( 200 \) \( \text{m} \). If the car moves with a constant speed of \( 80 \) \( \text{m/s} \),
Why do pilots sometimes black out while pulling out at the bottom of a dive?
Find the escape speed from a planet of mass \(6.89 \times 10^{25} \, \text{kg}\) and radius \(6.2 \times 10^{6} \, \text{m}\).
A 250 newton centripetal force acts on a car moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed v and the frequency f of the car?
27 years
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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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