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Objective 1: Calculate the altitude of the satellite orbiting Earth with a period of 132 minutes. Note, Kepler’s Law applies to both circular and elliptical orbits. You can also use centripetal motion in place of Kepler’s Law.
Step | Formula Derivation | Reasoning |
---|---|---|
1 | [katex] T = 132 , \text{min} = 132 \times 60 , \text{s} [/katex] | Convert period from minutes to seconds. |
2 | [katex] T^2 = \frac{4\pi^2 r^3}{GM} [/katex] | Kepler’s Third Law for orbital period ([katex] T [/katex]), where [katex] r [/katex] is the orbit radius, [katex] G [/katex] is the gravitational constant, and [katex] M [/katex] is Earth’s mass. NOTE: You can derive this equation using just circular motion and gravitation formulas. |
3 | [katex] r = \left( \frac{GMT^2}{4\pi^2} \right)^{\frac{1}{3}} [/katex] | Rearranging the formula to solve for [katex] r [/katex], the orbit radius. |
4 | [katex] \text{Altitude} = r – R_{\text{earth}} [/katex] | Altitude is the orbit radius minus Earth’s radius. |
5 | Orbit radius [katex] r \approx 8587604.79 , \text{m} [/katex] | Calculated orbit radius. |
6 | Altitude [katex] \approx 2216604.79 , \text{m} [/katex] | Calculated altitude above Earth’s surface. |
The altitude of the satellite is approximately [katex] \boxed{2216604.79 , \text{meters}} [/katex] (or about 2216.6 km).
Objective 2: Calculate the value of [katex] g [/katex] at the location of this satellite. This time using centripetal motion.
Step | Formula Derivation | Reasoning |
---|---|---|
1 | [katex] g = \frac{GM}{r^2} [/katex] | Gravitational acceleration formula, where [katex] r [/katex] is the distance from the center of Earth. |
2 | [katex] g_{\text{satellite}} [/katex] | Calculate [katex] g [/katex] using the orbit radius ([katex] r [/katex]) from Objective 1. |
3 | [katex] g_{\text{satellite}} \approx 5.40 , \text{m/s}^2 [/katex] | Calculated gravitational acceleration at the satellite’s orbit. |
The value of [katex] g [/katex] at the location of this satellite is approximately [katex] \boxed{5.40 , \text{m/s}^2} [/katex].
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Find the escape speed from a planet of mass 6.89 x 1025 kg and radius 6.2 x 106 m.
A [katex] 2.2 \times 10^{21} \, \text{kg}[/katex] moon orbits a distant planet in a circular orbit of radius [katex] 1.5 \times 10^8 \, \text{m}[/katex]. It experiences a [katex] 1.1 \times 10^{19} \, \text{N}[/katex] gravitational pull from the planet. What is the moon’s orbital period in earth days?
Two identical object rests on a platform rotating at constant speed. Object A is at distance of half the platform’s radius from the center. Object B lays at edge of the platform. Assuming the platform continues rotating at the same speed, how does the centripetal force of the two objects compare?
A rock is whirled on the end of a string in a horizontal circle of radius R with a constant period T. If the radius of the circle is reduced to R/3, while the period remains T, what happens to the centripetal acceleration (ac) of the rock?
A curve with a radius of 125 m is properly banked for a car traveling 40 m/s. What must be the coefficient of static friction (µs) for a car not to skid on the same curve when traveling at 53 m/s?
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Kinematics | Forces |
---|---|
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
Circular Motion | Energy |
---|---|
[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
[katex]KE_i + PE_i = KE_f + PE_f[/katex] |
Momentum | Torque and Rotations |
---|---|
[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
Simple Harmonic Motion |
---|
[katex]F = -k x[/katex] |
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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