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Part (a) – Finding the orbital speed.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] r = \frac{50}{2} = 25 \text{km} = 25000 \text{m} [/katex] | The radius [katex] r [/katex] of the orbit is half the diameter. Convert it to meters. |
| 2 | [katex] T = 11 \times 86400 [/katex] s | Convert period [katex] T [/katex] from days to seconds. There are 86400 seconds in a day. |
| 3 | [katex] v = \frac{2\pi r}{T} [/katex] | The orbital speed [katex] v [/katex] is calculated by dividing the circumference of the orbit by the orbital period. |
| 4 | [katex] v = \frac{2\pi \times 25000}{950400} [/katex] m/s | Substitute the values of [katex] r [/katex] and [katex] T [/katex] into the formula. Convert [katex] r [/katex] from km to m by multiplying by 1000. |
| 5 | [katex] v \approx 0.165 [/katex] m/s | Simplifying the expression gives the orbital speed [katex] v [/katex]. |
| 6 | [katex] \boxed{v \approx 0.165 \text{m/s}} [/katex] | This is the final value for the satellite’s orbital speed. |
Part (b) – Finding the comet’s mass.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] v^2 = \frac{GM}{r} [/katex] | The inwards gravitational force is equal to the centripetal force of the orbiting comet. In terms of Newtons law this can be expressed as [katex]\frac{GMm}{r^2} = \frac{mv^2}{r} [/katex], where [katex] M [/katex] is the mass of the comet, [katex] m [/katex] [katex] is the mass of the satellite, and G [/katex] is the gravitational constant. |
| 2 | [katex] M = \frac{rv^2}{G} [/katex] | Rearrange the formula to solve for the mass [katex] M [/katex] of the comet. |
| 3 | [katex] M = \frac{25000 \times (0.165)^2}{6.674 \times 10^{-11}} [/katex] kg | Substitute the values of [katex] r [/katex] and [katex] v [/katex] into the formula, remembering that [katex] r [/katex] is already converted to meters. |
| 4 | [katex] M \approx 1.02 \times 10^{13} [/katex] kg | Calculating the value gives the mass of the comet. |
| 5 | [katex] \boxed{M \approx 1.02 \times 10^{13} \text{kg}} [/katex] | This is the final value for the mass of the comet. |
Part (c) – Finding the landing speed.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] h = 25\,\text{km}\, -\,1.8 \, \text{km} = 23.2 \, \text{km} = 23200 \, \text{meters} [/katex] | The distance from the satellite to the center of the comet is 25km. Since the comet has an average diameter of 3.6 km (or a radius of 1.8 km). The distance from the satellite to the surface of comet is 23.2 km. Convert this to meters. |
| 2 | [katex] t = 7 \times 3600 [/katex] s | Convert the fall time [katex] t [/katex] from hours to seconds. |
| 3 | [katex] g = \frac{GM}{r^2} [/katex] | Calculate the acceleration [katex] g [/katex] due to the comet’s gravity using the values from the previous two parts. Set [katex] mg = \frac{GMm}{r^2} [/katex] and solve for [katex] g [/katex]. |
| 4 | [katex] g = \frac{(6.67\times 10^{-11}) \times (1.02 \times 10^{13})}{25000^2} [/katex] | Substitute values into the equation for gravitational acceleration. |
| 5 | [katex] g = 1.09 \times 10^{-6} \, \text{m/s}^2 [/katex] | Final value for [katex]g [/katex],the acceleration due to gravity of the comet. |
| 6 | [katex] v^2_{\text{final}} = v^2_{\text{initial}} + 2a\Delta x [/katex] | Now that we have [katex] v_{initial}, \, a, \, \Delta x [/katex] we can use a kinematic formula to solve for [katex] v_f [/katex]. |
| 7 | [katex] \boxed{v_{\text{final}} \approx .735\, \text{m/s}} [/katex] | Plug in all known values and solve for [katex] v_f [/katex]. |
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While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
A neighbor’s child wants to go to a carnival to experience the wild rides. The neighbor is worried about safety because one of the rides looks particularly dangerous. She knows that you have taken physics and so asks you for advice.
The ride in question has a 4 kg chair which hangs freely from a 10 m long chain attached to a pivot on the top of a tall tower. When the child enters the ride, the chain is hanging straight down. The child is then attached to the chair with a seat belt and shoulder harness. When the ride starts up, the chain rotates about the tower. Soon the chain reaches its maximum speed and remains rotating at that speed, which corresponds to one rotation about the tower every 3 seconds.
When you ask the operator, he says the ride is perfectly safe. He demonstrates this by sitting in the stationary chair. The chain creaks but holds, and he weighs 90 kg.
Which of the following do not affect the maximum speed that a car can drive in a circle? Choose both correct answers.
A person’s back is against the inner wall of spinning cylinder with no support under their feet. If the radius is R, find an expression for the minimum angular speed so the person does not slide down the wall. The coefficient of static friction is µs.
Note: If you haven’t studied angular velocity [katex] \omega [/katex] yet, just find the linear velocity v.
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
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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] | |
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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