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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]. |
Just ask: "Help me solve this problem."
A concrete highway curve of radius 60.0 m is banked at a 12.0 ° angle. What is the maximum speed with which a 1300 kg rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0).
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?
The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.
Two satellites of equal mass, \( S_1 \) and \( S_2 \), orbit the Earth. \( S_1 \) is orbiting at a distance \( r \) from the Earth’s center at speed \( v \). \( S_2 \) orbits at a distance \( 2r \) from the Earth’s centre at speed \( \dfrac{v}{\sqrt{2}} \). The ratio of the centripetal force on \( S_1 \) to the centripetal force on \( S_2 \) is
A delivery truck is traveling north. It then turns along a leftward circular curve. This the packages in the truck to slide to the RIGHT. Which of the following is true of the net force on the packages as they are sliding?
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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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