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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."
Two blocks, A and B, are connected by a light string that passes over a frictionless pulley. Block A, of mass \( 10 \) \( \text{kg} \), rests on a rough plane that makes an angle of \( 45^{\circ} \) with the horizontal, while block B, of mass \( 17 \) \( \text{kg} \), hangs vertically. Starting from rest, what is the minimum coefficient of static friction between block A and the plane required to keep the system in static equilibrium?
A truck is traveling at \(35 \, \text{m/s}\) when the driver realizes the truck has no brakes. He sees a ramp off the road, inclined at \(20^\circ\), and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction)?
An Olympic bobsled team goes through a horizontal curve at a speed of \( 120 \) \( \text{km/hr} \). If the radius of curvature is \( 10.0 \) \( \text{m} \), what is the apparent weight the crew experiences—express in terms of \( mg \)?
A spacecraft somewhere in between the Earth and the Moon experiences zero net force acting on it. This is because the Earth and the Moon pull the spacecraft in equal but opposite directions. Find the distance \(D\) away from Earth such that the spacecraft experiences zero net force. The distance between the Moon and Earth is \(\sim 3.844 \times 10^8 \, \text{m}\).
Note: You may need the mass of the Earth and the Moon. You can find this in the formula table.
A \(1.00 \, \text{kg}\) mass is attached to a \(0.800 \, \text{m}\) long string and spun in a vertical circle. The mass completes \(2.00\) revolutions in \(1.00 \, \text{s}\).
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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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