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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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| 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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