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Explanation of Tidal Forces
| Step | Explanation | Reasoning |
|---|---|---|
| 1 | Tidal force depends on the difference in gravitational pull on different parts of Earth. | Tides are caused by the differential gravitational force. |
| 2 | The Moon is much closer to Earth than the Sun. | Distance greatly affects the gradient of the gravitational field. |
| 3 | Gravitational force decreases with the square of the distance. | The inverse-square law dictates gravitational attraction. |
| 4 | Tidal forces are proportional to the inverse cube of the distance. | The gradient of the gravitational field decreases even more steeply. |
| 5 | The Moon’s proximity means a steeper gravitational gradient across Earth’s diameter. | This causes stronger tidal effects than the Sun, despite its greater overall force. |
| 6 | The Sun’s influence is more uniform across Earth. | The Sun’s further distance causes a less steep gradient, resulting in weaker tidal forces. |
| 7 | The combined effect of the Sun and Moon creates spring and neap tides. | When their forces align or oppose, the tides are respectively higher or lower. |
In summary, while the Sun exerts a greater overall gravitational force on Earth, the Moon’s closer proximity results in a greater differential in gravitational pull across the Earth, leading to more significant tidal forces.
Further Understanding: Gravitational Gradient
| Step | Explanation | Reasoning |
|---|---|---|
| 1 | Gravitational force follows the inverse-square law. | Gravitational force [katex]F[/katex] is proportional to [katex]\frac{1}{r^2}[/katex], where [katex]r[/katex] is the distance. |
| 2 | Gravitational gradient is the rate of change of force with distance. | It measures how much the force changes over a small change in distance ([katex]\frac{dF}{dr}[/katex]). |
| 3 | The gradient is steeper for closer objects. | Due to the inverse-square law, nearby objects (like the Moon) have a more rapidly changing force over a given distance. |
| 4 | Tidal forces depend on the difference in gravitational force across an object’s diameter. | For Earth, this means the difference in the Moon’s (or Sun’s) pull between the side facing the celestial body and the opposite side. |
| 5 | The Moon’s gradient has a greater effect than the Sun’s. | Although the Sun exerts a stronger overall force, the change in its force across Earth’s diameter is less than the change in the Moon’s force. |
The gradient is vital in explaining tidal phenomena because it’s not just the strength of the gravitational pull that matters, but how much this pull changes over the distance spanning the Earth. The Moon, being much closer to Earth, exerts a significantly varying force across Earth’s diameter compared to the more uniform force exerted by the distant Sun, leading to stronger tides despite its weaker overall gravitational pull.
Just ask: "Help me solve this problem."
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
Why is more fuel required for a spacecraft to travel from the Earth to the Moon than to return from the Moon to the Earth?
Find the net gravitational force on a 2.0 kg sphere midway between a 4.0 kg sphere and a 7.0 kg sphere that are 1.2 m apart.
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?
The magnitude of the gravitational field on the surface of a new planet is 20 N/kg. The planet’s mass half the mass of Earth. The radius of Earth is 6400 km. What is the radius of the new planet?
The Moon has a greater influence on Earth’s tides than the Sun due to the differential gravitational forces.
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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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