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The gravitational force that the moon exerts on Earth is often cited as the source for the tides we witness. However, the gravitational force the Sun exerts on Earth is over 100 times greater than the force the moon exerts on Earth.

Why is the force from the moon credited for the tides, and not the force from the sun?

*differential gravitational forces*.

Explanation

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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 F is proportional to \frac{1}{r^2}, where r 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 (\frac{dF}{dr}). |

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.

- Statistics

*v _{A}* . Satellite B has an orbital radius nine times that of satellite A. What is the speed of satellite B?

A satellite circling Earth completes each orbit in 132 minutes.

- Find the altitude of the satellite.
- What is the value of g at the location of this satellite?

^{6} m. What is the radius of a planet that has the same mass as earth but on which the free-fall acceleration is 5.50 m/s^{2}？

^{24}.

A 1.5 kg object is located at a distance of 1.7 x10^{6} m from the center of a larger object whose mass is 7.4 x 10^{22} kg.

- What is the magnitude of the force acting on the smaller object?
- What is the magnitude of the force acting on the larger object?
- What is the acceleration of the smaller object when it is released?

The International Space Station has a mass of 4.2 x10^{5} kg and orbits Earth at a distance of 4.0 x10^{2} km above the surface. Earth has a radius of 6.37 x10^{6} m, and mass of 5.97 x10^{24} kg. Calculate the following:

- The gravitational force between Earth and the ISS.
- The orbital speed of the ISS.
- The orbital period of the ISS (in minutes)

Imagine a hypothetical planet that has two moons. Moon #1 is in a circular orbit of radius R and has a mass M.

- Suppose Moon #2 has a mass M, equal to Moon #1 but has an orbital radius twice that of Moon #1, 2R. What is the ratio of the force the planet exerts on Moon #1 compared to the force the planet exerts on Moon #2.
- Moon 1 has a orbital radius of R. Suppose that Moon #2 has a mass of 3M (three times greater than Moon #1). Where must moon #2 be located (in terms of R) such that it experiences the same gravitational force as Moon #1?

*differential gravitational forces*.

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Kinematics | Forces |
---|---|

\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |

v = v_i + a \cdot t | F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu \cdot N |

R = \frac{v_i^2 \cdot \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{m \cdot v^2}{r} | KE = \frac{1}{2} m \cdot v^2 |

a_c = \frac{v^2}{r} | PE = m \cdot g \cdot h |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |

J = \Delta p | I = \sum m \cdot r^2 |

p_i = p_f | L = I \cdot \omega |

Simple Harmonic Motion |
---|

F = -k \cdot x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kg·m/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (N·m)} |

I (Moment of Inertia) | \text{kilogram meter squared (kg·m}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!