AP Physics Unit

Unit 3 - Circular Motion

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?

The Moon has a greater influence on Earth’s tides than the Sun due to the differential gravitational forces.

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.

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The Moon has a greater influence on Earth’s tides than the Sun due to the differential gravitational forces.

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\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2F = m \cdot a
v = v_i + a \cdot tF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m \cdot v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum m \cdot r^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k \cdot x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring constant, in \text{N/m}
VariableSI 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)}
VariableDerived 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

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

  1. Start with the given measurement: \text{5 km}

  2. 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}}

  3. 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}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}



Power of Ten




















(Base unit)


Deca- or Deka-


















  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. 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 .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. 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!
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