Circular Motion and Kepler's Laws

Circular Motion: Unbanked Curves

Free Body Diagram for a Car on a Curve: When a car moves at a steady speed around a curve, it experiences forces that can be represented in a free body diagram.
Centripetal Force on Unbanked Curves: The centripetal force that keeps the car moving on the curve is provided by static friction between the road and the tires.

Vertical Equilibrium: The gravitational force and the normal force balance each other out, resulting in no net motion in the vertical direction, provided there isn't any additional force acting vertically.
Radial Force (Centripetal Force): A radial force, directed towards the center of the circle, is necessary to keep the car moving along the circular path.
Static Friction as Centripetal Force: This radial force is supplied by static friction, not kinetic friction. Static friction is crucial because the tires are prevented from slipping in the radial direction.
Insufficient Friction: If the static frictional force is insufficient for the given speed and radius of the turn, the car will skid off the road.
Example of Icy Road: An icy road reduces the force of friction, limiting safe driving speeds.

Friction Coefficient Comparison: If \mus for a road on a dry day = 0.5, \mus for the same road on an icy day is < 0.5.

Equation Derivation: For a car traveling on a horizontal turn with radius R, we can define the centripetal force.
Maximum Speed Expression: Using the centripetal force equation, we derive the expression for the maximum speed the car can have without slipping towards the center of the curve:
v{max} = \sqrt{rg\mus}
Where:
- v_{max} is the maximum speed
- r is the radius of the curve
- g is the acceleration due to gravity
- \mu_s is the coefficient of static friction
Driving on the Moon vs. Earth: Other things being equal, it would be easier to drive at high speed around an unbanked horizontal curve on the Moon than on Earth because the gravitational force (and thus the required friction) would be less.

Tension Relationship: Consider two identical stones attached to cords being whirled on a tabletop at the same speed. The radius of the larger circle is twice that of the smaller circle. The tension T1 in the longer cord is related to the tension T2 in the shorter cord by the equation T1 = 2T2. This is because the tension is proportional to the radius in circular motion, given the mass and speed are constant.

Skid Assessment: A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s. To determine if the car will skid:
- On a dry day in Summer (\mu_s = 0.6)
- On an icy day in Winter (\mu_s = 0.25)
  Calculations are needed to compare the required centripetal force with the maximum static friction force in each scenario.

Eliminating Reliance on Friction: The reliance on friction to provide centripetal force can be eliminated completely for a given speed if the curve is banked at an angle relative to the horizontal, similar to how a plane is banked while making a turn.
Friction-Free Banked Curve: Consider a car going around a friction-free banked curve of radius r, where r is measured parallel to the horizontal.

Components:
- F_N \cos \theta: Vertical component of the normal force
- F_N \sin \theta: Horizontal component of the normal force
- mg: Gravitational force

Angle Calculation: For a car traveling with speed v along a curve of radius r, the formula for the angle at which a road should be banked so that no friction is required is derived as follows:
FN \cos \theta = mg FN \sin \theta = \frac{mv^2}{r}
Dividing the second equation by the first:
\tan \theta = \frac{v^2}{gr}

Angle Determination:
Determine this angle for a road which has a curve of radius 50 m with a design speed of 50 km/h:
v = 50 \frac{km}{h} = 50 \times \frac{1000 m}{3600 s} = 13.89 m/s
\theta = \arctan(\frac{v^2}{gr}) = \arctan(\frac{13.89^2}{9.8 \times 50}) \approx 21.5^{\circ}

Speed Calculation: The Daytona 500 is held at the Daytona International Speedway, where turns have a maximum radius of 316 m and are banked at an angle of 31°. If these turns were frictionless, the speed at which cars would have to travel around them can be calculated using:
v = \sqrt{gr \tan \theta} = \sqrt{9.8 \times 316 \times \tan(31^{\circ})} \approx 43.1 m/s

Prehistoric Observations: Our ancestors noticed that celestial objects follow a periodic and predictable path in the sky.
Timekeeping and Navigation: They used the motions of the celestial objects to keep time and determine directions.

Underpinnings:
- Earth at the center of the universe
- Heavens must be "perfect": Objects moving on perfect spheres or in perfect circles.
- But the Ptolemaic model couldn’t explain the motion of planets completely.

Heliocentric Belief: Kepler strongly believed in the heliocentric model.
Kepler's Laws of Planetary Motion: He proposed Kepler's laws of planetary motion.

Source: What produces the centripetal force on a planet due to the Sun?
Equation: Write down an equation for the centripetal force on the planet due to the Sun.

Calculation: Determine the mass of the Sun given the Earth’s distance from the Sun is 1.5 \times 10^{11} m. (Assume circular orbit, G = 6.67 \times 10^{-11} N \cdot m^2/kg^2)

Law of Ellipses: The path of each planet around the Sun is an ellipse, with the Sun at one focus.
Ellipse Properties:
- The longer axis = major axis
- The shorter axis = minor axis
- Semi-major axis (a) = major axis/2
- Semi-minor axis (b) = minor axis/2

Law of Equal Areas: Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal area in equal periods of time.

Law of Harmonies: The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit.
T^2 \propto a^3

Mars Orbital Distance: One year of Mars is equal to 687 Earth days. Determine the mean distance of Mars from the Sun using Earth as a reference (e.g., Mars is x times Earth’s distance from the Sun).