AP Physics Unit

Unit 3 - Circular Motion

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Mathematical

FRQ

The International Space Station has a mass of 4.2 x105 kg and orbits Earth at a distance of 4.0 x102 km above the surface. Earth has a radius of 6.37 x106 m, and mass of 5.97 x1024 kg. Calculate the following:

  1. (a) The gravitational force between Earth and the ISS. (2 points)
  2. (b) The orbital speed of the ISS. (2 points)
  3. (c) The orbital period of the ISS (in minutes) (3 points)
  1. 3.65 x 106 N
  2. 7,672 m/s
  3. 92.41 minutes

1. Gravitational Force Between Earth and the ISS

Step Formula Derivation Reasoning
1 F = G \frac{M_{\text{earth}} M_{\text{ISS}}}{r^2} Newton’s law of universal gravitation.
2 r = R_{\text{earth}} + h_{\text{ISS}} Distance r is Earth’s radius plus ISS’s altitude.

2. Orbital Speed of the ISS

Step Formula Derivation Reasoning
1 F_{\text{gravity}} = \frac{GM_{\text{earth}}M_{\text{ISS}}}{r^2} Gravitational force between Earth and ISS.
2 F_{\text{centripetal}} = \frac{M_{\text{ISS}}v^2}{r} Centripetal force required for circular orbit.
3 F_{\text{gravity}} = F_{\text{centripetal}} For a stable orbit, gravitational force equals centripetal force.
4 \frac{GM_{\text{earth}}M_{\text{ISS}}}{r^2} = \frac{M_{\text{ISS}}v^2}{r} Equating the two forces.
5 GM_{\text{earth}} = rv^2 Cancelling M_{\text{ISS}} and rearranging.
6 v = \sqrt{\frac{GM_{\text{earth}}}{r}} Solving for orbital velocity v. Note that is the total distance from the center of earth to ISS.

3. Orbital Period of the ISS

Step Formula Derivation Reasoning
1 T = \frac{\text{Circumference of Orbit}}{\text{Orbital Speed}} Orbital period T is the time to complete one orbit.
2 \text{Circumference} = 2\pi r Circumference formula for a circle.
3 T = \frac{2\pi r}{v} Substituting the circumference and orbital speed v.
4 T_{\text{minutes}} = \frac{T}{60} Converting period from seconds to minutes.

Let’s perform the calculations using the given values.

The calculations yield the following results:

  1. Gravitational force between Earth and the ISS: \boxed{3,651,338, \text{N}}
  2. Orbital speed of the ISS: \boxed{7,671.77, \text{m/s}}
  3. Orbital period of the ISS: \boxed{92.41, \text{minutes}}

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  1. 3.65 x 106 N
  2. 7,672 m/s
  3. 92.41 minutes

Nerd Notes

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
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}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
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 (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^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}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

  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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