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 r 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:
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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?
A person’s back is against the inner wall of spinning cylinder with no support under their feet. If the radius is R, find an expression for the minimum angular speed so the person does not slide down the wall. The coefficient of static friction is µs.
Note: If you haven’t studied angular velocity \omega yet, just find the linear velocity v.
A ball of mass M is attached to a string of length L. It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is 3 times the weight of the ball. Give all answers in terms of M, L, and g.
Two identical object rests on a platform rotating at constant speed. Object A is at distance of half the platform’s radius from the center. Object B lays at edge of the platform. Assuming the platform continues rotating at the same speed, how does the centripetal force of the two objects compare?
A 2 kg ball is swung in a vertical circle. The length of the string the ball is attached to is 0.7 m. It takes 0.4 s for the ball to travel one revolution ( assume ball travels at constant speed).
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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_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
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 |
Momentum | Torque and Rotations |
---|---|
p = m v | \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 |
---|
F = -k 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} |
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 |
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 (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.
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 |
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