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A 1.5 kg object is located at a distance of 1.7 x106 m from the center of a larger object whose mass is 7.4 x 1022 kg.
Two satellites are in circular orbits around Earth. Satellite A has speed vA . Satellite B has an orbital radius nine times that of satellite A. What is the speed of satellite B?
Find the net gravitational force on a 2.0 kg sphere midway between a 4.0 kg sphere and a 7.0 kg sphere that are 1.2 m apart.
The distance from earth to sun is 1.0 AU. The distance from Saturn to sun is 9 AU. Find the period of Saturn’s orbit in years. You can assume that the orbits are circular.
A satellite circling Earth completes each orbit in 132 minutes.
The Earth’s radius is 6.37 x 106 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/s2？
A 3300-m-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The earth’s radius is 6400 km and the earth’s mass is 5.97 x 1024.
Two identical satellites are placed in orbit of two different planets. Satellite A orbits Mars, and Satellite B orbits Jupiter. The orbital speeds of each satellite are the same. Which satellite has a greater orbital radius?
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:
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?
Imagine a hypothetical planet that has two moons. Moon #1 is in a circular orbit of radius R and has a mass M.
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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
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} | |
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} |
