| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | g = \frac{G M}{R^2} | The formula to calculate the gravitational field at the surface of a planet where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet. |
| 2 | g_{\text{Earth}} = \frac{G M_{\text{Earth}}}{R_{\text{Earth}}^2} | Apply the gravitational field formula to Earth. Given that the gravitational field on Earth (standard gravity) is approximately 9.8 \, \text{m/s}^2 and R_{\text{Earth}} is approximately 6400 km or 6.4 \times 10^6 m. |
| 3 | M_{\text{new planet}} = \frac{1}{2} M_{\text{Earth}} | The mass of the new planet is defined to be half that of Earth. |
| 4 | g_{\text{new planet}} = \frac{G (\frac{1}{2} M_{\text{Earth}})}{R_{\text{new planet}}^2} | Substitute the mass of the new planet into the gravitational field formula. |
| 5 | 20 = \frac{G (\frac{1}{2} M_{\text{Earth}})}{R_{\text{new planet}}^2} | Given that g_{\text{new planet}} = 20 \, \text{N/kg}, we set this equal to the derived expression for gravitational field on the new planet. |
| 6 | g_{\text{Earth}} \cdot R_{\text{Earth}}^2 = 9.8 \cdot (6.4 \times 10^6)^2 | Substitute the given values for Earth’s gravitational field and radius to calculate this product for further use. |
| 7 | \frac{1}{2} \cdot 9.8 \cdot (6.4 \times 10^6)^2 = 20 R_{\text{new planet}}^2 | Equating the expressions from Earth and the new planet to find the radius of the new planet. |
| 8 | R_{\text{new planet}} = \sqrt{\frac{1}{2} \cdot \frac{9.8 \cdot (6.4 \times 10^6)^2}{20}} | Solving for R_{\text{new planet}} using the simplified expression. |
| 9 | R_{\text{new planet}} \approx 3200 \, \text{km} | The radius of the new planet is equal to half that of Earth because of the proportional division of mass and the square of the radius in calculating gravitational field strength. |
Phy can also check your working. Just snap a picture!
Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the radius of a geosynchronous orbit about the neutron star? The mass of the sun can be found in the formula sheet above.
A satellite circling Earth completes each orbit in 132 minutes.
A child on Earth has a weight of 500N. Determine the weight of the child if the earth was to triple in both mass and radius (3M and 3r).
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?
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.
3200 km
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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} | |
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} |
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