Part (a) – Finding the orbital speed.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | r = \frac{50}{2} = 25 \text{km} = 25000 \text{m} | The radius r of the orbit is half the diameter. Convert it to meters. |
| 2 | T = 11 \times 86400 s | Convert period T from days to seconds. There are 86400 seconds in a day. |
| 3 | v = \frac{2\pi r}{T} | The orbital speed v is calculated by dividing the circumference of the orbit by the orbital period. |
| 4 | v = \frac{2\pi \times 25000}{950400} m/s | Substitute the values of r and T into the formula. Convert r from km to m by multiplying by 1000. |
| 5 | v \approx 0.165 m/s | Simplifying the expression gives the orbital speed v . |
| 6 | \boxed{v \approx 0.165 \text{m/s}} | This is the final value for the satellite’s orbital speed. |
Part (b) – Finding the comet’s mass.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | v^2 = \frac{GM}{r} | The inwards gravitational force is equal to the centripetal force of the orbiting comet. In terms of Newtons law this can be expressed as \frac{GMm}{r^2} = \frac{mv^2}{r} , where M is the mass of the comet, m is the mass of the satellite, and G is the gravitational constant. |
| 2 | M = \frac{rv^2}{G} | Rearrange the formula to solve for the mass M of the comet. |
| 3 | M = \frac{25000 \times (0.165)^2}{6.674 \times 10^{-11}} kg | Substitute the values of r and v into the formula, remembering that r is already converted to meters. |
| 4 | M \approx 1.02 \times 10^{13} kg | Calculating the value gives the mass of the comet. |
| 5 | \boxed{M \approx 1.02 \times 10^{13} \text{kg}} | This is the final value for the mass of the comet. |
Part (c) – Finding the landing speed.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | h = 25\,\text{km}\, -\,1.8 \, \text{km} = 23.2 \, \text{km} = 23200 \, \text{meters} | The distance from the satellite to the center of the comet is 25km. Since the comet has an average diameter of 3.6 km (or a radius of 1.8 km). The distance from the satellite to the surface of comet is 23.2 km. Convert this to meters. |
| 2 | t = 7 \times 3600 s | Convert the fall time t from hours to seconds. |
| 3 | g = \frac{GM}{r^2} | Calculate the acceleration g due to the comet’s gravity using the values from the previous two parts. Set mg = \frac{GMm}{r^2} and solve for g . |
| 4 | g = \frac{(6.67\times 10^{-11}) \times (1.02 \times 10^{13})}{25000^2} | Substitute values into the equation for gravitational acceleration. |
| 5 | g = 1.09 \times 10^{-6} \, \text{m/s}^2 | Final value for g ,the acceleration due to gravity of the comet. |
| 6 | v^2_{\text{final}} = v^2_{\text{initial}} + 2a\Delta x | Now that we have v_{initial}, \, a, \, \Delta x we can use a kinematic formula to solve for v_f . |
| 7 | \boxed{v_{\text{final}} \approx .735\, \text{m/s}} | Plug in all known values and solve for v_f . |
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A 2.00 x102 g block on a 50.0 cm long string swings in a circle on a horizontal, frictionless table at 75.0 rpm. What is the speed of block? What is the tension in the string?
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.
A discus is held at the end of an arm that starts at rest. The average angular acceleration of 54 \, \text{rad/s}^2 lasts for 0.25 s. The path is circular and has radius 1.1 m.
Note: A discuss is a heavy, flattened circular object for throwing.
A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius R.
Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:
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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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