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1. Gravitational Force Between Earth and the ISS
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F = G \frac{M_{\text{earth}} M_{\text{ISS}}}{r^2}[/katex] | Newton’s law of universal gravitation. |
| 2 | [katex]r = R_{\text{earth}} + h_{\text{ISS}}[/katex] | Distance [katex]r[/katex] is Earth’s radius plus ISS’s altitude. |
2. Orbital Speed of the ISS
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F_{\text{gravity}} = \frac{GM_{\text{earth}}M_{\text{ISS}}}{r^2}[/katex] | Gravitational force between Earth and ISS. |
| 2 | [katex]F_{\text{centripetal}} = \frac{M_{\text{ISS}}v^2}{r}[/katex] | Centripetal force required for circular orbit. |
| 3 | [katex]F_{\text{gravity}} = F_{\text{centripetal}}[/katex] | For a stable orbit, gravitational force equals centripetal force. |
| 4 | [katex]\frac{GM_{\text{earth}}M_{\text{ISS}}}{r^2} = \frac{M_{\text{ISS}}v^2}{r}[/katex] | Equating the two forces. |
| 5 | [katex]GM_{\text{earth}} = rv^2[/katex] | Cancelling [katex]M_{\text{ISS}}[/katex] and rearranging. |
| 6 | [katex]v = \sqrt{\frac{GM_{\text{earth}}}{r}}[/katex] | Solving for orbital velocity [katex]v[/katex]. 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 | [katex]T = \frac{\text{Circumference of Orbit}}{\text{Orbital Speed}}[/katex] | Orbital period [katex]T[/katex] is the time to complete one orbit. |
| 2 | [katex]\text{Circumference} = 2\pi r[/katex] | Circumference formula for a circle. |
| 3 | [katex]T = \frac{2\pi r}{v}[/katex] | Substituting the circumference and orbital speed [katex]v[/katex]. |
| 4 | [katex]T_{\text{minutes}} = \frac{T}{60}[/katex] | 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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A 2.2 kg ball on the end of a 0.35 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 5.3 m/s. Find the tension in the string.
A centripetal force of \( 5.0 \) newtons is applied to a rubber stopper moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed, \( v \), and the frequency, \( f \), of the stopper?

Refer to the diagram above and solve all equations in terms of \(R\), \(M\), \(k\), and constants.
A car is going through a dip in the road whose curvature approximates a circle of radius \( 200 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 20\% \) more than their normal weight \( (1.2\,W) \)?
A spacecraft somewhere in between the Earth and the Moon experiences zero net force acting on it. This is because the Earth and the Moon pull the spacecraft in equal but opposite directions. Find the distance \(D\) away from Earth such that the spacecraft experiences zero net force. The distance between the Moon and Earth is \(\sim 3.844 \times 10^8 \, \text{m}\).
Note: You may need the mass of the Earth and the Moon. You can find this in the formula table.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\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}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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