Step | Formula Derivation | Reasoning |
---|---|---|

1 | \frac{v_{A}^2}{r_A} = \frac{GM}{r_A^2}, \quad \frac{v_{B}^2}{9r_A} = \frac{GM}{(9r_A)^2} | Centripetal force equals gravitational force for both satellites. |

2 | v_{A}^2 = \frac{GM}{r_A}, \quad v_{B}^2 = \frac{GM}{9r_A} | Simplifying each equation. |

3 | \frac{v_{B}^2}{v_{A}^2} = \frac{\frac{GM}{9r_A}}{\frac{GM}{r_A}} | Ratio of the squared velocities of Satellite B and A. |

4 | \frac{v_{B}^2}{v_{A}^2} = \frac{1}{9} | Simplifying the ratio. |

5 | v_{B} = \frac{v_{A}}{3} | Taking the square root of the ratio. |

Thus, the speed of Satellite B is one-third the speed of Satellite A, or v_{B} = \frac{v_{A}}{3} . This result arises from the relationship between orbital speed and radius for satellites in circular orbits.

Phy can also check your working. Just snap a picture!

- Statistics

Intermediate

Mathematical

GQ

A communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be 6 \times 10^{24} \text{ kg} the the radius of Earth to be 6.4 \times 10^6 \text{ m}. What is the satellite’s velocity?

- Centripetal Acceleration, Circular Motion, Gravitation, Linear Forces

Intermediate

Mathematical

GQ

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 speed of a point on the equator of the star?

- Circular Motion

Intermediate

Mathematical

FRQ

A linear spring of negligible mass requires a force of 18.0 N to cause its length to increase by 1.0 cm. A sphere of mass 75.0 g is then attached to one end of the spring. The distance between the center of the sphere M and the other end P of the un-stretched spring is 25.0 cm. Then the sphere begins rotating at constant speed in a horizontal circle around the center P. The distance P and M increases to 26.5 cm.

- Circular Motion, Linear Forces

Beginner

Mathematical

GQ

A 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.

- Circular Motion

Advanced

Proportional Analysis

GQ

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.

- Circular Motion, Gravitation

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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

Conversion Example

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 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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