^{6} 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/s^{2}？

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

1 | g = \frac{GM}{r^2} | Gravitational acceleration formula, where G is the gravitational constant, M is mass, and r is radius. |

2 | g_{\text{earth}} = \frac{GM_{\text{earth}}}{r_{\text{earth}}^2} | Applying the formula to Earth. |

3 | g_{\text{planet}} = \frac{GM_{\text{planet}}}{r_{\text{planet}}^2} | Applying the formula to the unknown planet. |

4 | M_{\text{earth}} = M_{\text{planet}} | Given that the mass of the planet is the same as Earth’s mass. |

5 | g_{\text{earth}} r_{\text{earth}}^2 = g_{\text{planet}} r_{\text{planet}}^2 | Equating the two expressions since the masses are equal. |

6 | r_{\text{planet}} = r_{\text{earth}} \sqrt{\frac{g_{\text{earth}}}{g_{\text{planet}}}} | Solving for the radius of the planet. |

Let’s calculate the radius of the planet.

The radius of a planet that has the same mass as Earth but a free-fall acceleration of 5.50 m/s² is approximately \boxed{8.51 \times 10^6, \text{m}}.

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- Statistics

*g* on HD 69830b is most nearly

A spacecraft somewhere in between the earth and the moon experiences 0 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 ~3.844 x 10^{8} m.

NOTE: You may need the mass of the earth and moon. You can find this in the formula table.

*v _{A}* . Satellite B has an orbital radius nine times that of satellite A. What is the speed of satellite B?

A satellite circling Earth completes each orbit in 132 minutes.

^{24}.

^{6} m from the center of a larger object whose mass is 7.4 x 10^{22} kg.

^{5} kg and orbits Earth at a distance of 4.0 x10^{2} km above the surface. Earth has a radius of 6.37 x10^{6} m, and mass of 5.97 x10^{24} kg. Calculate the following:

Why is the force from the moon credited for the tides, and not the force from the sun?

R = 8.51 × 10^{6} m

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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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