The net gravitational force on the 2.0 kg sphere will be the vector sum of the forces exerted by the 4.0 kg and 7.0 kg spheres. Since the 2.0 kg sphere is midway, each force will act along the line connecting the centers of the spheres. See the FBD below.

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

1 | r_{\text{half}} = \frac{1.2 , \text{m}}{2} = 0.6 , \text{m} | Half the distance between the two spheres. |

2 | F_{\text{4 to 2}} = G \frac{4 \times 2}{0.6^2} | Gravitational force between the 4.0 kg and 2.0 kg spheres. |

3 | F_{\text{7 to 2}} = G \frac{7 \times 2}{0.6^2} | Gravitational force between the 7.0 kg and 2.0 kg spheres. |

4 | F_{\text{net}} = F_{\text{7 to 2}} – F_{\text{4 to 2}} | Net force is the difference between the two forces, as they are in opposite directions. |

Let’s calculate the net gravitational force.

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

4 | F_{\text{net}} \approx 1.11 \times 10^{-9} , \text{N} | Calculated net gravitational force. |

The net gravitational force on the 2.0 kg sphere, located midway between the 4.0 kg and 7.0 kg spheres (1.2 m apart), is approximately \boxed{1.11 \times 10^{-9} , \text{Newtons}} . This force is directed towards the 7.0 kg sphere due to its larger mass.

- The blue arrow represents the gravitational force ( F_{4 \to 2} ) exerted by the 4.0 kg sphere.
- The green arrow represents the gravitational force ( F_{7 \to 2} ) exerted by the 7.0 kg sphere.

These forces indicate the gravitational pull exerted on the 2.0 kg sphere from each of the other two spheres. In this scenario, the net force is the vector sum of these two forces, with the direction towards the 7.0 kg sphere due to its larger mass.

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- Statistics

Advanced

Mathematical

GQ

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.

- Gravitation, Linear Forces

Intermediate

Proportional Analysis

GQ

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).

- Gravitation, Linear Forces

Advanced

Proportional Analysis

MCQ

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

- Circular Motion, Gravitation

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

Intermediate

Proportional Analysis

FRQ

Imagine a hypothetical planet that has two moons. Moon #1 is in a circular orbit of radius R and has a mass M.

- Gravitation

*F _{net}* = 1.11 x 10

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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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