## Newton’s Law of Universal Gravitation

12.1 Newton’s Law of Universal Gravitation

### Learning Objectives

By the end of this section, you will be able to:

- List the significant milestones in the history of gravitation
- Calculate the gravitational force between two point masses
- Estimate the gravitational force between collections of mass

### Newton’s Law of Universal Gravitation

Newton’s law of gravitation can be expressed as

where is the force on object 1 exerted by object 2 and is a unit vector that points from object 1 toward object 2.

Practice 12.1.1 |
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Is Earth’s gravitational force on the moon larger than, smaller than, or equal to the moon’s gravitational force on Earth? |

(a) Earth’s gravitational force on the moon is larger than the moon’s gravitational force on Earth |

(b) Earth’s gravitational force on the moon is smaller than the moon’s gravitational force on Earth |

(c) Earth’s gravitational force on the moon is equal to the moon’s gravitational force on Earth |

**Practice!**

Practice 12.1.2 |
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The space shuttle orbits about 200 km above Earth’s surface. The shuttle’s “pre-launch weight” is its weight measured on the ground. (Neglecting the fuel) The magnitude of the force of gravity on the shuttle while it is in orbit is: |

(a) Slightly greater than pre-launch. |

(b) Equal to pre-launch |

(c) Slightly less than pre-launch |

(d) Lots less than pre-launch, almost (but not quite) zero. |

(e) Precisely zero. |

Practice 12.1.3 |
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An astronaut is floating around in the space shuttle’s cabin. Her acceleration, as measured from the earth’s surface, is |

(a) zero – she’s floating |

(b) very small, in some random direction |

(c) quite large, nearly g, directed towards the center of the earth |

(d) quite large, nearly g, directed along the line of travel of the shuttle |

She’s in orbit, so she’s in uniform circular motion. She has a large acceleration! We just said F_{(grav)} is almost the same up there, so her acceleration is almost the same as on earth in freefall, namely, nearly *g*, directed towards the center of the earth. She’s just like a ball that’s been tossed up in the air, accelerating straight down. The only difference is she has such a large sideways speed that when she has fallen 200 km, she’s gone so far to the side that the curved surface of the earth has moved 200 km away from her, so she’s still 200 km above the ground!!!

Practice 12.1.4 |
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Three 5.0-kg masses are located at points in the xy plane, as shown. What is the magnitude of the resultant force (caused by the other two masses) on the mass at x = 0, y = 0.30 m? |

(a) 2.6 × 10^{–8} N |

(b) 2.0 × 10^{–8} N |

(c) 2.9 × 10^{–8} N |

(d) 2.3 × 10^{–8} N |

(e) 2.1 × 10^{–8} N |

Practice 12.1.5 |
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At some instant in time, two asteroids in deep space are a distance r = 20 km apart. Asteroid 2 has 10 times the mass of asteroid 1. What is the ratio of their resulting acceleration (due to gravitational attraction to each other), a_{1} / a_{2} = |

(a) 10:1 |

(b) 1:1 |

(c) 1:10 |

(d) Not enough information |

F_{(1 on 2)} = F_{(2 on 1)} (in magnitude), that’s Newton’s 3rd Law.

(Or, use F = G M_{1}M_{2}/r^{2}, it’s the same if you reverse M_{1} and M_{2)} Since F_{net} = ma, and F_{net} is the same for each, the acceleration of the little one must be 10 times bigger than the acceleration of the big one, 10:1.

Practice 12.1.6 |
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A rock is released from rest, in deep space, far from the earth. At a certain point, as it falls towards us, it crosses the moon’s orbit. When the rock is the same distance from the earth as the moon, is the acceleration of the rock greater than, equal to, or smaller than the acceleration of the moon? |

(a) Greater than |

(b) Equal to |

(c) Smaller than |

(d) Not enough information |

The only force acting on either one is gravity, F_{(grav)} = GM_{(earth)}m /r^{2}. The acceleration is then F/m, and that’s the same for both!!

Practice 12.1.7 |
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A rock is released from rest, in deep space, far from the earth. At a certain point, as it falls towards us, it crosses the moon’s orbit. As the rock falls toward the Earth, its acceleration is… |

(a) constant. |

(b) not constant. |

As it approaches, *r* is getting smaller so GM/r^{2} is getting bigger. The acceleration is thus getting larger. It’s *not* constant.

(It’s only approximately constant when you’re NEAR the surface, because *r* isn’t changing very much as you go from 6000 km + 0 to 6000 km + 20 feet, for example.)

**Discuss!**

Estimate the force that the moon exerts on you when it is directly overhead.