As mentioned in chapter 2 (here), Henry Cavendish used Newton’s laws to calculate the mass of the earth. You can do it too! Here’s how.

According to Newton’s law of universal gravitation, the strength of the gravitational force between any two objects depends on their masses and the distance between their centers:

F =G × m_{1} × m_{2} |

d^{2} |

And according to Newton’s second law of motion, the net force on an object is equal to the object’s mass times its acceleration:

F = maWhen gravity is the only force acting on an object, the gravitational force is equal to the net force. For example, when an apple falls from a tree to the ground, the net force on the apple is simply the force of Earth’s gravity. Technically, there are many other forces acting on the apple too. For example, there is an upward force from air resistance, and there are gravitational forces pulling the apple toward every other object that has mass. But these other forces are negligible, because they are weak compared to the force of Earth’s gravity. Air resistance doesn’t become significant unless the apple falls much farther than the height of an apple tree, and Earth is the only object massive enough and close enough to exert a gravitational pull that is noticeable. If we use the variables “m_{1}” and “m_{2}” for the masses of the apple and Earth, respectively, and use “a” for the apple’s acceleration, then we can combine the above equations. The gravitational force between the apple and Earth is equal to the apple’s mass times its acceleration:

G × m_{1} × m_{2} |

d^{2} |

Since the apple’s mass (m_{1}) appears on both sides of the equation, we can divide both sides by m_{1} and cancel it out:

G × m_{2} | |

d^{2} |

Now we can solve for the earth’s mass (m_{2}). Multiply both sides of the equation by d^{2} and divide both sides by G, and we have:

a × d^{2} | |

G |

Next, let’s fill in the values of the known variables. As it falls from the tree, the apple accelerates toward the ground at approximately 9.8 m/s^{2}, so a = 9.8 m/s^{2}. The variable d represents the distance between the center of the apple and the center of Earth, which is approximately 6,370,000 meters, or 6.37 × 10^{6} m in scientific notation. And the universal gravitational constant G is equal to 6.67 × 10^{-11} N m^{2}/kg^{2}. So, the mass of the earth is:

m_{2} =

m_{2} =

m_{2} =

m_{2} =

9.8 m/s^{2} × (6.37 × 10^{6} m)^{2} | |

6.67 × 10^{-11} N m^{2}/kg^{2} |

m

1.47 m/s^{2} × (6.37 × 10^{6} m)^{2} | |

10^{-11} N m^{2}/kg^{2} |

m

1.47 m/s^{2} × 4.06 × 10^{13} m^{2} | |

10^{-11} N m^{2}/kg^{2} |

m

5.97 × 10^{24} m^{3}/s^{2} | |

N m^{2}/kg^{2} |

That looks messy, so let’s clean up the units. Multiply the numerator and denominator both by square meters (m^{2}), to cancel those out:

m_{2} =

5.97 × 10^{24} m/s^{2} | |

N/kg^{2} |

Let’s also get rid of the unit fractions that appear in the numerator and denominator of the main fraction. Multiply the numerator and denominator of the main fraction by s^{2} kg^{2}, and we have:

m_{2} =

5.97 × 10^{24} m kg^{2} | |

N s^{2} |

Also, remember that a newton is equal to 1 kg m/s^{2}, so we can replace N with kg m/s^{2}. Then all units will cancel except kg:

m_{2} =

m_{2} =

m_{2} = 5.97 × 10^{24} kg

5.97 × 10^{24} m kg^{2} | |

(kg m/s^{2}) s^{2} |

m

5.97 × 10^{24} m kg^{2} | |

kg m |

m

And that is the mass of the earth!