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Understanding Universal Gravity and Newton’s Law

Delve into the concept of universal gravity and Newton’s Law of Universal Gravitation, exploring how gravity affects objects on Earth and elsewhere in the universe. Learn about the relevant equations and calculations involved in determining gravitational force and acceleration due to gravity. Discover how gravitational force varies with distance and how it impacts weight on different celestial bodies like the moon.

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Understanding Universal Gravity and Newton’s Law

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  1. Outline Universal Gravity • Newton’s Law of Universal Gravitation • Course evaluations

  2. On Earth: Fg = mg, where m is the mass of an object and g is the acceleration due to gravity.This equation is valid only near the surface of the earth ! m Earth

  3. Anywhere else, Fg =GMm/r2, where M and m are the masses, r is their separation and G is the universal gravitational constant. M m r

  4. We can use F=GMm/r2 on the surface of the earth, but the equation is more complicated. We need to:- use G=6.67x10-11 Nm2/kg2- use the mass of the earth, M=5.98x1024kg- and know the distance of the object from the center of the earth, Re = 6370kmSmall differences in r (eg, 10, 100 or 1000m) will not make a large difference in the force.So, how do we get to F=mg ???

  5. The force of gravity is: F=GMm/r2and we also know that: F=maIf we equate these two we get: ma=GMm/r2 a =GM/r2Subbing in G, M (earth) and r (radius of earth) we get: a=9.8m/s2

  6. Because F=GMm/r2, we can see that F will decrease with r. Also, since g=GM/r2, g will decrease with r as well: at h=100km g=9.5m/s2 at h=1000km g=7.3m/s2 h r

  7. Eg: The mass of the moon is M=7.36x1022 kg, its radius is r=1.74x106m. What is the value of g on the moon? (6.67x10-11Nm2/kg2)(7.36x1022 kg) g=GM/r2 = ___________________________ (1.74x106 m)2= 1.6 m/s21.6/9.8 = 1/6, therefore, you would weight about 1/6 as much on the moon as you do on earth !

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