1. Law of Universal Gravitation
AP Physics
Sections 5-6 to 5-7
Law of Universal Gravitation

2. Newton’s Law of Universal Gravitation
If the force of gravity is being exerted on objects on Earth, what is the origin of that force?
Newton’s realization was that the force must come from the Earth. Fg
He further realized that this force must be what keeps the Moon in its orbit.

3. The gravitational force on you (and on the moon) is one-half of an action–reaction pair: the Earth exerts a downward force on you, and you exert an upward force on the Earth.
When there is such a disparity in masses as that between you and the Earth, the reaction force is undetectable, but for bodies more equal in mass it can be significant. Fg
Gravitational force of Earth on the moon.
Gravitational force of moon on the Earth.
barycenter: center of mass of Earth–moon system. Fg

4. Since the force is mutual to both objects, Newton realized that the gravitational force must be proportional to both masses.
Using the data for planetary orbits presented by Johannes Kepler, and using Kepler’s three laws of planetary motion, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses.
His law is known as the Law of Universal Gravitation. Fg
m1 m2 r2 = G
r is the distance between the two masses!
Where G is a proportionality constant called the universal gravitational constant.
Where G is a proportionality constant called the universal gravitational constant.

5. Universal gravitational constant
The value for the gravitational constant could be estimated roughly using Kepler’s data, but it was not until 1798 that Henry Cavendish performed his famous and ingenious experiment to calculate G.
Cavendish calculated a value of:
G = 6.67 × 10–11 Nm2/kg2

6. Gravity Near the Earth’s Surface Geophysical Applications
Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on the surface of the Earth the force on an object is:
m is the object mass
Fg = mg
The force between the Earth and the object is:
mE is the Earth’s mass Fg
m mE
rE2 = G
rE is the Earth’s radius
Setting these equal we get: mg
m mE
rE2 = G

7. g mE rE2 = G mE g rE2 G = g = 9.81 m/s2 rE = 6.38 × 106 m
Canceling the mass of the object we get the value for g, the acceleration due to gravity on the Earth’s surface: g mE
rE2 = G
Once again, we see g does not depend on the mass of the object!
We can rearrange this equation now to find the Earth’s mass: mE
g rE2 G =
g = 9.81 m/s2
rE = 6.38 × 106 m
G = 6.67 × 10–11 Nm2/kg2
mE = 5.98 × 1024 kg

8. The acceleration due to gravity, g
The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.
The earth is an “oblate spheroid.” In other words it’s flattened somewhat like an M&M, but not that much.