Law of Universal Gravitation
The force of attraction F between two objects with respective masses m1 and m2 separated by a center-to-center distance R, anywhere in the universe is described by Newton's Law of Universal Gravitation:
The two postulates of this law are:
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The force of attraction is directly proportional to the product of the objects' masses
The force of attraction is inversely proportional to the distance (taken from their respective centers) squared
If we combine the above statements mathematically, we obtain the following relation:
We translate the above into a definite equation by introducing a constant G
Therefore: 
Where:
F is the force of attraction between two masses m1 and m2 anywhere in the Universe, and
and, G is the Universal Gravitational Constant: G = 6.67 X 10-11 Nm2 /Kg2
Notes:
Observe proper units (mass in kg, distance in m, Force in N)
The distance R has to be center-to center; i.e. from the center of one mass to the center of the other
Don't forget to square the distance R
Remember the significance of the Inverse Square Law
The value of g
Another way to calculate the acceleration due to gravity g is by using Newton's Law of Universal Gravitation
Recall that the Force of gravity around the surface of the Earth is calculated with the formula Fg = m x g
But we can also use Newton's Law of Universal Gravitation to obtain the same result. After all this equation is UNIVERSAL.
Therefore we can say that 
where m is the mass of any object at the surface of the Earth. M is the mass of the Earth G is the constant of Universal Gravitation (6.67 x 10-11 Nm2/kg2).
** Note that the mass m drops out from each side of the equation therefore the value of g on Earth can be calculated if we know the mass of the Earth and its radius R. This fact should not surprise us after all we've always stated that the acceleration due to gravity on an object is independent of the object's mass
Exercise 1:
Use the information that the mass of the Earth is 5.98 x 1024 kg and that its radius is 6.38 x 103 km (6.38 x 106 m) to confirm that g on Earth is approximately 9.8 N/kg
Solution:
Using the formula above we have:
Exercise 2:
Given the following data use graphical techniques to confirm that Newton's Law of Universal Gravitation is an inverse square law.
| Average Earth -Moon Distance (m) | Radius of Moon (m) | Radius of Earth (m) | Mass of Earth (kg) | Mass of Moon (kg) | Mass of Sun (kg) | Average Earth -Sun Distance (m) | Radius of the Sun (m) |
| 3.84 x 10 6 | 1.74 x 106 | 6.38 x 106 | 5.98 x 1024 | 7.35 x 1022 | 1.99 X 10 30 | 1.496 x 10 11 | 6.96 x 108 |
| | | | | | | | |
Solution:
1. The center to center distance between the Moon and the Earth is the distance between the two celestial bodies plus their respective radii.
2. i.e. Moon -Earth System
R1 = 1.74 x 106 m + 6.38 x 106 m + 3.84 x 10 6 m = 1.196 x 10 7 m
Sun -Earth System
R2 = 1.496 x 1011 m + 6.38 x 106 m + 6.96 x 10 8 m = 1.50 x 10 11 m
Find the forces using
Moon -Earth System
F1 = 2.05 x 1023 N
Sun -Earth System
F2 = 3.53 x 1022 N
Compile the data in a chart and Plot the data
System Forces (N) Distances (R)
(m)Distances-squared (R)2 , (m2) Product of Masses
(kg2)Earth-Moon 2.05 x 1037 1.196 x 10 7 1.430 x 10 14 4.57 x 1047 Earth-Sun 3.53 x 1022 1.50 x 10 11 2.25 x 10 22 1.19 x 1055 Here is the same data and its analysis using spreadsheets
System Forces (N) Distances (R) Distances-squared (R)2, (m2) Product of Masses Inverse Distances-squared (1/R)2 , (1/m2) (m) (kg2) Earth-Moon 2.05E+37 1.20E+07 1.43E+14 4.57E+47 6.99E-15 Earth-Sun 3.53E+22 1.50E+11 2.25E+22 1.19E+55 4.44E-23
This shows the relationship F α 1/R2
To confirm plot F vs. 1/R2
Note: F vs. 1/R2 gives us a linear relationship
