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The given figure shows two circular objects, one with a larger mass M on the right side, and another with a smaller mass m on the left side. A point in the center of each object is shown, with both depicting the center of mass of the objects at these points. A line is drawn joining the center of the objects and is labeled as r. Two red arrows, one each from both the center of the objects, are drawn toward each other and are labeled as F, the magnitude of the gravitational force on both the objects.
Gravitational attraction is along a line joining the centers of mass of these two bodies. The magnitude of the force is the same on each, consistent with Newton’s third law.

Misconception alert

The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton’s third law.

The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass    (CM), which will be further explored in Linear Momentum and Collisions . For two bodies having masses m size 12{m} {} and M size 12{M} {} with a distance r size 12{r} {} between their centers of mass, the equation for Newton’s universal law of gravitation is

F = G mM r 2 , size 12{F=G { { ital "mM"} over {r rSup { size 8{2} } } } } {}

where F size 12{F} {} is the magnitude of the gravitational force and G size 12{G} {} is a proportionality factor called the gravitational constant . G size 12{G} {} is a universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be

G = 6 . 674 × 10 11 N m 2 kg 2 size 12{G=6 "." "673" times "10" rSup { size 8{ - "11"} } { {N cdot m rSup { size 8{2} } } over {"kg" rSup { size 8{2} } } } } {}

in SI units. Note that the units of G size 12{G} {} are such that a force in newtons is obtained from F = G mM r 2 size 12{F=G { { ital "mM"} over {r rSup { size 8{2} } } } } {} , when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of 6 . 674 × 10 11 N size 12{6 "." "673" times "10" rSup { size 8{ - "11"} } N} {} . This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of 6 × 10 24 kg size 12{6 times "10" rSup { size 8{"24"} } `"kg"} {} .

Recall that the acceleration due to gravity g size 12{g} {} is about 9.80 m /s 2 size 12{9 "." 8`"m/s" rSup { size 8{2} } } {} on Earth. We can now determine why this is so. The weight of an object mg is the gravitational force between it and Earth. Substituting mg for F size 12{F} {} in Newton’s universal law of gravitation gives

mg = G mM r 2 , size 12{ ital "mg"=G { { ital "mM"} over {r rSup { size 8{2} } } } } {}

where m size 12{m} {} is the mass of the object, M size 12{M} {} is the mass of Earth, and r size 12{r} {} is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See [link] . The mass m size 12{m} {} of the object cancels, leaving an equation for g size 12{g} {} :

g = G M r 2 . size 12{g=G { {M} over {r rSup { size 8{2} } } } } {}

Substituting known values for Earth’s mass and radius (to three significant figures),

g = 6 . 67 × 10 11 N m 2 kg 2 × 5 . 98 × 10 24 kg ( 6 . 38 × 10 6 m ) 2 , size 12{g= left (6 "." "67" times "10" rSup { size 8{ - "11"} } { {N cdot m rSup { size 8{2} } } over {"kg" rSup { size 8{2} } } } right ) times { {5 "." "98" times "10" rSup { size 8{"24"} } " kg"} over { \( 6 "." "38" times "10" rSup { size 8{6} } " m" \) rSup { size 8{2} } } } } {}

and we obtain a value for the acceleration of a falling body:

g = 9 . 80 m/s 2 . size 12{g=9 "." "80"" m/s" rSup { size 8{2} } } {}
The given figure shows two circular images side by side. The bigger circular image on the left shows the Earth, with a map of Africa over it in the center, and the first quadrant in the circle being a line diagram showing the layers beneath Earth’s surface. The second circular image shows a house over the Earth’s surface and a vertical line arrow from its center to the downward point in the circle as its radius distance from the Earth’s surface. A similar line showing the Earth’s radius is also drawn in the first quadrant of the first image in a slanting way from the center point to the circle path.
The distance between the centers of mass of Earth and an object on its surface is very nearly the same as the radius of Earth, because Earth is so much larger than the object.

This is the expected value and is independent of the body’s mass . Newton’s law of gravitation takes Galileo’s observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in fact, in terms of a universally existing force of attraction between masses.

Take-home experiment

Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations.

Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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