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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.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
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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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