
Newton’s Law of Universal Gravitation describes the gravitational force between any two objects with mass.
Proposed by Isaac Newton, this law explains:
The law states that every mass attracts every other mass through a gravitational force.
Any two objects with mass exert an attractive gravitational force on each other.
The magnitude of the force is:
FG=Gm1m2r2F_G = \frac{Gm_1m_2}{r^2}FG=r2Gm1m2
where:
The constant GGG has the value:
G=6.67×10−11 N⋅m2/kg2G = 6.67\times10^{-11} \ \text{N}\cdot\text{m}^2/\text{kg}^2G=6.67×10−11 N⋅m2/kg2
This extremely small number explains why gravitational forces between everyday objects are usually negligible.
Gravity is always attractive.
Objects pull toward one another.
There is no known gravitational repulsion between ordinary masses.
Objects do not need to touch.
Gravity acts through space.
According to Newton’s Third Law:
F12=F21F_{12} = F_{21}F12=F21
The forces on the two masses are equal in magnitude and opposite in direction.
The gravitational force varies inversely with the square of the distance.
FG∝1r2F_G \propto \frac{1}{r^2}FG∝r21
If the distance doubles:
r→2rr \rightarrow 2rr→2r
then:
FG=14FF_G = \frac{1}{4}FFG=41F
The force becomes one-fourth as large.
If the distance triples:
r→3rr \rightarrow 3rr→3r
then:
FG=19FF_G = \frac{1}{9}FFG=91F
The force becomes one-ninth as large.
The gravitational force is directly proportional to both masses.
FG∝m1m2F_G \propto m_1m_2FG∝m1m2
If one mass doubles:
m1→2m1m_1 \rightarrow 2m_1m1→2m1
then:
FG→2FGF_G \rightarrow 2F_GFG→2FG
The force doubles.
If both masses double:
FG→4FGF_G \rightarrow 4F_GFG→4FG
The force quadruples.
The force acts along the line connecting the centers of the two objects.
The direction is always toward the other mass.
Because gravity is attractive, both objects pull inward toward one another.
A mass creates a gravitational field around itself.
The gravitational field strength is:
g=GMr2g = \frac{GM}{r^2}g=r2GM
where:
This field describes the gravitational force per unit mass.
The weight of an object is the gravitational force exerted by Earth.
Near Earth’s surface:
FG=mgF_G = mgFG=mg
where:
Applying Newton’s Law of Gravitation to Earth:
FG=GMEmRE2F_G = \frac{GM_E m}{R_E^2}FG=RE2GMEm
Since:
FG=mgF_G=mgFG=mg
we obtain:
g=GMERE2g = \frac{GM_E}{R_E^2}g=RE2GME
where:
Because:
g=GMr2g = \frac{GM}{r^2}g=r2GM
gravity decreases as distance from Earth’s center increases.
At higher altitudes:
r↑r\uparrowr↑
therefore:
g↓g\downarrowg↓
Astronauts in orbit still experience gravity; they are simply in continuous free fall.
For a satellite orbiting Earth:
Gravity provides the required centripetal force.
FG=FCF_G = F_CFG=FC GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}r2GMm=rmv2
The satellite remains in orbit because gravity continuously changes the direction of its velocity.
From:
GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}r2GMm=rmv2
the satellite mass cancels.
Solving for speed:
v=GMrv = \sqrt{\frac{GM}{r}}v=rGM
This equation gives the orbital speed for a circular orbit.
Using:
v=2πrTv = \frac{2\pi r}{T}v=T2πr
and the orbital-speed equation:
T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}T=2πGMr3
This result leads directly to Kepler’s Third Law.
Two masses:
m1=5 kgm_1=5\,kgm1=5kg m2=10 kgm_2=10\,kgm2=10kg
are separated by:
r=2 mr=2\,mr=2m
Find the gravitational force between them.
Using:
FG=Gm1m2r2F_G = \frac{Gm_1m_2}{r^2}FG=r2Gm1m2
Substitute values:
FG=(6.67×10−11)(5)(10)(2)2F_G = \frac{(6.67\times10^{-11})(5)(10)} {(2)^2}FG=(2)2(6.67×10−11)(5)(10) FG=8.34×10−10 NF_G = 8.34\times10^{-10}\,NFG=8.34×10−10N
The force is extremely small.
When multiple masses are present, gravitational forces add vectorially.
The net gravitational force is:
F⃗net=∑F⃗\vec{F}_{\text{net}} = \sum \vec{F}Fnet=∑F
Each force must be calculated separately before combining them as vectors.
Using surface gravity equations when the object is far from Earth.
Use:
FG=GMmr2F_G = \frac{GMm}{r^2}FG=r2GMm
for general situations.
Using the radius of Earth instead of the distance between centers.
Always measure rrr from center to center.
Assuming gravity disappears in orbit.
Gravity remains significant and provides the centripetal force for orbital motion.
Newton’s Law of Gravitation explains all three of Kepler’s Laws.
It provides the physical reason why:
Newton’s Law of Universal Gravitation:
FG=Gm1m2r2F_G = \frac{Gm_1m_2}{r^2}FG=r2Gm1m2
Gravitational field strength:
g=GMr2g = \frac{GM}{r^2}g=r2GM
Weight near Earth’s surface:
FG=mgF_G=mgFG=mg
Orbital speed:
v=GMrv = \sqrt{\frac{GM}{r}}v=rGM
Orbital period:
T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}T=2πGMr3
Key ideas:
Newton’s Law of Gravitation is the foundation of celestial mechanics and connects forces, orbits, satellites, and planetary motion within a single universal framework.
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