Universal Law of Gravitation
Honors Physics
Newton’s Law of Universal Gravitation
The greatest moments in science are when two phenomena that were considered
completely separate suddenly are seen as just two different versions of the same thing.
Newton’s explanation that the motion of the moon around the Earth (and the planets
around the sun) and of an apple falling to the ground are two expressions of the same
phenomenon, gravity, represents a wonderful instance of this sort of synthesis.
Newton conjectured that objects near the surface of the Earth fall towards the center of the
Earth as a result of a universal attraction of all matter towards all other matter. He then
considered that the circular motion of the moon around the Earth could then be explained if
that same force of gravity were supplying the force towards the center of the moon’s
circular orbit, the Earth. That force of gravity would provide the unbalanced force necessary
for circular motion.
Universal Gravity
Newton postulated that all matter in the universe is attracted to all other matter. He named
this force of attraction “gravity” and he formulated the following expression to describe
how the force of gravity depends on the mass of each object and the distance between their
centers.
In this equation, “G” is a constant that would need to be discovered by experiment, “m1” and
“m2” are the masses of the two objects and “r” is the distance between their centers. It
doesn’t matter which mass you call “m1” and which one you call “m2”: That’s a result of
Newton’s third law which states that the force on one object will be equal in size to the force
on the other.
Newton had to invent calculus in order to show that the distance between two spherical
objects, the “r” in his equation, should be measured between the centers of the objects. He
was able to show that spherical objects act, from the perspective of gravitational force, as if
all of their mass were located at a point at their center. As a result, the distance from a
spherical mass, such as the Earth or the sun, must always be measured from its center.
It took more than a hundred years before Henry Cavendish was able to directly measure the
value of “G”. The modern accepted value of “G” is: G = 6.67 x 10-11 N∙m2/kg2
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Example 1
What is the force between a 5.0kg spherical object and a 10 kg spherical object whose
centers are located 2.5m apart?
𝐹𝐺=𝐺𝑚1𝑚2/𝑟2
𝐹𝐺 = (6.67 x 10-11 N∙m2/kg2)(5.0kg)(10kg)/(2.5m)2
𝐹𝐺 = (6.67 x 10-11 N∙m2/kg2)(5.0kg)(10kg)/(2.5m)2
𝐹𝐺 = 5.3 x 10-10 N
Example 2
What is the average gravitational force between the Earth and the sun? The mass of
the Earth is 6.0 x 1024 kg, the mass of the sun is 2.0 x 1030 kg and the average distance
between their centers is 1.5 x 1011m.
𝐹𝐺=𝐺𝑚1𝑚2/𝑟2
𝐹𝐺 = (6.67 x 10-11 N∙m2/kg2)( 2.0 x 1030)( 6.0 x 1024 kg)(1.5 x 1011m)2
𝐹𝐺 = (6.67 x 10-11 N∙m2/kg2)( 2.0 x 1030)( 6.0 x 1024 kg)/( 1.5 x 1011m)2
𝐹𝐺 = 3.6 x 1022 N
Example 3
What is the distance between two objects whose masses are 28,000 kg and 50,000 kg if
the gravitational force between them is 50N?
𝐹𝐺=𝐺𝑚1𝑚2/𝑟2
To solve for r, first multiply both side by r2
(FG)(r2)= 𝐺𝑚1𝑚2
Then divide both sides by FG
2
And take the square root of both sides
r = 𝐺𝑚1𝑚2/ FG
r= (Gm1m2/FG)1/2
r= ((6.67 x 10-11 N∙m2/kg2)(2.8 x 104 kg)(5.0 x 104 kg)/(50N))1/2
r = (1.86 x 10-3)
r = 18.6 x 10-4
r = 4.3 x 10-2 m
r = 4.3 cm
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Universal Gravitation:
, G = 6.67 x10-11 Nm2/kg2
Examples
1. Two spherical objects have masses of 200 kg and 500 kg. Their centers are
separated by a distance of 25 m. Find the gravitational attraction between them.
(1.067x10-8 N)
2. Two spherical objects have masses of 1.5 x 105 kg and 8.5 x 102 kg. Their centers
are separated by a distance of 2500 m. Find the gravitational attraction between
them. (1.361x10-9 N)
3. Two spherical objects have masses of 3.1 x 105 kg and 6.5 x 103 kg. The
gravitational attraction between them is 65 N. How far apart are their centers?
(0.045 m)
4. A 1 kg object is located at a distance of 6.4 x106 m from the center of a larger
object whose mass is 6.0 x 1024 kg.
a. What is the size of the force acting on the smaller object? 9.77 N
b. What is the size of the force acting on the larger object? 9.77 N
c. What is the acceleration of the smaller object when it is released? 9.77
m/s2
d. What is the acceleration of the larger object when it is released? 1.63 x10-24
m/s2
Class Work
5. Two spherical objects have masses of 8000 kg and 1500 kg. Their centers are
separated by a distance of 1.5 m. Find the gravitational attraction between them.
0.000356 N or 3.56 x10-4 N
6. Two spherical objects have masses of 7.5 x 105 kg and 9.2 x 107 kg. Their centers
are separated by a distance of 2.5 x 103 m. Find the gravitational attraction
between them.
0.736
7.36 x10-4 N
7. Two spherical objects have masses of 8.1 x 102 kg and 4.5 x 108 kg. The
gravitational attraction between them is 1.9 x 10-3 N. How far apart are their
centers? 113.229 m
8. Two spherical objects have equal masses and experience a gravitational force of
85 N towards one another. Their centers are 36mm apart. Determine each of
their masses. 40640 kg
9. A 1 kg object is located at a distance of 7.0 x108 m from the center of a larger
object whose mass is 2.0 x 1030 kg.
a. What is the size of the force acting on the smaller object? 272 N
b. What is the size of the force acting on the larger object? 272 N
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c. What is the acceleration of the smaller object when it is released? 272 m/s2
d. What is the acceleration of the larger object when it is released? 1.36 x10-28
m/s2
Homework
1. Two spherical objects have masses of 8000 kg and 5.0 kg. Their centers are
separated by a distance of 1.5 m. Find the gravitational attraction between them.
1.19 10-6 N
2.
Two spherical objects have masses of 9.5 x 108 kg and 2.5 kg. Their centers
are separated by a distance of 2.5 x 108 m. Find the gravitational attraction
between them. 2.53 x10-18 N
3. Two spherical objects have masses of 6.3 x 103 kg and 3.5 x 104 kg. The
gravitational attraction between them is 6.5 x 10-3 N. How far apart are their
centers? 1.50 m
4. Two spherical objects have equal masses and experience a gravitational force of
25 N towards one another. Their centers are 36 cm apart. Determine each of their
masses. 220400 kg
5. A 1 kg object is located at a distance of 1.7 x106 m from the center of a larger
object whose mass is 7.4 x 1022 kg.
a. What is the size of the force acting on the smaller object? 1.71 N
b. What is the size of the force acting on the larger object? 1.71 N
c. What is the acceleration of the smaller object when it is released? 1.71
m/s2
d. What is the acceleration of the larger object when it is released? 2.31 x10-23
m/s2
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Orbital Motion
Newton based much of his theory on the work of Johannes Kepler, which was based
on the astronomical observations of Tycho Brahe. Newton concluded that in order
for the planets to proceed in their orbits around the sun there must be a force
attracting them towards the sun. Kepler writings could be summarized into three
laws:
•
Kepler’s first law states that the orbits of the planets are elliptical, with the
sun at one focus.
Sun
•
An ellipse has two “focal points” and no real defined center like a
circle. (Actually… if Kepler didn’t study the data concerning Mars, he
probably would never have discovered this fact! Mars has an
extremely elliptical orbit, while the other planets are nearly circular!)
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•
Kepler’s second law states that if you extend an imaginary line between the
sun and a planet, that that line would sweep out equal areas in equal times. In
other words, Kepler found that the planets move faster when they are closer
to the Sun and slower when they are farther away from the Sun.
This means that the earth is closer to the sun in winter! (strange…but true!)
The following diagram shows this:
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•
Kepler’s third law states that there is a specific mathematical relationship
between the orbital periods of planets (the time to go around the sun) and
their mean (average) distance from the sun. Specifically, for all planets
orbiting the sun the ratio of their orbital period squared divided by their
orbital radius cubed, T2/r3, yields the same number.
Thus, if the periods of the planets are TA and TB and the average distances
from the Sun are rA and rB, the third law can be expressed as follows:
TA
TB
2
=
rA
rB
3
The following data table can be useful for using this equation:
Planetary Data
planets
not
shown to
scale >>
Mercur
Venus Earth
y
Mars
Jupite Satur Uranu Neptun
r
n
s
e
Pluto
Mean
Distance
to the
Sun
(AU)
0.3871
0.806
1
1.524
5.203
9.539
19.19
30.06
39.48
Orbital
Period
(years)
0.24
0.72
1
1.88
11.86
29.46
84.01
164.79
248.54
Period
Squared
.0576
.5184
1
3.534 140.65 867.89
7057.68
27155.744
61772.13
Mean
Distance
Cubed
.058
.523
1
3.53
140.85
867.97
7066.83
27162.32
61536.3
T2/r3
0.99.0
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
Equatori
al Radius
(km)
2439
6052
6378
3397
71490 60268
25559
25269
1160
AU=Astronomical Unit
1AU=distance from earth to sun=146,600 km
This data can be used for any of Kepler’s 3rd law problems that compare orbits and
periods of planets.
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This concept can also be used for orbits and periods of satellites (moons) of planets!
The orbits of moons around planets are also ellipses.
Example:
Galileo discovered four moons of Jupiter. Io, which he measured to be 4.2 units form
the center of Jupiter, has a period of 1.8 days. He measured the radius of
Ganymede’s orbit as 10.7 units. Use Kepler’s third law to find the period of
Ganymede.
Given:
Unknown:
Ganymede’s period, TA
For Io
Period TB = 1.8 days
Radius rB = 4.2 units
For Ganymede
Radius rA = 10.7 units
Basic Equation:
TA
TB
2
Solution:
TA2 =
=
rA
rB
TB2
rA
rB
3
3
TA2 = (1.8 days)2 10.7 units 3
4.2 units
TA2 = 52.8 days2
TA = 7.3 days
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Free Response
1.
As shown in the diagram below, a 5.0 kg space rock is located 2.5x107 m from the center
of the earth. The mass of the earth is 6.0x1024 kg.
Earth
Space Rock
a. Determine the force of gravity acting on the space rock, due to the earth. Calculate the
magnitude and state the direction. (3.2016 N)
b. Compare your answer in a) to the force of gravity acting on the earth, due to the space
rock. Indicate that force on the diagram above. (3.2016 N)
c. On the diagram above, indicate the direction the space rock would accelerate if released.
Label that vector “a”. (←)from rock towards earth
d. Calculate the acceleration the rock would experience. 0.64 m/s2
e. **If instead of falling, the object were in a stable orbit, indicate on the diagram above a
possible direction of its velocity. Label that vector “v”. (↑or↓) up or down from rock
f.
**Calculate the velocity the rock needs to be in a stable orbit. 4000 m/s
g. **Calculate the period of the rock orbiting the earth. 39260 s
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2. As shown in the diagram below, a 2000 kg spacecraft is located 9.2x106 m from the
center of the earth. The mass of the earth is 6.0x1024 kg.
Earth
Spacecraft
a. Determine the force of gravity acting on the spacecraft, due to the earth. Calculate the
magnitude and state the direction. (9456.5 N)
b. Compare your answer in a) to the force of gravity acting on the earth, due to the
spacecraft. Indicate that force on the diagram above. (9456.5 N)
c. On the diagram above, indicate the direction the spacecraft would accelerate if released.
Label that vector “a”. (←)from spacecraft towards earth
d. Calculate the acceleration the spacecraft would experience. 4.73 m/s2
e. **If instead of falling, the spacecraft were in a stable orbit, indicate on the diagram above
a possible direction of its velocity. Label that vector “v”. (↑or↓) up or down from
spacecraft
f.
**Calculate the velocity the spacecraft needs to be in a stable orbit. 6595 m/s
g. **Calculate the period of the spacecraft orbiting the earth. 8764 s
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3.
As shown in the diagram below, a 1000 kg asteroid is located 6.8x106 m from the center of
the Mars. The mass of the Mars is 6.4x1023 kg.
Mars
Asteroid
a. Determine the force of gravity acting on the asteroid, due to the Mars. Calculate the
magnitude and state the direction. 923 N left
b. Compare your answer in a) to the force of gravity acting on the Mars, due to the asteroid.
Indicate that force on the diagram above. 923 N right
c. On the diagram above, indicate the direction the asteroid would accelerate if released.
Label that vector “a”. (←)from asteroid towards mars
Calculate the acceleration the asteroid would experience. 0.92 m/s2
d.
e. **If instead of falling, the asteroid were in a stable orbit, indicate on the diagram above a
possible direction of its velocity. Label that vector “v”. (↑or↓) up or down from asteroid
f.
**Calculate the velocity the asteroid needs to be in a stable orbit. 2505 m/s
g. **Calculate the period of the asteroid orbiting the earth. 17052 s
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