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Every object on the
surface of the Earth
feels the force of
gravity, and no matter
where the object is, the
force is directed toward
the center of the Earth.

Newton’s reasoning:
“If gravity acts at the tops of the trees,
and even at the tops of the mountains,
then perhaps it acts all the way to the
moon.”
Newton’s Law of Universal Gravitation

When moon’s nearly circular orbit about
the earth has a radius of about 385,000 km
and a period of 27.3 days, the centripetal
acceleration of the moon (nearly uniform) is
ac = 2.73 x 10-3 m/s2
= (2.78 x 10-4) g
𝑎𝑐 ≈
1
3600
𝑔

The Moon is 384,000 km away from the
center of the Earth, which is about 60 times the
Earth’s radius of 6380 km. That is, the moon is 60
times farther from the Earth’s center than are
objects at the Earth’s surface.
Then, taking the square of that radius, it
equals to the proportionality constant of the
centripetal acceleration of the moon and the
acceleration due to gravity at the Earth’s surface.

Thus, Newton concluded that the
gravitational force exerted by the
earth on any object decreases with
the square of its distance from the
earth’s center.
𝐹
𝑔 ∝
1
𝑟2

Also, the force of gravity depends on the object’s
mass
According to 3rd law,
When the earth exerts its
gravitational force on an object,
such as the moon, the object exerts
an equal and opposite (in direction)
force on the earth. Newton
reasoned, that the magnitude of
the force of gravity must be
proportional to both masses**. **the earth's mass is greater than yours by a factor of about 1023

Law of Universal Gravitation states
“Every particle in the universe attracts every
other particle with a force that is proportional to
the product of their masses and inversely
proportional to the square of the distance
between them. This force acts along the line
joining the two particles.”
𝐹 = 𝐺
𝑚1𝑚2
𝑟2
Where
G is the Gravitational Constant

Henry Cavendish
He determined the value
of G by an experiment (using
a torsion balance) and
confirmed Newton’s
hypothesis.

Example:
A 50-kg person and a 75-kg
person are sitting on a bench, 50
cm away from each other.
Estimate the magnitude of the
gravitational force each exerts on
the other.

Example:
What is the force of gravity
acting on a 2000-kg spacecraft when
it orbits two Earth radii from the
earth’s center? The mass of the
earth is 5.98 x 1024 kg.

Example:
Determine the net force
on the moon (mM = 7.36 x 1022
kg) due to the gravitational
attraction of both the Earth
(mE = 5.98 x 1024 kg) and the
sun (mS = 1.99 x 1030 kg),
assuming they are at right
angles to each other (See the
figure given).

Gravity Near the Earth’s Surface
When the equation of the law of
gravitation is applied at the surface of the
Earth, between the Earth and an object, we
can have the following equation:

Then,
We can now solve for g, the acceleration of
gravity of the Earth in terms of the Earth’s mass
and radius. This is also called the effective value
of g and can be applied to other planets.

Example:
Estimate the effective value of g on top of
Mt. Everest, 8,850 m (29,035 ft) above sea level.
That is, what is the acceleration due to gravity
of objects allowed to fall freely at this altitude?

Satellites
A satellite is put into
orbit by accelerating it to
a sufficiently high
tangential speed with
the use of rockets.

If the speed is too
high, the spacecraft will
not be confined by the
Earth’s gravity and will
escape, never to return. If
the speed is too low, it
return to Earth

Satellites are usually
put into circular or
nearly circular orbits,
because such orbits
require the least
takeoff speed.

If it weren’t for the
gravitational force of the
Earth pulling it into orbit, it
would quickly fly out into
space. In fact, a satellite is
falling or accelerating
toward the Earth but its high
tangential speed keeps it
from hitting Earth.

Example:
A geosynchronous satellite is one that stays
above the same point on the Earth, which is possible
only if it is above a point on the equator. Such
satellites are used for TV and radio transmission, for
weather forecasting, and as communication relays.
Determine the height above the Earth’s surface such
a satellite must orbit, and such a satellite’s speed.

Apparent Weightlessness
People and other objects in a
satellite circling the Earth are said to
experience apparent weightlessness.
This can also be experience even
near the Earth’s surface.

In an elevator is at rest (figure a) with a
bag hanging from a spring scale, the scale
reads the true mass of the hanging bag.

If the elevator has an acceleration with
an upward direction, the scale reads more
than the true of the hanging bag.

If the elevator accelerates downward, the acceleration
will be negative and the scale reads lower than the true
mass of the hanging bag. This is the apparent mass of
the object.

In the figure c, that phenomenon is called
apparent weightlessness because in the
reference frame of the person, objects don’t
fall or seem to have weight – yet gravity does
not disappear. The objects seem weightless
only because the elevator is in free fall, and
there is no contact force to make us feel the
weight.

In a situation where a spacecraft
is far from planets, moons and other
attracting objects, the force of gravity
due to them will be quite small due
to long distances and the spacecraft
will experience real weightlessness

Kepler’s Laws
German astronomer Johannes Kepler worked out a
detailed description of the motion of the planets about the
Sun. Three empirical findings by Kepler is now refer to as
Kepler’s laws of planetary motion. They are summarized as
follows:
•Kepler’s first law: The path
of each planet about the
Sun is an ellipse with the
Sun at one focus

•Kepler’s second law: Each planet moves so that
an imaginary line drawn from the Sun to the
planet sweeps out equal areas in equal periods
of time.

•Kepler’s third law: The ratio of the squares of the
periods of any two planets revolving about the Sun is
equal to the ratio of the cubes of their mean distances
from the Sun.
𝑇1
𝑇2
2
=
𝑠1
𝑠2
3

Example:
Mar’s period was noted by Kepler
to be about 687 days (Earth days),
which is
687 𝑑
365 𝑑
= 1.88 𝑦𝑟. Determine
the distance of Mars from the Sun
using the Earth as a reference.

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