Tuesday, July 10, 2018
Unizor - Physics4Teens - Mechanics - Dynamics - Free Falling
Notes to a video lecture on http://www.unizor.com
Free falling is a movement of an object on a surface of a planet
relative to this surface, when the only force acting on this object is
the gravitational force of a planet.
Our task is to describe this movement in mechanical terms of force, mass and acceleration.
In this task we will assume that
(a) an object in question is a point-object of mass m,
(b) a planet has a spherical form and its mass M is uniformly distributed within its volume,
(c) a planet has a radius R,
(d) [an important assumption that can be justified by complex
calculations] we can model the combined forces of gravitation between
all microscopic particles inside a planet and our object in question as a
gravitational force of a point-object of mass M positioned at the center of a planet.
In this case the one and only force of attraction acting on an object
and directed towards the center of a planet can be expressed using the
Law of Gravitation as follows:
F = G·M·m /R²
where G is a gravitational constant,
G = 6.674·10−11 N·m²/kg²
Knowing the force of gravity F and mass of an object m, we can determine the acceleration using the Newton's Second Law:
a = F/m = G·M /R²
Notice that this acceleration does not depend on m - mass of an object, which means that all objects fall on the surface of a planet with the same acceleration.
An interesting aspect of this formula is that we can imagine how to
measure an acceleration (easy) and radius of a planet (more difficult,
but possible), while we have no idea how to measure the mass of a
So, this formula is used exactly for this purpose - to determine the mass of a planet, resolving the formula above for M:
M = a·R² /G
Experiments show that on the surface of our planet Earth the acceleration caused by gravitational force is approximately 9.8 m/sec².
The radius of Earth is approximately 6.4·106 m.
From this we can calculate the mass of Earth (in kilograms - units of mass in SI):
The result of this calculation is
M ≅ 6·1024 kg
Let's solve a different problem now. We'd like to launch a satellite
around the Earth that circulates around the planet at height H. What linear speed should a satellite have to stay on a circular orbit?
We know from Kinematics that an object rotating along a circular trajectory of radius r and angular speed ω has acceleration a=r·ω².
In terms of linear speed V=r·ω along an orbit this formula looks like
a = V²/r
Since the radius of an orbit is the radius of Earth R plus the height above its surface H, we should replace r in this formula with R+H.
The force of gravity is the only force acting on a satellite and the
only source of its acceleration towards the Earth, so the acceleration
above must be equal to acceleration of a free fall of a satellite. Here
we will take into consideration already known mass of Earth and use
distance from the center of the Earth to satellite as R+H, where R is the radius of Earth and H is a height above the Earth's surface.
The acceleration of a free fall to Earth at height H above the surface, using its radius R and already calculated mass of Earth M, is:
a = G·M/(R+H)²
Therefore, equating the acceleration of free fall to acceleration of an
object rotating along a circular orbit, we come to the following
V²/(R+H) = G·M/(R+H)²
from which we derive the value of required linear speed V:
V = √G·M/(R+H)
For example, International Space Station rotates around our planet on a height of about 400 kilometers (4·105 meters).
That means that, to stay on an orbit, it should have linear speed of
V ≅ 0.78·104 m/sec
which is about 28,000 km/hour.