Unizor - Physics4Teens - Mechanics - Dynamics - Free Falling

Notes to a video lecture on http://www.unizor.com



Free Falling



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
planet.

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):

M≅9.8· 6.4²·10¹²/(6.674·10⁻¹¹)

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
equation:

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.