*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

**is uniformly distributed within its volume,**

*M*(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

**positioned at the center of a planet.**

*M*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

**is a gravitational constant,**

*G*

*G = 6.674·10*^{−11}N·m²/kg²Knowing the force of gravity

**and mass of an object**

*F***, we can determine the acceleration using the Newton's Second Law:**

*m*

*a = F/m = G·M /R²*Notice that this acceleration does not depend on

**- mass of an object, which means that all objects fall on the surface of a planet with the same acceleration.**

*m*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

**.**

*6.4·10*^{6}mFrom this we can calculate the mass of Earth (in kilograms - units of mass in SI):

*M≅9.8· 6.4²·10*^{12}/(6.674·10^{−11})The result of this calculation is

*M ≅ 6·10*^{24}kgLet's solve a different problem now. We'd like to launch a satellite

around the Earth that circulates around the planet at height

**. What linear speed should a satellite have to stay on a circular orbit?**

*H*We know from Kinematics that an object rotating along a circular trajectory of radius

**and angular speed**

*r***has acceleration**

*ω***.**

*a=r·ω²*In terms of linear speed

**along an orbit this formula looks like**

*V=r·ω*

*a = V²/r*Since the radius of an orbit is the radius of Earth

**plus the height above its surface**

*R***, we should replace**

*H***in this formula with**

*r***.**

*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

**, where**

*R+H***is the radius of Earth and**

*R***is a height above the Earth's surface.**

*H*The acceleration of a free fall to Earth at height

**above the surface, using its radius**

*H***and already calculated mass of Earth**

*R***, is:**

*M*

*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 (4·10

^{5}meters).

That means that, to stay on an orbit, it should have linear speed of

*V ≅ 0.78·10*^{4}m/secwhich is about 28,000 km/hour.

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