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THERMO Spoken Here! ~ J. Pohl © ~ 2017 | www.ThermoSpokenHere.com (A9800-060) |

We need unit vectors that rotate.

Communications rockets routinely place telecasting satellites in Earth orbit such that they maintain a constant position above and relative to Earth. Such orbits, called geostationary, are only possible only within the plane of the Earth equator. Rockets place the electronic packages at the proper altitude and with the proper circular velocity. Orbit established, the satellite will remain stationary (relative to Earth) and directly above some point on the equator of earth. This permits Earth-surface communication satellite dishes to be "pointed" in a specific directions to constantly receive satellite telecasted information.

The position of the satellite (above ground in the plane of the equator) can be selected. Physics of Earth motion prescribe the altitude and velocity.

**Calculate** the altitude of a geo-stationary orbit. The figure (right) depicts a satellite orbiting in the plane of the Earth equator. **0XY** coordinates are drawn. The **X**-axis is locked (at mid-night) to point at a far distant star. The **Y**-axis is perpendicular to **X**.
This coordinate-pair is assumed inertial. Put otherwise, Earth is assumed to move in a straight line.

The time dependent position of the satellite, P(t), is written as:

(1) 1 |

The velocity is the derivative of position:

(2) 2 |

Next multiply Eqn-2 by the satellite mass. Place the factor, dθ(t)/dt to the right. Eqn-3 is the satellite momentum.

(3) 3 |

The above is the momentum of the satellite. We write Newton's Second Law:

(4) 4 |

Now enter momentum (Eqn-3) into Eqn-4. Also write the gravity force.

(5)
A longer equation needs more paper. |

As we differentiation above, left of equality, keep in mind that d**θ**(t)/dt is a constant. This angular speed of the satellite is 2**π**rad/24hr. Differentiate then gather terms.

(6) Write more compactly, as you go. |

(7) Things got smaller? Is this right? Check everything above. |

(8) 8 |

Our answer, 26,261 miles is referenced to the Earth Center. Accounting for the diameter of Earth, changes the above answer to 22,298 miles (above Earth surface).

**In Brief Review:** Proper, inertial coordinates were established (otherwise Newton's 2nd Law could not be applied). A sketch of the situation was made. The position vector was written using sine and cosine functions with the vector basis locked to Earth. The unit vector I is directed at a distant star (as is J). The Z-axis aligns with the Earth-axis of rotation; that unit vector, K is perpendicular to the plane of the equator. The only calculus needed is that one be able to differentiate the sine and cosine functions.

**About Being "Weightless"**: The above conclusion draws attention to the fanciful idea of "weightlessness" as a state of complete independence of the Earth pull. Every physical entity in the universe experiences a perpetual force in consequence of the mass of Earth. There is never an instant when that pull "lets go." Newton's Methods and Law of Gravitation define the force for a mass to equal the magnitude of its mass times the magnitude of the acceleration of gravity (caused by Earth) at the location of the mass." In Equation form:

Definition of gravity force in equation form is Eqn-i. Force is a vector; **e**_{r} is directed from Earth center toward the mass; gravity force (on the mass) acts oppositely, i.e., **- e _{r}**. Weight is the magnitude of the gravity force (at the location of the mass in question) - Eqn-ii. Finally, the weight of a physical object is made quantitative in accord with Eqn-iii.

What might "weightless" mean. Weight is a force. Suppose a condition existed where the force was zero, a location which was "weightless."

From basic math a product of three things can equal zero only if one of the three is zero. Conclusion: "Weight (as we have defined it) is never zero."

Since the force of gravity is never zero,

Weightless never occurs.

Corrections requested!