The satellites | POINT classical mechanics
The satellites
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Space exploration has been a dream that has haunted human minds for centuries. Achieving this dream involved the development of monitoring devices and rockets that launch a spacecraft to orbit Earth or reach another planet such as Mars, until the dream was realized on October 4, 1957, when the Sputnik satellite was sent into space as the first satellite of planet Earth. This was followed by the sending of other satellites and the successful landing on the surface of the moon, and space exploration continues with great success.
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| Replica of ''Sputnik 1'' [first artifcial earth satellite] in the Museum of space and missile technology (saint petersburg) |
The Idea of Launching a Satellite:
In its orbit, a satellite represents a body freely falling towards Earth because its motion is affected only by gravity. Despite this, it never comes close to Earth. Isaac Newton explained this when he imagined that when a cannonball is fired from the top of a mountain horizontally (ignoring air resistance).
• The cannonball travels a horizontal distance before falling freely and taking a curved path towards Earth
• As the speed at which the projectile is launched increases, the horizontal distance it travels before reaching the ground increases, and it follows a less curved path.
• If the projectile's launch speed reaches a certain limit such that the curvature of the projectile's path equals the curvature of the Earth's surface, it revolves in a fixed, quasi-circular path around the Earth and becomes a satellite of the Earth, like a natural satellite. Therefore, it is called a satellite. This speed is called the satellite's orbital speed.
Satellite's Orbital Speed: The speed that makes the satellite revolve in a quasi-circular path such that its distance from the Earth's surface remains constant
Deduction Deduction of the orbital velocity of the satellite (v)
A satellite of mass m is assumed to move around a planet of mass M at a constant velocity of v in a circular orbit of radius r As shown in the figure, then:
- The gravitational force between the planet and the satelliteis given by the relation: F = GMm/r²
- The force of attraction between the planet and the satellite is perpendicular to the path of the satellite's motion, causing it to move in a circular path: F = mv²/r
that: The force of attraction between the planet and the satellite is the same as the centripetal force
∴ GMm/r² = mv²/r
∴ v² = GM/r
∴ v = √(GM/r)
If the height at which the satellite was launched into space from the surface of the planet is h and the radius of the planet is R, then:
r = R+h
v = √(GM/R+h)
(1) If we imagine a sudden stop of a satellite orbiting the Earth (its speed becomes zero), it moves in a straight line toward the Earth under the influence of gravity and falls to its surface.
(2) If we imagine a sudden disappearance of the lateral force between the Earth and a satellite orbiting it, the satellite moves in a straight line tangent to the circular path, away from the Earth.
(3) The period of a geosynchronous satellite is equal to the period of the Earth's rotation around its axis, i.e., one Earth day (24 hours). Therefore, the satellite remains above a fixed point on the Earth's surface.
(4) The period (T) of a satellite orbiting a planet can be calculated from this equation: T = 2Ï€r/v
(5) The relationship between the radius of the orbit of a satellite (1) orbiting a planet and the period of its motion (T) can be deduced as follows:
∵ v = √(GM/r) = 2Ï€r/T
∴ GM/r = 4Ï€²r²/T²
∴ T² = 4Ï€²r³/GM
∴ T² ∝ r³
(6) The orbital speed of the satellite around the Earth is inversely proportional to the square root of the radius of the circular orbit according to this equation [v = √(GM/r)] and it cannot be said that:
- is directly proportional to the radius of the circular orbit according to this equation [v = 2Ï€r/T] because the period also depends on the radius of the orbit according to this equation [T² = 4Ï€²r³/GM].
- is directly proportional to the square root of the radius of the circular orbit according to this equation (v = √gr ) because the intensity of the gravitational field also depends on the radius of the orbit according to the relationship
(g = GM/r²)
(7) The greater the mass of the satellite to be sent into space, the more we need a rocket that can exert a greater force on the satellite.So that it can gain the speed necessary to orbit the Earth.