Table of Contents
GEOSTATIONARY SATELLITE AND PARKING ORBIT
The Earth rotates on its axis (the straight line connecting the North and South Poles) once every 24 hours—this is the Earth’s Diurnal Motion. Suppose an artificial satellite is placed in an orbit such that:
- The orbit of the satellite is circular.
- The orbit is in the equator.
- The satellite orbits the Earth once in 24 hours in the direction of the Earth’s Diurnal Motion (i.e. west to east).
Then the relative angular velocity of the satellite with respect to the angular momentum of the Earth (around its own axis) is zero. That is, when viewed from the surface, it seems as if the artificial satellite is fixed in the same place in the sky above the equator. Such an artificial satellite is called a geostationary satellite and its orbit is called a geostationary orbit. Obviously, the center of such an orbit is at the center of the earth.
WHAT’S A GEOSTATIONARY SATELLITES?
If the relative angular velocity of an artificial satellite with respect to the Earth’s angular momentum is zero and the satellite is always at the equator, it appears from the surface that the satellite is fixed in the same place in the sky. Such satellites are called geostationary satellites.
ORBITAL ALTITUDE AND ORBITAL VELOCITY OF SATELLITES AT EARTH LEVEL
Let mass of earth = M, radius of earth = R, mass of geostationary satellite = m, distance of satellite from center of earth = r, orbital velocity of satellite = v, orbital period = T.
Since the force of gravity provides the necessary centripetal force in a circular orbit, therefore,
mv²/r = GMm/r² or, v² = GM/r
If the gravitational acceleration at the surface is g,
g = GM/R² or, GM = gR²
So, v² = gR²/r or, v = √gR²/r …….(1)
Now, orbital period, T = 2πr/v = 2πr√r/gR² = 2π√r³/gR²
That is, T² = 4π²r³/gR²
or, r = (gR²T²/4π²)1/3 ……….(2)
For geostationary satellite T = 24h = 24 X 60 X 60 s, g = 9.8 m/s², R = 6400 km = 6.4 X 10⁶ m
Substituting these values and calculating gives,
r = 4.234 X 10⁴ km = 42340 km
Hence the altitude of the geostationary satellite in the equatorial region of the Earth is,
h = r – R = 42340 – 6400 ≈ 36000 km
That is, each geostationary satellite is about 36,000 km above the Earth’s equatorial region. Here also it should be remembered that this altitude of geostationary orbit does not depend at all on the orbit of the artificial satellite.
The orbital speed of a geostationary satellite is,
v = 2πr/T = 3079 m/s = 3.079 km/s
Obviously, this orbital velocity value is much lower than the orbital velocity relative to Earth or the orbital velocity of a near-surface artificial satellite.
WEIGHTLESSNESS IN ARTIFICIAL SATELLITES
We know that an object of mass m weighs mg, that is, with the help of this force the earth pulls the object towards its center. An object placed on a non-accelerating surface exerts a force mg on the surface.
Again, the floor also exerts an upward perpendicular force (n = mg) on the object. This upward perpendicular force is the apparent weight of the object, i.e. the object only experiences this upward perpendicular force as its own weight. Therefore, if there is no upward perpendicular force on a person (or object) he cannot feel his weight, that is, he feels weightless. Suppose, the object along the floor has an acceleration in the vertical direction. In this case the upward perpendicular force is not equal to the weight of the object. That is why a person in an elevator moving with acceleration feels that his weight has decreased or increased (according to Newton’s laws of motion). Although in some cases the existence of a floor (on which the object rests) has nothing to do with the apparent weight of an object, in all cases imagining the existence of such a floor helps to measure the apparent weight of an object.
ASTRONAUT'S SENSE OF WEIGHT IN SPACECRAFT BEFORE LAUNCH INTO ORBIT
After launching from the surface, the speed of the spacecraft is increased at a very fast rate with the help of rockets. Its upward acceleration is sometimes up to 15 times the gravitational acceleration. If the mass of the astronaut is m, the upward perpendicular force of the spacecraft floor on the astronaut is n, and the acceleration of the spacecraft is a = 15g,
n – mg = ma = m · 15g
or, n = 16mg
This is the astronaut’s apparent weight, that means he feels 16 times heavier than his actual weight.
ASTRONAUT’S SENSE OF WEIGHT IN SPACECRAFT DURING ORBIT
Let us assume that the spacecraft is placed in a circular orbit of radius r with respect to the center of the Earth. Considering an astronaut of mass m, the gravitational force of the Earth on the astronaut and the upward perpendicular force (n) of the spacecraft floor provide the centripetal force necessary to rotate the astronaut into orbit. that is,
mg’ – n’ = mv²/r ……..(1)
Here, g’ = value of gravitational acceleration at position of astronaut and v = velocity of astronaut.
Considering a spacecraft of mass M’, only the gravitational force of the Earth on the spacecraft provides the centripetal force necessary to spin the spacecraft into orbit. that is,
M’g’ = M’v²/r or, g’ = v²/r ………(2)
From (1) and (2) we got,
mg’ – n = mg’ or, n = 0
Since the perpendicular force is zero, the astronaut cannot feel his weight. As a result, he feels himself floating. This is called the weightlessness of man or any object.
Similarly, if the motion of the spacecraft including the ground is considered, assuming the spacecraft is on an imaginary floor, the spacecraft is also weightless. Since the surface is only an illusion, it can be said that any object orbiting under the influence of gravitational force (such as planets, satellites etc.) does not experience the centripetal attraction force of the orbit, i.e. all such objects are weightless.
Just as an astronaut feels many times his actual weight during the launch of a spacecraft, he feels weightless as soon as the spacecraft is placed in orbit.
Just as an object stored on an artificial satellite has no gravity due to Earth, an object stored on the lunar surface has no gravitational weight due to Earth. However, no object on the lunar surface is weightless, because the moon’s own gravitational force acts on each object. That is why every object on the lunar surface has weight. This weight is about 1/6 of the weight of the object on Earth. Notably, the satellite or spacecraft’s own gravitational force also acts on every object placed inside the spacecraft. But that force is so insignificant that its existence is not understood.
Incidentally, the gravitational force of the earth acting on the artificial satellite or the astronaut inside it can never be zero. If this force were zero there would be no question of the satellite orbiting the Earth; because in that case it would not be possible to provide the centripetal force required for circular rotation. So, an object being weightless means that the perpendicular force acting on it is zero, but the gravitational force is not zero.