I really need you answered this to my email. It's very important, for this may decide my result in the school exams, and I really need a good mark.
Unfortunately, Portuguese system of education is full of problems...
The question is...
The altitude of a geostationary satellite depends on:
a) … the satellite’s mass.
b) … the absolute value of its speed
(Considering this is a circular motion)
c) … the Earth’s mass.
d) … the launching speed of the satellite (speed which he was thrown)
Consider while in orbit, there are one force operating on the satellite: Gravity. To be in orbit, gravity must equal exactly the "force" expected in centripetal acceleration.
Gravity Force= centripetalforce
GMeM/r^2= M v^2/r
dividing out M, the mass of the satellite, and r , one has
GMe/r= v^2 or
r= v^2/GMe so the factors are velocity in orbit, and the mass of the Earth.
The altitude of a geostationary satellite depends on:
a) … the satellite’s mass.
b) … the absolute value of its speed
(Considering this is a circular motion)
c) … the Earth’s mass.
d) … the launching speed of the satellite (speed which he was thrown)
How do you determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the earth to make one complete rotation?
The force exerted by the earth on the satellite derives from
...................................................F = GMm/r^2
where G = the universal gravitational constant, M = the mass of the earth, m = the mass of the satellite and r = the radius of the satellite from the center of the earth.
GM = µ = 1.407974x10^16 = the earth's gravitational constant.
The centripetal force required to hold the satellite in orbit derives from F = mV^2/r.
Since the two forces must be equal, mV^2/r = µm/r^2 or V^2 = µ/r.
The circumference of the orbit is C = 2Pir.
A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, requires 23hr-56min-4.09sec. to rotate 360º, not 24 hours. Therefore, the time to complete one orbit is 23.93446944 hours or 86,164 seconds
Squaring both sides, 4Pi^2r^2 = 86164^2
But V^2 = µ/r
Therefore, 4Pi^2r^2/(µ/r) = 86164^2 or r^3 = 86164^2µ/4Pi^2
Thus, r^3 = 86164^2(1.407974x10^16)/4Pi^2 = 2.647808686x10^24
Therefore, r = 138,344,596 feet. = 26,201.6 miles.
Subtracting the earth's radius of 3963 miles, the altitude for a geosynchronous satellite is ~22,238 miles.
To determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the Earth to make one complete rotation, you can follow the following steps:
1. Recognize that in order for a satellite to stay in orbit, the force exerted by the Earth on the satellite (gravity force) must be equal to the centripetal force required to hold the satellite in orbit.
2. Write down the formula for the force exerted by the Earth on the satellite: F = GMm/r^2, where G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the satellite from the center of the Earth.
3. Recognize that the centripetal force required to hold the satellite in orbit is given by the formula: F = mV^2/r, where V is the speed of the satellite in its circular orbit.
4. Set these two forces to be equal: GMm/r^2 = mV^2/r.
5. Simplify the equation by dividing out the mass of the satellite and canceling out the r term, leaving you with V^2 = GM/r.
6. Recognize that the circumference of the orbit is given by C = 2πr, where C is the circumference and r is the radius of the orbit.
7. Determine the time period required for the satellite to complete one orbit. In this case, it is 23.93446944 hours or 86,164 seconds.
8. Square both sides of the equation V^2 = GM/r.
9. Substitute C = 2πr into the equation and rearrange to solve for r^3:
4π^2r^2 = (86,164)^2.
V^2 = GM/r.
4π^2r^2/(GM/r) = (86,164)^2.
10. Calculate the value of r using the known values of G, M, and the period. In this case, it will give you r = 138,344,596 feet or 26,201.6 miles.
11. Subtract the radius of the Earth (3963 miles) from the calculated radius to obtain the altitude for a geosynchronous satellite, which is approximately 22,238 miles.
Therefore, the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the Earth to make one complete rotation is approximately 22,238 miles.