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Question: A geosynchronous orbit is one in which the satellite makes one revolution around the Earth in \[24hr...

A geosynchronous orbit is one in which the satellite makes one revolution around the Earth in 24hrs24hrs.How far above the surface of the Earth does this satellite have to orbit?
Assume the radiusofearth=0.637×107mradius of earth=0.637\times {{10}^{7}}m and massofearth=5.98×1024kgmass of earth=5.98\times {{10}^{24}}kg

& A.3.59\times {{10}^{7}}m \\\ & B.4.23\times {{10}^{7}}m \\\ & C.2.76\times {{10}^{6}}m \\\ & D.6.37\times {{10}^{6}}m \\\ & E.1.27\times {{10}^{7}}m \\\ \end{aligned}$$
Explanation

Solution

Hint : This question is related to the topic binding energy and satellites. In this question we have given the mass of the earth and radius of the earth. Using the given information we have to find the distance between the surface of the earth and orbit in which satellites revolve. Using the formula for finding the time period of a satellite moving in an orbit of radius R. First we will find radius and then after that we will find height by getting the difference of radius R and earth’s radius.

Complete step-by-step solution:
HAROLD ROSEN was the first person to design a geosynchronous satellite. He designed it while working on an aircraft. Earth’s moon is a natural satellite . Artificial satellites are the ones which are launched into space and move around earth or any other planet in the universe to gather information like whether there are essential things for living that are present on mars or not?. Data which comes from Satellites also helps scientists to predict weather and climate changes to help farmers what crops to plant for profit and spread warning and alertness for emergencies. Satellites have different shapes and sizes but there are two different parts which are very important are antenna and power source through which sending and receiving information took place. There are various types of satellites such as communication satellite, drone satellite, ground satellite, navigation satellite etc.
Satellites which are in geosynchronous orbit are very commonly used for various types of communication purposes like radio, television network, and direct broadcasting of any channel.
There are many other orbits such as super synchronous orbit, sub synchronous orbit, geostationary orbit , graveyard orbit ,lunar orbit which is around earth’s moon, Heliocentric orbit is around the sun in which there are all planets , comets and asteroids present.
In this question we have to find distance between surface of the earth and orbit in which satellite revolve
We have given
t=24hrst=24hrs
Now we will convert into seconds:

& t=24\times 60\times 60 \\\ & t=86400\sec onds.....Equation1 \\\ & massofearth=5.98\times {{10}^{24}}kg.....Equation2 \\\ & radius of earth=0.637\times {{10}^{7}}m.....Equation3 \\\ \end{aligned}$$ Let the mass of earth be $$m$$& radius of earth be$$r$$. $$Gravitationalcons\tan t=G=6.67\times {{10}^{-11}}\dfrac{N{{m}^{2}}}{k{{g}^{2}}}.....Equation4$$ Now, we have a formula for finding the time period of a satellite moving in orbit of radius R. $$t=2\pi \sqrt{\dfrac{{{R}^{3}}}{G\times m}}......Equation5$$ Using equation 1, 2, 3and 4 we put the value on equation 5 then we get: $$86400=2\pi \sqrt{\dfrac{{{R}^{3}}}{\left( 6.67\times {{10}^{-11}} \right)\times \left( 5.98\times {{10}^{24}} \right)}}$$ Hence, on simplifying above equation we get the value of R as: $$R=4.23\times {{10}^{7}}m$$. Thus height of satellite above the surface is: $$h=R-r$$ Putting the value of R and r on the above equation and on simplifying that we get: $$h=\left( 4.23-0.637 \right)\times {{10}^{7}}\approx 3.59\times {{10}^{7}}m$$ Hence,$$3.59\times {{10}^{7}}m$$ above the surface of earth satellites have to orbit. **The correct option is A.** **Note:** The United States has the largest number of satellites. There are some separate geosynchronous satellites which are only used for military purposes such as PAN; it also moves with commercial satellites. Satellites usually have zero inclination to make an orbit over the equator all the time.