Question

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Calculate the height from the Earth's surface that a satellite must attain in order to be in geosynchronous orbit and the satellite's velocity?

Given,
G=
$6.67 \times {10}^{ - 11} \: newton \: {metre}^{2} {kg}^{ - 2}$
T=24 hours
$mass \: of \: earth = 6 \times {10}^{24} kg$
$radius \: of \: earth = 6.4 \times {10}^{6} m$
$\pi = 3.14$




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Answer 1

Answer:

Geosynchronous means that the satellite has same period as the earth, back to the same place in 24 hours.

T =24hrs = 86400 s

And let

h = height of the satellite from the surface of the earth.

r = radius of the satellite from the center of the Earth

RE = earth radius

ME = mass of the earth

The gravitational pull from the earth causes the satellite to go in orbit (otherwise it flies away. Hence the gravity is the cause of the centripetal force

FG=msv2r

⇒GmEmsr2=msv2r

rv2=GME

Because the orbital speed v=2πrT

⇒r(2πrT)2=GME

⇒r3=GME4π2T2

r=(GME4π2T2)⅓

r=RE+h

h=r−RE=(GME4π2T2)⅓−RE

Explanation:

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