Question

derive an expression for the time period of a satellite ​

Answer 1

Answer:

Time period of Revolution of the satellite about the Earth and Kepler's third law derivation. Say, period of revolution is The and the radius of the orbit is r. And

r =R + h = radius of the earth + hight of the satellite from the surface of the earth.

Please mark me as a brainlist answer.

Answer 2

Let  T = time period of revolution
      R = radius of the earth
       h = height of satellite from earth's surface
       g = acceleration due to gravity
       v = speed
       V = speed of satellite
       G = Gravitational constant
      M = Mass of the earth

Distance traveled by satellite in one revolution = $2\times\pi\times (R + h)$

∴ v = $\frac{distance}{time}$ = $\frac{2\times\pi\times (R + h)}{T}$

We know that the velocity of a body rotating around the earth is given by

$V = \sqrt{\frac{GM}{R+h} }$

∴ $\frac{2\times\pi\times (R + h)}{T} = \sqrt{\frac{GM}{R+h} }$
$\implies T = 2\pi \times \frac{(R+h)^{3/2} }{\sqrt{GM} }$
$\implies T^{2} = 4\times\pi^{2} \times\frac{(R+h)^{3} }{GM}$

if, h<<R,

$T^{2} = 4\times\pi^{2} \times\frac{R^{3} }{GM}$ $= 4\times\pi^{2} \times\frac{R }{\frac{GM}{R^{2} } }$$= 4\times\pi^{2} \times\frac{R }{{g} }$

$(g = \frac{GM}{R^{2} })$
$\implies T = 2\times\pi \sqrt{\frac{R}{g} }$

∴ The time period  of the satellite is $\implies T = 2\times\pi \sqrt{\frac{R}{g} }$