Question

Solve the following numerical problem. If the height of satellite completing one revolution around the earth in T seconds in h1 meters, then what would be the height of a satellite taking 2√2 T seconds for one revolution​

Answer 1

pls refer to the above photo

Answer 2

SoluTion :-

${\fbox {\sf {Given:-}}$

The height of the satellite completing one revolution around the earth in T seconds.let the height be x of a satellite taking 2√2 T seconds for one revolution.

$\fbox {From\ Kelpers\ third\ law\ of\ motion:-}$

The square of the time period of revolution of the planet around the sun is directly proportional to the cube semi-major axis of its elliptical orbit.

$\sf {T \propto r^{\frac {3}{2}}}$

Where

$\fbox {T = Time\ period}$

$\fbox {r = Radius\ of\ the\ satellite}$

$\sf {Thus\ we\ can\ say}\\\\T = K\ r^{\frac{3}{2}}----1}$

We get,

$\sf {k = \frac{T}{r^{\frac{3}{2}}} }$

$\sf {Then,\ if\ T2\ = 2\sqrt{2r}}$

$\sf {Putting\ the\ value\ in\ the\ equation (i), we\ get\ 2T = K\ {r_^1}^ {\frac{3}{2}}}$

Putting the value of K in the above equation, we get

$\sf {2\sqrt{2T} = \frac{T}{r^{\frac{3}{2} }} {r_1}^{\frac{3}{2} }}$

On solving the above equation, we get

$\implies \sf {{r_1}^{\frac {3}{2}}} = 2\sqrt{2}r ^{\frac {3}{2}}$

$\fbox {Squaring\ both\ side\ and\ then\ taking\ cube-root}$

$\sf {r_1 = 2r} -----2$

Now r₁= R + x (radius of earth + height above earth) and

r = R + h₁

Using these relations we get x = R + 2h₁

$\rule {130}{2} \ Be\ Brainly \ \star$