Question

If the height of satellite completing one revolution around the earth in T seconds is h1 meters,then what would be the height of a satellite taking
$2 \sqrt{2}$
T seconds for one revolution

Answer 1

Hi there!
Here's the answer:

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¶ POINTS TO REMEMBER:

•Kepler's Third Law:

The cube of the mean distance of a planet from the sun is directly proportional to the square of time taken to move around the sun.

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

where , r is the mean distance from the sun

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¶¶¶ SOLUTION:

Given,

height of satellite = $h_{1}$ m

time taken by satellite for one revolution around earth = T sec

When satellite takes $2\sqrt{2}T$ seconds for one revolution, height of satellite = ?

$T_{1}= T$
$T_{2}= 2\sqrt{2}\\ T$
$r_{1} = R+h_{1}$
$r_{2} = R + h_{2}$

$\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{T_{1}^{2}}{T_{2}^{2}}$

$\implies \frac{(R+h_{1})^{3}}{(R+h_{2})^{3}}=\frac{T^{2}}{(2\sqrt{2}T)^{2}}$

$\implies \frac{(R+h_{1})^{3}}{(R+h_{2})^{3}}=\frac{T^{2}}{(8T^{2})}$

$\implies \frac{(R+h_{1})^{3}}{(R+h_{2})^{3}}=\frac{1}{8}$

Cross Multiply

$8(R+h_{1})^{3} = (R+h_{2})^{3}$

$\implies 2^{3}(R+h_{1})^{3} = (R+h_{2})^{3}$

$\implies (2(R+h_{1}))^{3} = (R+h_{2})^{3}$

$\implies (2R+2h_{1})^{3} = (R+h_{2})^{3}$

Applying cuberoot on both sides, we get

$\implies 2R+2h_{1} = R+h_{2}$

$\implies h_{2} = 2R+2h_{1}-R$

$\implies h_{2} = R+ 2h_{1}$

•°• If satellite takes $2\sqrt{2}T$ seconds for one revolution, then height of satellite = $R+2h_{1}$

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