Question

an artificial satellite revolves around the earth in circular orbit of radius r with time period T. the satellite made to stop in the orbit which makes it fall onto earth. the time of fall of the satellite onto earth is given by

Answer 1

Answer:

$\frac{\sqrt{2}}{8}T$

Explanation:

The satellite will follow the elleptical path where the radius r of the original circular path will be the semi-major axis of the satellite

Thus using Kepler's law

$T^2\propto r^3$

or, $T^2=Kr^3$

For the elleptical path if the time period (twice the period of falling) is T' then

$T'^2=K(\frac{r}{2})^3$

$T'^2=K\frac{r^3}{8}$

Thus,

$\frac{T'^2}{T^2} =\frac{1}{8}$

$\implies \frac{T'}{T}=\frac{1}{2\sqrt{2}}$

$\implies {T'}=\frac{T}{2\sqrt{2}}$

The period of falling = T'/2

Therefore,

$T'=\frac{T}{4\sqrt{2}}$

or, $T'=\frac{T\sqrt{2}}{8}=\frac{\sqrt{2}}{8}T$

Hope this helps.