Question

An antificial satellite revolves around earth in
circular orbit of radius r with time period of orbit T.
The satellite made to stop in the orbit which makes
it fall onto earth. 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.