Question

Why do artificially launched satellite lose height with time?
please answer my question..​

Answer 1

Explanation:

the height is decreased time period will decrease

Answer 2

The artificially launched satellites lose height with time because the height of the satellite is directly proportional to time.

Explanation:

Let us consider a satellite of mass $M$ orbiting at radius $R$ from the center of Earth with height $H$.

The total distance from the center of the earth to the satellite's height is $r$.

                                       $r\;=\;R+H$

We know that, $T\;=\;\frac{2\mathrm{\pi r}}v$

Substituting, $r\;=\;R+H$

                      $T\;=\;\frac{2\mathrm\pi(\mathrm R+\mathrm H)}v$ -------------- ($1$)

With respect to the equation of speed,

                       $v=\;\frac{\sqrt{GM}}r\\\\\therefore\;v\;=\;\frac{\sqrt{GM}}{(R+H)}$

On substituting in the equation in ($1$),

                         $T\;=\;\frac{2\mathrm\pi(\mathrm R+\mathrm H)}v$

                        $T\;=\;\frac{2\mathrm\pi(\mathrm R+\mathrm H)}{\sqrt{\displaystyle\frac{GM}{(R+H)}}}$

                        $T\;=\;\frac{2\mathrm\pi(\mathrm R+\mathrm H)\;{(\mathrm R+\mathrm H)}^{\displaystyle\frac12}}{\sqrt{GM}}\\\\\\T\;=\;\frac{2\mathrm\pi\;{(\mathrm R+\mathrm H)}^\frac32}{\sqrt{GM}}\\\\\\T\;=\;2\mathrm\pi\sqrt{\frac{{(\mathrm R+\mathrm H)}^3}{\mathrm{GM}}}$

• From this derived equation, the time period $T$ and height of the satellite above the earth's surface $H$ are directly proportional to each other.
• Therefore, the height of the satellite decreases with a decrease in the time period.