Question

4. Why do artificially launched satellite lose height with time?​

Answer 1

Answer:

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 MM orbiting at radius RR from the center of Earth with height HH .

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

r\;=\;R+Hr=R+H

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

v

2πr

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

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

v

2π(R+H)

-------------- (11 )

With respect to the equation of speed,

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

v=

r

GM

∴v=

(R+H)

GM

On substituting in the equation in (11 ),

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

v

2π(R+H)

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

(R+H)

GM

2π(R+H)

\begin{gathered}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}}}\end{gathered}

T=

GM

2π(R+H)(R+H)

2

1

T=

GM

2π(R+H)

2

3

T=2π

GM

(R+H)

3

From this derived equation, the time period TT and height of the satellite above the earth's surface HH are directly proportional to each other.

Therefore, the height of the satellite decreases with a decrease in the time period.