91影视

The given data can be listed below,

According to the third law of Kepler, slower speeds and longer durations are associated with larger orbits, demonstrating that in a circular orbit, the period of a satellite or planet is proportionate to the 3/2 power of the orbit radius.

The satellite moves 15.65 revolutions in$8.64脳{10}^4鈥�s$ (24 hr), so the time for 1.00 revolution is given by,

$$\begin{gathered} T=\frac{8.64脳{10}^{4}s}{15.65} \\ =5.52脳{10}^{3}s \end{gathered}$$

Kepler鈥檚 third law is given by,

$T=\frac{2{\mathrm{蟺谤}}^{3/2}}{\sqrt{G{M}_E}}$

Here,$M_E$ is the mass of the earth whose value is $5.97脳{10}^{24}\mathrm{kg}$, G is the constant of gravitation whose value is $6.673脳{10}^{-11}N路m^2/k{g}^2$,r is the radius of the orbit.

Solve the above equation for r,

$r={\left(\frac{G{M}_E{T}^2}{4{\mathrm{蟺}}^2}\right)}^{1/3}$

Substitute values in the above equation.

localid="1668309309538" $$\begin{gathered} r={\left[\frac{\left(6.673脳{10}^{鈭�11}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.97脳{10}^{24}\mathrm{kg}\right){\left(5.52脳{10}^3\mathrm{s}\right)}^2\left(\frac{1\mathrm{kg}鈭�\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}{4(3.14{)}^2}\right]}^{1/3} \\ ={\left[\frac{\left(39.81脳{10}^{13}\right)\left(30.47脳{10}^6\right)\mathrm{m}}{39.438}\right]}^{1/3} \\ =6.75脳{10}^6\mathrm{m} \end{gathered}$$

The height h of the orbit above the surface of the earth is related to the orbit radius r by,

$\begin{array}{l}r=h+R\\ h=r-R\end{array}$

Here, ris the orbital radius, and R is the radius of the earth whose value is $6.37x{10}^{6}m$.

Substitute all values in the above,

localid="1668309344066" $$\begin{gathered} h=6.75脳{10}^6\mathrm{m}鈭�6.37脳{10}^6\mathrm{m} \\ =.38脳{10}^6\mathrm{m}\left(\frac{1\mathrm{km}}{1000\mathrm{m}}\right) \\ =380\mathrm{km} \end{gathered}$$

Thus, the height h of the orbit above the surface of the earth is 380 km.