91影视

Mass of satellite is $200\text{鈥�}\mathrm{kg}$

Circular orbit is $240\mathrm{km}$

Average energy loss is $1.4脳{10}^5\text{鈥�}\mathrm{J}$ revolution

Fromthegiven data, we can find the speed and period ofthesatellite. Moreover, from the energy formula, we can find the initial energy ofthesatellite as well astheenergy at the 1500^th orbit. After that, by using that energy, we can findthealtitude, speed, and period ofthesatellite.

Formula:

$\text{V}=\sqrt{\frac{\text{GM}}{\text{R}}}$

$\text{T}=\frac{\text{2蟺谤}}{\text{V}}$

${\text{E}}_{\text{0}}=-\frac{\text{GMm}}{\text{2r}}$

$\text{r}=-\frac{\text{GMm}}{\text{2E}}$

$\text{E}={\text{E}}_{\text{0}}-\text{nC}$

The radius of the orbit is the sum of the earth鈥檚 radius and the altitude of the satellite; it is

$\text{r}=(6.37脳{10}^6+640脳{10}^3)\mathrm{m}$

$\text{r}=7.01脳{10}^6\text{鈥�}\mathrm{m}$

We know that

$\begin{array}{c}\text{v}=\sqrt{\frac{\text{GM}}{\text{R}}}\\ =\sqrt{\frac{(6.67脳{10}^{鈭�11}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2)(5.98脳{10}^{24}\text{鈥�}\mathrm{kg})}{7.01脳{10}^6\text{鈥�}\mathrm{m}}}\\ =\sqrt{56.85脳{10}^6}\text{鈥尘/蝉}\\ =7.54脳{10}^3\text{m/s}\end{array}$

The period is given by

$\begin{array}{c}\text{T}=\frac{\text{2蟺谤}}{\text{v}}\\ =\frac{2\mathrm{蟺}(7.01脳{10}^3\text{鈥�}\mathrm{m})}{7.54脳{10}^3\text{鈥�}\mathrm{m}/\mathrm{s}}\\ =\frac{2\mathrm{蟺}(7.01脳{10}^3\text{鈥�}\mathrm{m})}{7.54脳{10}^3\text{鈥�}\mathrm{m}/\mathrm{s}}\\ =5.84脳{10}^3\mathrm{s}\end{array}$

The loss of energy per orbit

$\text{E}={\text{E}}_{\text{0}}-\text{nC}$

${\mathrm{E}}_0$be the initial kinetic energy.

Therefore,

${\text{E}}_{\text{0}}=鈭�\frac{\text{GMm}}{\text{2r}}$

$\text{r}=鈭�\frac{\text{GMm}}{\text{2E}}$

$\begin{array}{c}{\text{E}}_{\text{0}}=鈭�\frac{(6.67脳{10}^{鈭�11}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2)(5.98脳{10}^{24}\text{鈥�}\mathrm{kg})\left(220\text{鈥�}\mathrm{kg}\right)}{2(7.01脳{10}^6\text{鈥�}\mathrm{m})}\\ =鈭�6.26脳{10}^9\text{鈥�}\mathrm{J}\end{array}$

The energy after 1500 orbit,

$\begin{array}{c}\text{E}={\text{E}}_{\text{0}}\text{-nC}\\ =鈭�6.26脳{10}^9\text{鈥�}\mathrm{J}鈭�\left(1500\right)(1.4脳{10}^5\text{鈥�}\mathrm{J})\\ =鈭�6.47脳{10}^9\text{鈥�}\mathrm{J}\end{array}$

The radius after 1500 orbits,

$\begin{array}{c}\text{r}=鈭�\frac{\text{GMm}}{\text{2E}}\\ =鈭�\frac{(6.67脳{10}^{鈭�11}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2)(5.98脳{10}^{24}\text{鈥�}\mathrm{kg})\left(220\text{鈥�}\mathrm{kg}\right)}{2(鈭�6.47脳{10}^9\text{鈥�}\mathrm{J})}\\ =6.78脳{10}^6\mathrm{m}\end{array}$

Therefore, the altitude is

$\begin{array}{c}\mathrm{h}=\mathrm{r}鈭�\mathrm{R}\\ =(6.78脳{10}^6\text{鈥�}\mathrm{m}鈭�6.37脳{10}^6\text{鈥�}\mathrm{m})\\ =4.1脳{10}^5\text{鈥�}\mathrm{m}\end{array}$

$\begin{array}{c}\text{v}=\sqrt{\frac{\text{GM}}{\text{r}}}\\ =\sqrt{\frac{(6.67脳{10}^{鈭�11}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2)(5.98脳{10}^{24}\text{鈥�}\mathrm{kg})}{(6.78脳{10}^6\text{鈥�}\mathrm{m})}}\\ =7.67脳{10}^3\text{鈥尘/蝉}\end{array}$

$\begin{array}{c}\text{T}=\frac{\text{2蟺谤}}{\text{v}}\\ =\frac{2\mathrm{蟺}(6.78脳{10}^6\text{鈥�}\mathrm{m})}{7.67脳{10}^3\text{鈥�}\mathrm{m}/\mathrm{s}}\\ =5.6脳{10}^3\text{鈥�}\mathrm{s}\end{array}$

Work done by the satellite is the force on the satellite times distance traveled by it. In addition, the distance traveled by it is a circular orbit, sothedistance is given by $\text{s}=\text{2蟺谤}$

Work done by the satellite is given by a change in energy.

$\begin{array}{c}\text{W}=\text{E}\\ =鈭�\text{F}鈰�\text{s}\end{array}$

$\begin{array}{c}\text{F}=\frac{\text{E}}{\text{s}}\\ =\frac{1.4脳{10}^5\text{鈥�}\mathrm{J}}{4.40脳{10}^7\text{鈥�}\mathrm{m}}\\ =3.2脳{10}^{鈭�3}\text{鈥�}\mathrm{N}\end{array}$

The resistive force exerts a torque on the satellite therefore angular momentum is not conserved.

The satellite鈭抏arth system is essentially isolated, so its momentum is nearly conserved.