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Obtain an expression of total energy of satellite revolving around the earth.
Solution
Suppose, a satellite of mass $m$ revolves around the earth at a distance $r$ from the centre of earth. Its orbital velocity is $v_{0}$.
Satellite is in gravitational field hence its potential energy,
$\mathrm{V}=\frac{\mathrm{GM}_{\mathrm{E}} m}{r}$
Kinetic energy of satellite,
$\mathrm{K} =\frac{1}{2} m v_{0}^{2}$
$\therefore \mathrm{K} =\frac{1}{2} m\left(\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{r}}\right)^{2}\left(\because v_{0}=\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{r}}\right)$
$\therefore \mathrm{K}=\frac{\mathrm{GM}_{\mathrm{E}} m}{2 r} \text { where } r=\mathrm{R}_{\mathrm{E}}+h$
Total energy of satellite,
$\mathrm{E}=\mathrm{V}+\mathrm{K}$
$\mathrm{E}=-\frac{\mathrm{GM}_{\mathrm{E}} m}{2 r}$
Negative sign indicates that satellite is in bound state to the gravitational field of earth. Kinetic energy and potential energy at infinity distance is zero. Hence total energy will be zero. To escape to initial energy equal to total energy or greater than this positive energy is given to the satellite, otherwise it rotating constantly in orbit. This indicates this satellite is in the bound state with gravitational field of earth.
Similar Questions
Match list-$I$ with list-$II$:
| List-$i$ | List-$2$ |
| $(A)$Kinetic energy of plant | $(1)$ $-\frac{\mathrm{GMm}}{\mathrm{a}}$ |
| $(B)$ Gravaitatioin potentiyal energy of sun -plant system | $(2)$ $\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
| $(C)$Total mecaniacal energy of palnt | $(3)$ $\frac{\mathrm{Gm}}{\mathrm{r}}$ |
| $(D)$Escap energyat the surface of plant for unit mass object | $(4)$ $-\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
(Where $\mathrm{a}=$ radius of planet orbit, $\mathrm{r}=$ radius of planet, $M=$ mass of Sun, $m=$ mass of planet)
Choose the correct answer from the options given below:
