Time Period And Energy Of A Satellite - Practice Questions & MCQ

 The time period of satellite-

  It is the time taken by satellite to go once around the earth.         

And the time period (T) of the satellite is given by 

$
\begin{array}{ll}
T=\frac{2 \pi r}{v}=2 \pi r \sqrt{\frac{r}{G M}} & {\left[\text { As } v=\sqrt{\frac{G M}{r}}\right]} \\
T=2 \pi \sqrt{\frac{r^3}{G M}}=2 \pi \sqrt{\frac{r^3}{g R^2}} & {\left[\text { As } G M=g R^2\right]} \\
T=2 \pi \sqrt{\frac{(R+h)^3}{g R^2}}=2 \pi \sqrt{\frac{R}{g}}\left(1+\frac{h}{R}\right)^{3 / 2} & {[\text { As } r=R+h]}
\end{array}
$

Where
$r=$ radius of orbit
$T \rightarrow$ Time period
$M \rightarrow$ Mass of planet
- If the satellite is very close to the earth's surface, i.e., $h \ll \ll R$,
then

$
\begin{aligned}
& T=2 \pi \sqrt{\frac{R}{g}} \cong 84.6 \text { minutes } \\
& \text { hen } \\
& \text { or } T \simeq 1.4 \mathrm{hr}
\end{aligned}
$

- The time period of a satellite in terms of density

$
T=\sqrt{\frac{3 \pi}{G \rho}}
$

$\rho \rightarrow$ Density of planet
$T \rightarrow$ Time period
$G \rightarrow$ Gravitational constant

$
\rho=5478.4 \mathrm{Kg} / \mathrm{m}_{\text {for earth }}^3
$
 

• For a satellite, the time interval between the two consecutive appearances overhead

    If a satellite in the equilateral planes moves from west to east Angular velocity of the satellite  with respect to an observer on earth will be 

$\left(\omega_S-\omega_E\right)$
$\omega_S \rightarrow$ Satellite angular velocity
$\omega_E \rightarrow$ earth angular velocity
So $T=\frac{2 \pi}{\omega_S-\omega_E}=\frac{T_S T_E}{T_E-T_S}$
if $\quad \omega_S=\omega_E, T=\infty$
means satellite will appear stationary relative to earth.
Height of Satellite-
As we know, time period of satellite $T=2 \pi \sqrt{\frac{r^3}{G M}}=2 \pi \sqrt{\frac{(R+h)^3}{g R^2}}$ By squaring and rearranging both sides $\frac{g R^2 T^2}{4 \pi^2}=(R+h)^3$

$
\Rightarrow \quad h=\left(\frac{T^2 g R^2}{4 \pi^2}\right)^{1 / 3}-R
$

Putting the value of time period in the above formula we can calculate the height of the satellite from the surface of the earth.

The energy of Satellite-
When a satellite revolves around a planet in its orbit, it possesses both kinetic energy (due to orbital motion) and potential energy (due to its position against the gravitational pull of earth).

And these energies are given by
Potential energy: $U=m V=\frac{-G M m}{r}=\frac{-L^2}{m r_2^2}$
Kinetic energy : $K=\frac{1}{2} m v^2=\frac{G M^T m}{2 r}=\frac{L_{L^2}^2}{2 m r^2}$
Total energy : $E=U+K=\frac{-G M m}{r}+\frac{G M m}{2 r}=\frac{-G M m}{2 r}=\frac{-L^2}{2 m r^2}$
Where
$M \rightarrow$ mass of planet

$m \rightarrow$ mass of satellite
And

$
\begin{aligned}
& K=-E \\
& U=2 E \\
& U=-2 K
\end{aligned}
$
 

•     Energy Graph of satellite

Where

$E \rightarrow$ Energy of satellite
$K \rightarrow$ Kinetic energy
$U \rightarrow$ Potential energy

• Energy distribution in an elliptical orbit

In this Total Energy

$
E=-\frac{G M m}{2 a}=\text { const. }
$

Where $a=$ semi - major axis
- Binding Energy (B.E.)-

The minimum energy required to remove the satellite from its orbit to infinity is called Binding Energy.
And It is given by

$
B \cdot E=\frac{G M m}{2 r}
$

where
B. $E \rightarrow$ Binding energy
$M \rightarrow$ mass of planet
$m \rightarrow$ mass of satellite
- Work done in changing the orbit-

When the satellite is transferred to a higher orbit i.e $\left(r_2>r_1\right)$ as shown in the figure.

$
\begin{aligned}
W & =E_2-E_1 \\
W & =\frac{G M m}{2}\left[\frac{1}{r_1}-\frac{1}{r_2}\right]
\end{aligned}
$

Where
$W \rightarrow$ work done
$r_1 \rightarrow$ radius of 1 st orbit
$r_2 \rightarrow$ radius of 2 nd orbit