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Time Period And Energy Of A Satellite MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Time period and energy of a satellite is considered one the most difficult concept.

  • 32 Questions around this concept.

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A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has a radius R (R<<L). A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its center. If the period of the star is T and its distance from the Galaxies axis is r, then :

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

 

The time period of an earth satellite in circular orbit is independent of :

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A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity \mathrm{\frac{3\pi }{Gd}} represents:

Concepts Covered - 1

Time period and energy of a satellite

 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}}} & {\text { [As } v=\sqrt{\frac{G M}{r}} ]} \\ {T=2 \pi \sqrt{\frac{r^{3}}{G M}}=2 \pi \sqrt{\frac{r^{3}}{g R^{2}}}} & {\text { [As } G M=g R^{2} ]} \\ {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<<<R,

             then  T=2\pi \sqrt\frac{R}{g}\cong 84.6\ minutes
               or T\simeq1.4hr

  • 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= 5478.4Kg/m^{3} for earth

  • 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     \omega _{S}=\omega _{E}, T= \infty

means satellite will appear stationary relative to earth.

 Height of Satellite-

\begin{array}{l}{\text { 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}}}} \\ {\text { 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}\end{array}

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 

\begin{array}{l}{\text { Potential energy: } U=m V=\frac{-G M m}{r}=\frac{-L^{2}}{m r^{2}}} \\ {\text { Kinetic energy : } K=\frac{1}{2} m v^{2}=\frac{G M m}{2 r}=\frac{L^{2}}{2 m r^{2}}} \\ {\text { 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}}}\end{array}

Where 

M\rightarrow mass\: of\: planet

m\rightarrow mass\: of\: satellite

   And 

K=-E

U=2E

U=-2K

  •     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{GMm}{2a}=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.E=\frac{GMm}{2r}

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.

W=E_{2}-E_{1}

W=\frac{GMm}{2}\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]

Where 

W\rightarrow work\: done

r_{1}\rightarrow radius\: of\: 1st\: orbit

r_{2}\rightarrow radius\: of\: 2nd\: orbit

 

 

 

 

 

 

 

 

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Time period and energy of a satellite

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