Tag: geostationary orbits

Questions Related to geostationary orbits

Height of geostationary satellite is

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. $16000km$

  2. $22000km$

  3. $28000km$

  4. $36000km$

Show answer
Correct Option: D
Explanation:

$Answer:-$ D

Approximately $35,786$ km.
A geostationary equatorial orbit (GEO) is a circular geosynchronous orbit in the plane of the Earth's equator with a radius of approximately $42,164$ km ($26,199$ mi) (measured from the center of the Earth). A satellite in such an orbit is at an altitude ofapproximately $35,786$ km ($22,236$ mi) above mean sea level.

Two planets, $X$ and $Y$, revolving in a orbit around star. Planet $X$ moves in an elliptical orbit whose semi-major axis has length $a$. Planet $Y$ moves in an elliptical orbit whose semi-major axis has a length of $9a$. If planet $X$ orbits with a period $T$, Find out the period of planet $Y$'s orbit?

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. $729T$

  2. $27T$

  3. $3T$

  4. ${T}/{3}$

  5. ${T}/{27}$

Show answer
Correct Option: B
Explanation:

Kepler's third law of planetary motion gives    $T^2\propto r^{3}$

    where $T=$ time period of revolution  ,  $r=$ length of semi major axis of elliptical orbit  

by this relation we get  ,  $\dfrac{T _{X}}{T _{Y}}=\dfrac{r _{X}^{3}}{r _{Y}^{3}}$

                                   $\dfrac{T _{X}^2}{T _{Y}^2}=\dfrac{a^{3}}{\left({9a}\right)^{3}}$

but given  $T _{X}=T$

                therefore   $\dfrac{T^2}{T _{Y}^2}=\dfrac{a^{3}}{729a^{3}}$

               or              $T _{Y}=27T$


An object is released from rest at a distance of ${2r} _{e}$ from the center of the Earth, where ${r} _{e}$ is the radius of the Earth. Find out the velocity of the object when it hits the Earth in terms of the gravitational constant $\left(G\right)$, the mass of the Earth $\left(M\right)$, and ${r} _{e}$.

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. $\sqrt{{GM}/{{r} _{e}}}$

  2. ${GM}/{{r} _{e}}$

  3. $\sqrt{{GM}/{2{r} _{e}}}$

  4. ${GM}/{2{r} _{e}}$

  5. $2{GM}/{{r} _{e}}$

Show answer
Correct Option: A
Explanation:

Given:

Initial velocity, $u=0$  ,  $h=2r _{e}$
Now, by using: $v^{2}=u^{2}+2gh$

                          $v^{2}=0+2g\times2r _{e}$

By putting  $g=GM/{r _{e}}^{2}$ in above equation 
We get, $v^{2}=2GM/{r _{e}}^{2}\times2r _{e}$
             $v^{2}=GM/2r _{e}$
             $v=\sqrt{GM/2r _{e}}$

A geosynchronous orbit is one in which the satellite makes one revolution around the Earth in 24 hrs.
How far above the surface of the Earth does this satellite have to orbit?
Assume the radius of the Earth is $6.37 \times {10}^{6} m$, and the mass of the Earth is $5.98 \times {10}^{24} kg$.

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. $3.59 \times {10}^{7} m$

  2. $4.23 \times {10}^{7} m$

  3. $2.76 \times {10}^{6} m$

  4. $6.37 \times {10}^{6} m$

  5. $1.27 \times {10}^{7} m$

Show answer
Correct Option: A
Explanation:

Given :   $T = 24$ hrs $ = 84400$  s               $M _e = 5.98 \times 10^{24}$ kg            $R _e  = 0.637 \times 10^7$  m

Time period of satellite moving in orbit of radius $r$            $T = 2\pi \sqrt{\dfrac{r^3}{GM _e}}$

$\therefore$    $86400 = 2\pi \sqrt{\dfrac{r^3}{(6.67 \times 10^{-11}) \times (5.98 \times 10^{24})}}$                         $\implies r  =4.23 \times 10^7$ m

Thus height of satellite above the surface        $h = r - R _e = (4.23 - 0.637) \times 10^7  \approx 3.59 \times 10^7$  m

Imagine a geostationary satellite of earth which is used as an inter continental telecast station. At what height will it have to be established?

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. at $10^{3}m$

  2. at $6.4\times 10^{3}m$

  3. at $35.94\times 10^{6}m$

  4. at infinity

Show answer
Correct Option: C
Explanation:

Time period for a geostationary satellite is 24 hour, so $t=24$

And $t=2\pi \sqrt{\dfrac{(R+h)^2}{GM}}$
So, $h=35.94\times 10^6m$

Which of the followings are correct uses of satellite placed in an orbit around Earth?

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. For observing Earth from a distance

  2. For communication purposes

  3. To forecast weather

  4. All of the above

Show answer
Correct Option: D
Explanation:
Satellite revolves around the earth with fixed a velocity in a circular orbit of certain radius. During its motion, satellite collects informative signals from earth and then sends those signal back to the other part of earth for interpretation. Satellites are used for weather forecasting, communication purposes and observing earth from a distance, GPS etc. 

For a satellite to be geostationary, which of the following are essential conditions?

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. it must always be stationed above the equator

  2. it must be rotate from west to east

  3. it must be about $36,000 km$ above the earth surface

  4. it's orbit must be circular, and not elliptical

Show answer
Correct Option: A,B,C,D
Explanation:

Geostationary satellites revolve in the equatorial plane.
It has to revolve along Earth's rotation direction that is from west to east.
For the time period to be close to $24$ hours, the height of the satellite has to be $36,000\ km$ from earth's surface.
And the orbit is circular.
All options.

Given that the universal gravitational constant, $G = 6.7 10^{-11} Nm^{2} kg^{-2}$ and that the mass,
M of the earth is $6.0 10^{24} kg$, find the speed of a satellite that is fixed to permanently
focus on the city of Abuja for broadcast of the 2010 IJSO competition.

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. $ 3.08 \times 10^{3} ms^{-1}$

  2. $24 ms^{-1}$

  3. $40 ms^{-1}$

  4. $3.66 10^{3} ms^{-1}$

Show answer
Correct Option: A
Explanation:

Centripetal force $=$ gravitational force
,$ \Rightarrow   mv _{2}r=GmMr _{2} , from     where    v _{2}=GMr$
where v is the velocity. Period, $T=2\pi rv$ giving
$r=Tv2\pi$
Hence $v _{3}=2\pi GMT   With   T = 24 hr = 8.64 \times  104 s$
$v=32\pi\times 6.7 10^{-11} \times 6.0 \times 10248.64\times 104 = 3.08 10^{3} ms$

For geo stationary satelites,

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. Time period depends on the mass of the satelite

  2. The orbit radius is independent of the mass of earth.

  3. The period is equal to that of the rotation of earth about its axis

  4. None of these

Show answer
Correct Option: C
Explanation:
For geostationary satellites is a circular geosynchronous orbit above earth's equator follows direction of earths rotation ship period is equal to rotation of earth about its axis.

A geo-stationary satellite orbits around the earth in a circular orbit of radius $36000\ km$. Then, the time period of a spy satellite orbiting a few $100\ km$ above the earth's surface ($R _{earth}=6400\ km$) will approximately be -

evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

  1. $1/2\ hr$

  2. $1\ hr$

  3. $2\ hr$

  4. $4\ hr$

Show answer
Correct Option: C
Explanation:
Satellite orbits in a radius of $36000㎞\quad \quad { R } _{ 1 }=36000㎞$
${ R } _{ earth }=6400㎞,\quad { R } _{ 2 }=6400$
By Lepler's third law, we know that
${ T }^{ 2 }\propto { R }^{ 3 }$
Also we know that time period of geostationary satellite is $24h$
${ T } _{ 1 }=24h$
$\therefore { \left( \cfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 2 }={ \left( \cfrac { { R } _{ 1 } }{ { R } _{ 2 } }  \right)  }^{ 3 }$
$\Rightarrow \cfrac { { \left( 24 \right)  }^{ 2 } }{ { \left( { T } _{ 2 } \right)  }^{ 2 } } ={ \left( \cfrac { 36000 }{ 6400 }  \right)  }^{ 3 }$
${ T } _{ 2 }=24{ \left( \cfrac { 6400 }{ 36000 }  \right)  }^{ 3/2 }$
${ T } _{ 2 }=2h$