Tag: artificial satellite

Questions Related to artificial satellite

Orbital decay, a process of prolonged reduction in the attitude of a satellites orbit is caused by 
A) Atmospheric drag   B) Gravitational Pull      C) Tides

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

  1. A only

  2. C only

  3. A and C only

  4. A, B and C

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Correct Option: D

If the length of the day is $T$ , the height of that TV satellite above the earth's surface which always appears stationary from earth, will be:


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

  1. $h = \left[ \dfrac { 4 x ^ { 2 } G m } { T ^ { 2 } } \right] ^ { - 6 }$

  2. $h = \left[ \dfrac { 4 x ^ { 2 } G M } { T ^ { 2 } } \right] ^ { - 1 / 2 } - R$

  3. $h = \left[ \dfrac { G M T ^ { 2 } } { 4 \pi ^ { 2 } } \right] ^ { 1/3 } - R$

  4. $h = \left[ \dfrac { G M T ^ { 2 } } { 4 \pi ^ { 2 } } \right] ^ { 2 } + R$

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Correct Option: C

A planet of small mass m moves around the sun of mass M along an elliptical orbit such that its minimum and maximum distance from the sun are r and R respectively. Its period of revolution will be: 

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

  1. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{6GM}}} $

  2. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{3GM}}} $

  3. $\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{2GM}}} $

  4. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{GM}}} $

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Correct Option: D

A geostationary 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 $\displaystyle { R } _{ earth }={ 6400 } \quad km$ will approximately be

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

  1. $\cfrac { 1 }{ 2 } { h }$

  2. ${ 1h }$

  3. ${ 2h }$

  4. ${ 4h }$

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Correct Option: C
Explanation:

Given :   $R _{G}= 36000    km                              R _{S}= 6400+ 100= 6500     km$

Time period, $T= 2\pi\sqrt{\dfrac{r}{GM}}$
Thus   $(\dfrac{T _S}{T _G})^2= (\dfrac{R _S}{R _G})^3$ 
 $(\dfrac{T _S}{24 })^2= (\dfrac{6500}{36000})^3$ 
$\implies     T _S= 1.84    h        \approx  2  h$

A space shuttle is revolving around the earth in circular orbit. A certain point pilot fires forward pointing thruster to decrease shuttle's mechanical energy. Then orbital time period $T$ of shuttle

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

  1. Will increase

  2. Will decrease

  3. Will remain constant

  4. Will first decrease and then increase.

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Correct Option: B

Geo-stationary satellite is one which

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

  1. Remains stationary at a fixed height from the eath's surface

  2. Revolves like other satellites but in the opposite direction of eath's rotation

  3. Revolves round the earth at a suitable height with same angular velocity and in the same direction as earth does about its own axis

  4. None of these

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Correct Option: C
Explanation:

Geostationary satellites are also called synchronous satellite. They always remain about the same path on equator, i.e., it has a period of exactly one day $\displaystyle(86400)$
So orbit radius $\displaystyle \left [ T=2\pi \sqrt {\frac{r^3}{GM}} \right ]$ comes out to be $42400$ km, which is nearly equal to the circumference of earth. So height of Geostationary satellite from the earth surface is $42,400-6400=36,000$km.

The relay satellite transmits the $TV$ programmed continuously from one part of the world to another because its

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

  1. Period is greater than the period of rotation of the earth

  2. Period is less than the period of rotation of the eath about its axis

  3. Period has no relation with the period of the earth about its axis

  4. Period is equal to the period of rotation of the earth about its axis

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Correct Option: D
Explanation:

The period of satellite is equal to period of rotation of earth about its own axis and it seems to be at one point about the equator and so is able to transmit the signals from one part to other

For a geostationary satellite orbiting around the earth identify the necessary condition

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

  1. it must lie in the equatorial plane of earth

  2. its height from the surface of earth must be $36000 km $

  3. it period of revolution must be $\displaystyle 2\pi \sqrt{\frac{R}{g}}$ where R is the radius of earth

  4. its period of revolution must be $24 hrs$

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Correct Option: A,B,D
Explanation:

A geostationary satellite must satisfy the following requirements:

$1$. Its orbit must lie on an equatorial plane.
$2$. It must appear stationary when viewed from a point on earth which means its time period of revolution is $24 hrs$.
$3$. Its height above the surface of the earth must be $36000 km$.
So options A, B and D are correct.

A satellite is seen every $6$ hours over the equator. It is known that it rotates opposite to that of earth's direction. Then the angular velocity (in radian per hour) of satellite about the centre of earth will be :

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

  1. $\displaystyle\dfrac{\pi}{2}$

  2. $\displaystyle\dfrac{\pi}{3}$

  3. $\displaystyle\dfrac{\pi}{4}$

  4. $\displaystyle\dfrac{\pi}{8}$

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Correct Option: C
Explanation:
Let $w _1$ and $w _2$ be the angular speed of the satellite and earth respectively.
Angular speed of rotation of earth  $w _2 = \dfrac{2\pi}{24}$ $rad/hr$ 
Since, both are revolving in opposite direction. Thus sum of angle rotated by them in time $t = 6 $ hr is $2\pi$.   
$\therefore$  $\omega _{1}t+\omega _{2}t=2\pi$
$(\omega _{1})6+\left (\displaystyle\dfrac{2\pi}{24} \right ){6}=2\pi$

or   $6\omega _{1}=\displaystyle\dfrac{3\pi}{2}$

 $\implies \omega _{1}=\left (\displaystyle\dfrac{\pi}{4} \right )\displaystyle\dfrac{rad}{hr}$

Geostationary satellite

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

  1. is situated at a great height above the surface of the Earth.

  2. moves in the equatorial plane.

  3. have time period of $24$ hours.

  4. have time period of $24$ hours and moves in the equatorial plane.

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Correct Option: D
Explanation:
Geostationary satellite is used for communication purpose.
Thus, geostationary satellites are placed at an altitude of 36,000 kms and moves in the equatorial plane in the same direction as the Earth rotates. Hence, its time period is 24 hours.