Tag: geostationary orbits

Questions Related to geostationary orbits

Moon can be classified as a

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

  1. Asteroid

  2. Comet

  3. Satellite

  4. Star

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Correct Option: C
Explanation:

Moon revolves around earth with certain fixed velocity in a circular orbit of certain radius. Moon is a natural satellite of earth whose time period of revolution is $27.3$ days.

It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi.

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

  1. True

  2. False

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Correct Option: A

Parking orbit for a geostationary satellite is

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

  1. at 45 degrees north west of equator

  2. is in equatorial plane of earth

  3. in west-east direction.

  4. along the north direction

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Correct Option: B
Explanation:

Parking orbit for a geostationary satellite is in equatorial plane of earth, since it has to be at rest always with respect to equator

The correct option is option (b)

A geostationary satellite is at rest, relative to earth only for points in the 

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

  1. equitorial plane

  2. plane passing through the poles of the earth

  3. plane along magnetic north and south of the earth

  4. planes that are isoclinic

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Correct Option: A
Explanation:

A geostationary satellite is at rest, relative to earth only for points in the equitorial plane.

The correct option is (a)

The speed of a geostationary satellite relative to a person on the earth is 

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

  1. Zero

  2. 2 km/s

  3. 10 km/s

  4. 8 km/s

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Correct Option: A
Explanation:

A geostationary satellite is a satellite which always appears to be stationary from an earth's surface.

The correct option is option (a)

Time period of a simple pendulum inside a satellite orbiting earth is

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

  1. Zero

  2. $\infty$

  3. $T$

  4. $2T$

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Correct Option: B
Explanation:

As we know, $T\alpha \dfrac { 1 }{ \sqrt { g }  } $

On a artificial satellite, orbiting the earth, neg gravity is zero. As,
$\Rightarrow \quad g=0\Rightarrow T\rightarrow \infty $
$\Rightarrow$  Option B is the correct answer.

Two satellites of masses $m _{1}$ and $m _{2} (m _{1} > m _{2})$ are revolving round the earth in circular orbits of radii $r _{1}$ and $r _{2}(r _{1} > r _{2})$ respectively. Which of the following statements is true regarding their speeds $v _{1}$ and $v _{2}$?

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

  1. $v _{1} = v _{2}$

  2. $v _{1} < v _{2}$

  3. $v _{1} > v _{2}$

  4. $(v _{1}/r _{1}) = (v _{2}/r _{2})$

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Correct Option: B
Explanation:

Velocity of satellite $=\sqrt { \dfrac { GM }{ R+h }  } $

Now, 
        ${ V } _{ 1 }=\sqrt { \dfrac { GM }{ { r } _{ 1 } }  } $
        ${ V } _{ 2 }=\sqrt { \dfrac { GM }{ { r } _{ 2 } }  } $
As ${ r } _{ 1 }>{ r } _{ 2 }$
So  ${ V } _{ 1 }<{ V } _{ 2 }$

An artificial satellite is moving around earth in a circular orbit with speed equal to one fourth the escape speed of a body from the surface of earth. The height of satellite above earth is : ($R$ is radius of earth)

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

  1. $3R$

  2. $5R$

  3. $7R$

  4. $8R$

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Correct Option: D
Explanation:
Let's consider,

$r=$ height  of the satellite above the earth

$R=$ radius of the earth

$G=$ gravitational constant

$g=$ acceleration due to gravity

The escape velocity of a body from the surface of earth,

$v _e=\sqrt{\dfrac{2GM}{R}}$

$\dfrac{1}{4}.v _e=\dfrac{1}{4}\sqrt{\dfrac{2GM}{R}}$. . . . . .(1)

The orbital velocity of the satellite is given by

$v _o=\sqrt{\dfrac{GM}{r}}$. . . . . .(2)

Equation equation (1) and (2), we get

$\sqrt{\dfrac{GM}{r}}=\dfrac{1}{4}\sqrt{\dfrac{2GM}{R}}$

$\dfrac{GM}{r}=\dfrac{1}{16}.\dfrac{2GM}{R}$

$r=8R$

The correct option is D.

A stationary object is released from a point $P$ at a distance $3R$ from the centre of the moon which has radius $R$ and mass $M$. Which of the following gives the speed of the object on hitting the moon?

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

  1. $\left (\dfrac {2GM}{3R}\right )^{1/2}$

  2. $\left (\dfrac {4GM}{3R}\right )^{1/2}$

  3. $\left (\dfrac {GM}{3R}\right )^{1/2}$

  4. $\left (\dfrac {GM}{R}\right )^{1/2}$

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Correct Option: C

Suppose that the moon travels in a circle about the earth at a distance $ 3.84 \times 10^8 m$ once in every 28.3 days and that has a mass of $7.4 \times 10^{22}$ . Then the speed of the moon is most nearly:

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

  1. $10 m/s $

  2. $10^3 m/s$

  3. $10^5 m/s$

  4. $10^7 m/s$

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Correct Option: B