Tag: geostationary orbits

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Questions Related to geostationary orbits

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A geostationary satellite must remain directly above a fixed point on the earth's equator, which requires its orbital plane to be the equatorial plane. Since New Delhi is located at a northern latitude and not on the equator, no satellite can remain directly above it at all times.

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

Parking orbit for a geostationary satellite is

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

The correct option is (a)

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

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

  1. Zero
  2. $\infty$
  3. $T$
  4. $2T$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 $v _{2}$?

  1. $v _{1} = v _{2}$
  2. $v _{1} < v _{2}$
  3. $v _{1} > v _{2}$
  4. $(v _{1}/r _{1}) = (v _{2}/r _{2})$
Reveal answer Fill a bubble to check yourself
B Correct answer
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  ${ V } _{ 1 }<{ V } _{ 2 }$

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

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)

  1. $3R$
  2. $5R$
  3. $7R$
  4. $8R$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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?

  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. $\left (\dfrac {GM}{R}\right )^{1/2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using conservation of mechanical energy, the initial energy at distance 3R from the center equals the final energy at the surface of the moon (distance R). Setting initial kinetic energy to zero, we get minus G M m over 3R equals half m v squared minus G M m over R. Solving for v yields the square root of (4 G M over 3R), making option B correct.

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

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:

  1. $10 m/s $
  2. $10^3 m/s$
  3. $10^5 m/s$
  4. $10^7 m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Speed v = 2*pi*r / T. r = 3.84 * 10^8 m, T = 28.3 days = 28.3 * 24 * 3600 seconds approx 2.44 * 10^6 s. v = 2 * 3.14 * 3.84 * 10^8 / 2.44 * 10^6 approx 1000 m/s.