Tag: geostationary orbits

Questions Related to geostationary orbits

Select the correct statement from the following

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

  1. The orbital velocity of a satellite increase with the radius of the orbit

  2. Escape velocity of a particle from the surface of the earth depends on the speed with which it is fired

  3. The time period of a satellite does not depend on the radius of the orbit

  4. The orbital velocity is inversely proportional to the square root of the radius of the orbit

Show answer
Correct Option: B

The total energy of a satellite is-

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

  1. Always positive

  2. Always negative

  3. Always zero

  4. +ve or -ve depending upon radius of orbit.

Show answer
Correct Option: B
Explanation:

For every bounded system the total energy is always negative because if $k=x$ then $u$ will be $-2r$ so that $E=-x.$

So, the total energy of a satellite is negative.
Hence, the answer is negative.

Two identical satellites are at distance R and 7R from the surface of the earth of radius R. Which is the wrong statement from the following ?

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

  1. The ration of their total energies will be 4 but the ration of their potential and kinetic energies will be 2

  2. The ration of their potential energies will be 4

  3. The ration of their kinetic energies will be 4

  4. The ration of their total energies will be 4

Show answer
Correct Option: A

A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth R being the radius of the earth. What will be the time period of Another satellite at a height 2.5 R from the surface of the earth?

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

  1. 6 $\sqrt { 2 } $ hours

  2. 6 $\sqrt { 2.5 } $ hours

  3. 6 $\sqrt { 3 } $ hours

  4. 12 hours

Show answer
Correct Option: A

Which of the following is true ? 

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

  1. A polar molecule is one in which "centre of gravity" of positive nuclei and revolving electrons coincide.

  2. In polar dielectric material the different tiny electric dipoles are oriented in only one direction in the absence of electric field.

  3. For non polar dielectric material net dipole moment is nonzero in absence of electric field.

  4. Dielectric material develops a net dipole moment in presence of external electric field.

Show answer
Correct Option: A

At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?

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

  1. $\dfrac{2R}{5}$

  2. $\dfrac{2R}{3}$

  3. $\dfrac{3R}{7}$

  4. $\dfrac{R}{3}$

Show answer
Correct Option: B
Explanation:

We have,

$\begin{array}{l} \dfrac { { GM } }{ { { { \left( { R+h } \right)  }^{ 2 } } } } =\dfrac { { 36 } }{ { 100 } } \dfrac { { Gm } }{ { { R^{ 2 } } } }  \ \Rightarrow 100{ R^{ 2 } }=36{ \left( { R+h } \right) ^{ 2 } } \ \Rightarrow 25{ R^{ 2 } }=9\left( { { R^{ 2 } }+{ h^{ 2 } }+2Rh } \right)  \ \Rightarrow 25{ R^{ 2 } }=9{ R^{ 2 } }+9{ h^{ 2 } }+18Rh \ \Rightarrow 16{ R^{ 2 } }=9{ h^{ 2 } }+18Rh \ \Rightarrow 9{ h^{ 2 } }+18Rh-16{ R^{ 2 } }=0 \ R=\dfrac { { -18+\sqrt { 324+576 }  } }{ { 18 } } R \ =\dfrac { { -18+30 } }{ { 18 } } R \ =\dfrac { { 12R } }{ { 18 } } =\dfrac { { 2R } }{ 3 }  \ h=\dfrac { { 2R } }{ 3 }  \end{array}$
Then,
Option $B$ is correct answer.

What is the nature of relation betweenthe kinetic energy $\left( \mathrm { E } _ { \mathrm { k } } \right)$ and their orbitalradius $( \mathrm { r } )$ of the satellites revolvingaround the Earth?

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

  1. $E _ { k } \propto 1$

  2. $E _ { k } \propto \frac { 1 } { r }$

  3. $E _ { k } \propto r ^ { 2 }$

  4. $E _ { k } \propto \frac { 1 } { r ^ { 2 } }$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} \dfrac { { GMm } }{ { { r^{ 2 } } } } =\dfrac { { m{ v^{ 2 } } } }{ r }  \ \Rightarrow \dfrac { { m{ v^{ 2 } } } }{ 2 } =\dfrac { { GMm } }{ { 2r } }  \ \therefore K _E\propto \dfrac { 1 }{ r }  \end{array}$

$\therefore $ Option $B$ is correct .

An object weighs 10$\mathrm { N }$ at the north pole of the Earth. In a geostationary satelite at a distance of 7R from the centre of the Earth (of radius $\mathrm { R } )$ , the true weight and the apparent weight are respectively.-

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

  1. 0,0

  2. $0.2 \mathrm { N } , 0$

  3. $0.2 \mathrm { N } , 9.8 \mathrm { N }$

  4. $0.2 N , 0.2 \mathrm { N }$

Show answer
Correct Option: A

Two artificial satellite of masses $ m _1 $ and $ m _2 $ are moving with speed $ v _1 $ and $ v _2 $ in orbits of radii$ r _1 $ and $ r _2 $ respectively. if $ r _ 1>r _2 $ then which of the following statements in true:-

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 $

Show answer
Correct Option: C

A particle is projected upward from the surface of earth (radius  $= R$ ) with a speed equal to the orbital speed of a satellite near the earth's surface. The height to which it would rise is

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

  1. $\sqrt { 2 } R$

  2. $\dfrac { R } { \sqrt { 2 } }$

  3. $R$

  4. $2 R$

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} V=\sqrt { \dfrac { { GM } }{ R }  }  \ \frac { { -GMm } }{ R } +\dfrac { 1 }{ 2 } m\left( { \dfrac { { GM } }{ R }  } \right) =\dfrac { { -GMm } }{ { \left( { R+h } \right)  } } +0 \ \Rightarrow \dfrac { { -GMm } }{ R } +\dfrac { { GMm } }{ { 2R } } =\dfrac { { -GMm } }{ { \left( { R+h } \right)  } }  \ \Rightarrow \dfrac { { -GMm } }{ { 2R } } =\dfrac { { -GMm } }{ { \left( { R+h } \right)  } }  \ \Rightarrow 2R=R+h \ \Rightarrow r=R \ Hence, \ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$