Tag: artificial satellite

Quizzes (9) Questions (109)

Questions Related to artificial satellite

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

The total energy of a satellite is-

  1. Always positive

  2. Always negative

  3. Always zero

  4. +ve or -ve depending upon radius of orbit.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 ?

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

Total energy E = -GMm / (2r). For r1 = 2R and r2 = 8R (distances from center), the ratio of energies is 8/2 = 4. The potential energy U = -GMm/r, ratio is 8/2 = 4. Kinetic energy K = GMm/2r, ratio is 8/2 = 4. Option A is confusingly worded.

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

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?

  1. 6 $\sqrt { 2 } $ hours
  2. 6 $\sqrt { 2.5 } $ hours
  3. 6 $\sqrt { 3 } $ hours

  4. 12 hours
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Kepler's Third Law: T^2 is proportional to r^3. r1 = 7R, r2 = 3.5R. (T2/T1)^2 = (3.5R/7R)^3 = (1/2)^3 = 1/8. T2 = T1 / sqrt(8) = 24 / (2*sqrt(2)) = 12 / sqrt(2) = 6*sqrt(2) hours.

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

Which of the following is true ? 

  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.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Polar molecules have centers of positive and negative charges that do not coincide, resulting in a permanent dipole moment. In the presence of an external electric field, any dielectric material develops a net induced dipole moment due to alignment or stretching of molecular dipoles. Thus, option D correctly describes the behavior of dielectrics in an electric field.

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

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

  1. $\dfrac{2R}{5}$
  2. $\dfrac{2R}{3}$
  3. $\dfrac{3R}{7}$
  4. $\dfrac{R}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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?

  1. $E _ { k } \propto 1$
  2. $E _ { k } \propto \frac { 1 } { r }$
  3. $E _ { k } \propto r ^ { 2 }$
  4. $E _ { k } \propto \frac { 1 } { r ^ { 2 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 .

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

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.-

  1. 0,0
  2. $0.2 \mathrm { N } , 0$
  3. $0.2 \mathrm { N } , 9.8 \mathrm { N }$
  4. $0.2 N , 0.2 \mathrm { N }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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:-

  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
C Correct answer
Explanation

Orbital velocity v = sqrt(GM/r). As r increases, v decreases. Since r1 > r2, v1 < v2.

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

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

  1. $\sqrt { 2 } R$
  2. $\dfrac { R } { \sqrt { 2 } }$
  3. $R$
  4. $2 R$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  1. It mu always be stationed above the equator.

  2. It must rotate from west to east.

  3. It must be about 36,000 km above the earth.

  4. Its orbit must be circular, and not elliptical.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Since the satellite rotates in a plane which passes through the centre of earth, for it to be stationary relative to the earth, its angular velocity must be same as that of earth (direction also). And hence it must rotate in the equitorial plane.


Since the angular velocity is same as that of earth, its direction must be west to east.

Balancing forces,

$\dfrac { GMm }{ { (R+h) }^{ 2 } } =m{ \omega  }^{ 2 }(R+h)$

$\omega =\sqrt { \dfrac { GM }{ R+h }  } =\dfrac { 2\pi  }{ 3600X24 } $

This gives $h=36000km$

For constant $\omega$, orbit must be circular.

Answer is ABCD.