Tag: moon and stars in sky

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Questions Related to moon and stars in sky

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

Select the correct statement from the following

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

Escape velocity v_e = sqrt(2GM/R) is a property of the planet and the starting point, not the speed at which it is fired (though it must be fired at least at that speed to escape). The other statements are physically incorrect.

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
A Correct answer
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

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