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Moon and stars in sky - class-XI

Description: moon and stars in sky
Number of Questions: 90
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Tags: evs physics sunita in space gravitation: planets and satellites gravitational fields artificial satellite gravitation search of life outside earth
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LANDSAT series of satellite move in near polar orbits at an altitude of

  1. 3600km

  2. 3000km

  3. 918km

  4. 512km


Correct Option: C
Explanation:

LANDSAT 1 through $3$ operated in a near-polar orbit at an altitude of $920 km$ with an $18-day$ repeat coverage cycle. These satellites circled the earth every 103 min completing $14$ orbits a day

Which one of the following statements is correct?

  1. The energy required to rocket an orbiting satellite out of earth's gravitational influence is more than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.

  2. If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of potential energy.

  3. The first artificial satellite Sputnik I was launched in the year 1950.

  4. The orbital speed of the SYNCOMS (synchronous communications satellite) is $3.07\times10^{2}m s^{-1}$


Correct Option: D
Explanation:

The energy required to rocket an orbiting satellite out of earth's gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.
Hence, option (a) is incorrect statement.


If zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
Hence, option (b) is an incorrect statement.

The first artificial satellite was launched by Soviet scientists in the year 1957.
Hence, option (c) is an incorrect statement.

The orbital speed of the SYNCOMS is $v=\dfrac{2\pi{r}}{T}=\dfrac{2\times3.14\times4.22\times10^{7}}{8.64\times10^{5}}$
$=3.07\times10^{2}m s^{-1}$
Hence, option (d) is a correct statement.

Out of the following, the only correct statement about satellites is

  1. A satellite cannot move in a stable orbit in a plane passing through the earth's centre

  2. Geostationary satellites are launched in the equatorial plane

  3. Satellites are rectangular in shape. 

  4. Geostationary satellites are launched in axial plane


Correct Option: B
Explanation:

Geostationary satellites are launched to the equitorial plane, so the it could have a time period of 24 hours exact and can have maximum reach, i.e on both heminpheres for information transfer.

State whether the given statement is True or False :

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


Correct Option: B
Explanation:

Geostationary satellite appears to be stationary above particular point on the Earth. We can have geostationary satellites only above equator. If it is not above the equator then it will change its position relative to particular position of the Earth which we do not want. 

Since New Delhi is not on the equatorial plane and geostationary satellite is launched on the equatorial plane, so the satellite will not remain directly over New Delhi. 
Therefor, the given statement is false.

The time period of an orbiting satellite is (assuming spherical earth)

  1. Directly proportional to the density of earth

  2. Directly proportional to the square of density of earth

  3. Inversely proportional to the square root of density of earth

  4. Inversely proportional to the density of earth


Correct Option: D

A satellite is orbiting around the earth in an orbit in equatorial plane, of radius 2$R _{e}$ , where $R _{e}$ is theradius of the earth. Find the area on the earth, this satellite covers for communication purpose in its complete revolution 

  1. $4\pi R _{e}^{2}$

  2. $2\pi\sqrt{3} R _{e}^{2}$

  3. $2\pi(2-\sqrt{3}) R _{e}^{2}$

  4. $2\pi(4+\sqrt{3}) R _{e}^{2}$


Correct Option: A

Identify the correct full form of IRNSS from the following

  1. Independent  regional navigation satellite system

  2. International regional national satellite system

  3. Indian regional national service system

  4. indian regional nevigation satellite system.


Correct Option: A

For a satellite to be geostationary, which of the following are not essential conditions?

  1. It must always be stationed above the equator

  2. It must be rotated from west to east

  3. It must be about $36000 \mathrm { km }$ above the Earth

  4. Its orbit must be circular, and not elliptical


Correct Option: A

To have an earth synchronous satellites it should be launched at the proper height moving from

  1. North to South in a polar plane

  2. East to West in an equatorial plane

  3. South to North in a polar plane

  4. West to East in an equatorial plane


Correct Option: D
Explanation:

The sense of synchronous satellite must be same as the sense of rotation of earth i.e., from west to east

The time period of geostationary satellite is :

  1. Zero

  2. $24 h$

  3. $12 h$

  4. $48 h$


Correct Option: B
Explanation:

Answer is B.

A geostationary orbit can only be achieved at an altitude very close to $35,786$ km $(22,236 mi)$, and directly above the Equator. This equates to an orbital velocity of $3.07 km/s (1.91 mi/s)$ or an orbital period of $1,436$ minutes, which equates to almost exactly one sidereal day or $23.934461223$ hours, which is approximately $24$ hours.

The period of rotation of the geostationary satellite is ______ hours.

  1. 12

  2. 24

  3. 6

  4. 48


Correct Option: B
Explanation:

A geostationary satellite is an earth-orbiting satellite, placed at an altitude of approximately 35,800 kilometers (22,300 miles) directly over the equator, that revolves in the same direction the earth rotates (west to east). 

At this altitude, one orbit takes $24 hours $, the same length of time as the earth requires to rotate once on its axis. 
The term geostationary comes from the fact that such a satellite appears nearly stationary in the sky as seen by a ground-based observer. 

Hence correct answer is option $B$ 

Two stones are projected with the same speed but making different angles with horizontal. Their ranges are equal. If the ranges of projection of one is $\pi/3$ and its maximum height is ${h} _{1}$ then the maximum height of the other will be

  1. $\dfrac{Rg}{\omega^{2}}$

  2. $\dfrac{R^{2}g}{\omega^{}}$

  3. $\dfrac{R^{2}g}{\omega^{2}}$

  4. $\dfrac{R^{2}\omega}{g^{}}$


Correct Option: C
Explanation:

Assuming this to be a geo-stationary satellite, the angular speed of the satellite is the same as the angular speed of rotation of the earth. 

Hence, we have:
$GMm/r^2=mw^2r$ and $GM/R^2=g$
Hence, $r^3=R^2g/w^2$

The orbital angular velocity vector of a geostationary satellite and the spin angular velocity vector of the earth are

  1. always in the same direction

  2. always in opposite direction

  3. always mutually perpendicular

  4. inclined at 23 1/2 to each other


Correct Option: A
Explanation:

For the satellite to be 'synchronous' with the Earth, its period of revolution should be equal to Earth's rotation. For this, the satellite has to revolve along the direction of Earth's rotation thus their angular velocity vector always point in the same direction.

A synchronous satellite should be at a proper height moving

  1. From West to East in equatorial plane

  2. From South to North in polar plane

  3. From East to West in equatorial plane

  4. From North to South in polar plane


Correct Option: A
Explanation:

For the satellite to be 'synchronous' with the Earth, its period of revolution should be equal to Earth's rotation. For this, the satellite has to revolve along the direction of Earth's rotation that is from West to East.

The orbital period of revolution of an artificial satellite revolving in a geostationary orbit is ...

  1. 24 Hrs

  2. 48Hrs

  3. 12Hrs

  4. 6 Hrs


Correct Option: A
Explanation:

For the satellite to be geostationary at equatorial plane, its time period should be equal to Earth's rotation time period i.e 24 hours.

Communication satellites are also referred to as:

  1. Polar satellites

  2. Geostationary satellites

  3. Launch vehicles

  4. Projectiles


Correct Option: A
Explanation:

Communication satellites are also referred as Polar satellites. Polar satellites are those satellites that passes above or nearly above both the poles of the body orbited on each revolution and hence it has very close inclination of 90 degree's to the body's equator.

What are the general uses of satellite?

  1. Navigation

  2. Live telecast of programmer

  3. Both A and B

  4. None of these


Correct Option: C
Explanation:

Satellite revolves around the earth with fixed a velocity in a circular orbit of certain radius. During its motion, satellite collects informative signals from earth and then sends those signal back to the other part of earth for interpretation. Satellites are generally used for navigation, live telecast of programmer, weather forecasting, communication purposes, GPS etc. 

A satellite placed in an orbit around Earth with certain speed will revolve in_______ as seen from Earth

  1. A helical path

  2. A circular path

  3. Straight line

  4. A parabolic path


Correct Option: B
Explanation:

The centrifugal force acting on the satellite, revolving around the earth with certain velocity, is balanced by the gravitational force acting on it due to earth. The satellite keeps revolving around the earth in circular path of certain radius as seen from the earth.

Moon can be classified as a

  1. Asteroid

  2. Comet

  3. Satellite

  4. Star


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.

  1. True

  2. False


Correct Option: A

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


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 

  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


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 

  1. Zero

  2. 2 km/s

  3. 10 km/s

  4. 8 km/s


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

  1. Zero

  2. $\infty$

  3. $T$

  4. $2T$


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 $v _{2}$?

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

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

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

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


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)

  1. $3R$

  2. $5R$

  3. $7R$

  4. $8R$


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?

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


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:

  1. $10 m/s $

  2. $10^3 m/s$

  3. $10^5 m/s$

  4. $10^7 m/s$


Correct Option: B

An earth satellite X is revolving around earth in an orbit whose radius is one- fourth the radius of orbit of a communication satallite. Time period of revolution of X is 

  1. $3$ hrs

  2. $6$ hrs

  3. $4$ days

  4. $72$ days


Correct Option: B

The time period of a geostationary satellite at a height $36000\ km$, is $24\ h$. A spy satellite orbits very close to earth surface ($R=6400\ km$). What will be its time period ?

  1. $4\ h$

  2. $1\ h$

  3. $2\ h$

  4. $1.5\ h$


Correct Option: C

Satellite is revolving around the earth. If it's radius of orbit is increased to $4$ times the radius of geostationary satellite, what will become its time period ?

  1. $8\ days$

  2. $4\ days$

  3. $2\ days$

  4. $16\ days$


Correct Option: A
Explanation:
Given,
$r _1=r$ 
$r _2=4r$
$T _1=1day$
The time period of the geostationary satellite is given by
$T=2\pi \sqrt{\dfrac{r^3}{Gm _E}}$. . . . . .(1)
where, $G=$ gravitational constant 
From equation (1),
$T\propto \sqrt{r^3}$
$\dfrac{T _2}{T _1}=\sqrt{\dfrac{r _2^3}{r _1^3}}$
$T _2=T _1\sqrt{\dfrac{4r\times 4r\times 4r}{r^3}}=8T _1 days$
$T _2=8\times 1=8days$
The correct option is A.

The distance between the centre of the earth and moon is 384000 km. If the mass of the earth is $6 \times 10^{24} kg$ and $G=6.66\times 10^{-11}$ units,the speed of the moon is nearly

  1. 1 km/s

  2. 4 km/s

  3. 8 km/s

  4. 11.2 km/s


Correct Option: A

A satellite orbiting close to the earth's surface will escape if 

  1. its speed is increased by 41.4%

  2. its KE is made 1.5 times the original value

  3. its original speed is increased $\sqrt{1.5}$ times

  4. its stops moving in the orbit


Correct Option: A

A particle is projected with a velocity $\sqrt{\dfrac{4gR}{3}}$ vertically upward from the surface of the earth. R is the radius of the earth & g being the acceleration due to gravity on the surface of the earth. The velocity of the particle when it is at half the maximum height reached by it is:

  1. $\sqrt{\dfrac{gR}{2}}$

  2. $\sqrt{\dfrac{gR}{3}}$

  3. $\sqrt{gR}$

  4. $\sqrt{\dfrac{2gR}{3}}$


Correct Option: B

The relay satellite transmits the television programme continuously from one part to another because its : 

  1. Period is greater than the period of rotation of the earth about its axis

  2. Period is less than the period of rotation of the earth about its axis

  3. Period is equal to the period of rotation of the earth about its axis

  4. Mass is less than the mass of earth


Correct Option: C
Explanation:
Relay satellite has to forces on the are unstartly suppose if the satellite is stable in space of a certain position, then as the earth rotate satellite can show once one part and later another part which is absorb movement. Therefore the satellite has to revolve around the earth with same angular speed. Therefore the time period of satellite revolution should be same as time period of the rotation of earth and rotation of the satellite should be about axis of earth's rotation. So its period is equal to period of rotation of the earth about its axis.

If mass of earth is $5.98\times 10^{24}$ kg and earth moon distance is $3.8\times 10^5$ km, the orbital period of moon, in days is

  1. 27 days

  2. 2.7 days

  3. 81 days

  4. 8.1 days


Correct Option: A
Explanation:

$M=5.98 \times 10^{24}kg$


$R=3.8\times 10^8km$


The orbital time period of moon,

$T=2\pi \sqrt{\dfrac{R^3}{GM}}$

$T=2\times 3.14\sqrt{\dfrac{(3.8)^3\times 10^{24}}{6.67\times 10^{-11}\times 5.48 \times 10^{24}}}$

$T=23.2427 \times 10^5 sec$

In days,

Time period, $T=\dfrac{23.2427\times 10^5}{60\times 60\times 24}day$

$T=27\ days$

The correct option is A.

If the angular velocity of a planet about its own axis is halved, the distance of geostationary satellite of this planet from the cent of the planet will become :

  1. $(2)^{1/3}$ times

  2. $(2)^{3/2}$ times

  3. $(2)^{2/3}$ times

  4. 4 times


Correct Option: B
Explanation:
Lets consider, 

$\omega _1=\omega $ angular velocity

$\omega _2=\dfrac{\omega }{2}$

$r=$ distance between geostationary satellites.

The Time period, $T\propto r^{3/2}$

$\dfrac{2\pi}{\omega }\propto r^{3/2}$

$\omega \propto r^{-3/2}$. . . . .(1)

$\dfrac{\omega _2}{\omega _1}\propto(\dfrac{r _2}{r _1})^{-3/2}$

$\dfrac{1}{2}\propto (\dfrac{r _2}{r _1})^{-3/2}$

$r _2=(2)^{3/2} r _1$

The correct option is B.

A satellite has to revolve round the earth in a circular orbit of radius 8 x $10^3$km. The velocity of projection of the satellite in this orbit will be -

  1. 16 km/sec

  2. 8 km/sec

  3. 3 km/sec

  4. 7.08 km/sec


Correct Option: D

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


Correct Option: B

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.


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 ?

  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


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?

  1. 6 $\sqrt { 2 } $ hours

  2. 6 $\sqrt { 2.5 } $ hours

  3. 6 $\sqrt { 3 } $ hours

  4. 12 hours


Correct Option: A

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.


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 ?

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

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

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

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


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?

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

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

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

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


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.-

  1. 0,0

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

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

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


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:-

  1. $ v _1 = v _2 $

  2. $ v _1 > v _2 $

  3. $ v _1 < v _2 $

  4. $ v _1/r _1 = v _2/r _2 $


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

  1. $\sqrt { 2 } R$

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

  3. $R$

  4. $2 R$


Correct Option: C
Explanation:

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

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.


Correct Option: A,B,C,D
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.

Orbital decay, a process of prolonged reduction in the attitude of a satellites orbit is caused by 
A) Atmospheric drag   B) Gravitational Pull      C) Tides

  1. A only

  2. C only

  3. A and C only

  4. A, B and C


Correct Option: D

If the length of the day is $T$ , the height of that TV satellite above the earth's surface which always appears stationary from earth, will be:


  1. $h = \left[ \dfrac { 4 x ^ { 2 } G m } { T ^ { 2 } } \right] ^ { - 6 }$

  2. $h = \left[ \dfrac { 4 x ^ { 2 } G M } { T ^ { 2 } } \right] ^ { - 1 / 2 } - R$

  3. $h = \left[ \dfrac { G M T ^ { 2 } } { 4 \pi ^ { 2 } } \right] ^ { 1/3 } - R$

  4. $h = \left[ \dfrac { G M T ^ { 2 } } { 4 \pi ^ { 2 } } \right] ^ { 2 } + R$


Correct Option: C

A planet of small mass m moves around the sun of mass M along an elliptical orbit such that its minimum and maximum distance from the sun are r and R respectively. Its period of revolution will be: 

  1. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{6GM}}} $

  2. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{3GM}}} $

  3. $\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{2GM}}} $

  4. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{GM}}} $


Correct Option: D

A geostationary satellite orbits around the earth in a circular orbit of radius $36000 km$. Then, the time period of a spy satellite orbiting a few $100 km$ above the earth's surface $\displaystyle { R } _{ earth }={ 6400 } \quad km$ will approximately be

  1. $\cfrac { 1 }{ 2 } { h }$

  2. ${ 1h }$

  3. ${ 2h }$

  4. ${ 4h }$


Correct Option: C
Explanation:

Given :   $R _{G}= 36000    km                              R _{S}= 6400+ 100= 6500     km$

Time period, $T= 2\pi\sqrt{\dfrac{r}{GM}}$
Thus   $(\dfrac{T _S}{T _G})^2= (\dfrac{R _S}{R _G})^3$ 
 $(\dfrac{T _S}{24 })^2= (\dfrac{6500}{36000})^3$ 
$\implies     T _S= 1.84    h        \approx  2  h$

A space shuttle is revolving around the earth in circular orbit. A certain point pilot fires forward pointing thruster to decrease shuttle's mechanical energy. Then orbital time period $T$ of shuttle

  1. Will increase

  2. Will decrease

  3. Will remain constant

  4. Will first decrease and then increase.


Correct Option: B

Geo-stationary satellite is one which

  1. Remains stationary at a fixed height from the eath's surface

  2. Revolves like other satellites but in the opposite direction of eath's rotation

  3. Revolves round the earth at a suitable height with same angular velocity and in the same direction as earth does about its own axis

  4. None of these


Correct Option: C
Explanation:

Geostationary satellites are also called synchronous satellite. They always remain about the same path on equator, i.e., it has a period of exactly one day $\displaystyle(86400)$
So orbit radius $\displaystyle \left [ T=2\pi \sqrt {\frac{r^3}{GM}} \right ]$ comes out to be $42400$ km, which is nearly equal to the circumference of earth. So height of Geostationary satellite from the earth surface is $42,400-6400=36,000$km.

The relay satellite transmits the $TV$ programmed continuously from one part of the world to another because its

  1. Period is greater than the period of rotation of the earth

  2. Period is less than the period of rotation of the eath about its axis

  3. Period has no relation with the period of the earth about its axis

  4. Period is equal to the period of rotation of the earth about its axis


Correct Option: D
Explanation:

The period of satellite is equal to period of rotation of earth about its own axis and it seems to be at one point about the equator and so is able to transmit the signals from one part to other

For a geostationary satellite orbiting around the earth identify the necessary condition

  1. it must lie in the equatorial plane of earth

  2. its height from the surface of earth must be $36000 km $

  3. it period of revolution must be $\displaystyle 2\pi \sqrt{\frac{R}{g}}$ where R is the radius of earth

  4. its period of revolution must be $24 hrs$


Correct Option: A,B,D
Explanation:

A geostationary satellite must satisfy the following requirements:

$1$. Its orbit must lie on an equatorial plane.
$2$. It must appear stationary when viewed from a point on earth which means its time period of revolution is $24 hrs$.
$3$. Its height above the surface of the earth must be $36000 km$.
So options A, B and D are correct.

A satellite is seen every $6$ hours over the equator. It is known that it rotates opposite to that of earth's direction. Then the angular velocity (in radian per hour) of satellite about the centre of earth will be :

  1. $\displaystyle\dfrac{\pi}{2}$

  2. $\displaystyle\dfrac{\pi}{3}$

  3. $\displaystyle\dfrac{\pi}{4}$

  4. $\displaystyle\dfrac{\pi}{8}$


Correct Option: C
Explanation:
Let $w _1$ and $w _2$ be the angular speed of the satellite and earth respectively.
Angular speed of rotation of earth  $w _2 = \dfrac{2\pi}{24}$ $rad/hr$ 
Since, both are revolving in opposite direction. Thus sum of angle rotated by them in time $t = 6 $ hr is $2\pi$.   
$\therefore$  $\omega _{1}t+\omega _{2}t=2\pi$
$(\omega _{1})6+\left (\displaystyle\dfrac{2\pi}{24} \right ){6}=2\pi$

or   $6\omega _{1}=\displaystyle\dfrac{3\pi}{2}$

 $\implies \omega _{1}=\left (\displaystyle\dfrac{\pi}{4} \right )\displaystyle\dfrac{rad}{hr}$

Geostationary satellite

  1. is situated at a great height above the surface of the Earth.

  2. moves in the equatorial plane.

  3. have time period of $24$ hours.

  4. have time period of $24$ hours and moves in the equatorial plane.


Correct Option: D
Explanation:
Geostationary satellite is used for communication purpose.
Thus, geostationary satellites are placed at an altitude of 36,000 kms and moves in the equatorial plane in the same direction as the Earth rotates. Hence, its time period is 24 hours.

The distance of a geostationary satellite from the centre of earth (radius R = 6400 Km) is nearly.

  1. 18 R

  2. 10R

  3. 7R

  4. 5R


Correct Option: C
Explanation:

A geostationary satellite is placed at an altitude of about 36000 km  from the earth surface.

So, its distance from centre of earth, $r= 36000+6400   km=  42400    km$ 
$d= \dfrac{42400}{6400}  R  \approx  7R$

A satellite launching station should be

  1. near the equatorial region.

  2. near the polar region.

  3. on the polar axis.

  4. all locations are equally good.


Correct Option: A
Explanation:
At equator, $g$ is minimum relative to all the other points on the surface of the Earth. When a spacecraft is launched into the orbit, it should end up spinning around the Earth quickly enough not to be pulled back in by the Earth's gravity. If a ship is launched from the equator, it goes up into space, and it is also moving around the Earth at the same speed it was moving before launching. This is because of inertia. This speed will help the spacecraft keep up a good enough speed to stay in the orbit.

A communication satellite of earth which takes $24 hr$. to complete one circular orbit eventually has to be replaced by another satellite of double mass. If the new satellites also has an orbital time period of $24 hrs$, then what is the ratio of the radius of the new orbit to the original orbit ?

  1. $1 : 1$

  2. $2 : 1$

  3. $\sqrt 2 : 1$

  4. $1 : 2$


Correct Option: A
Explanation:
Time period of revolution of satellite,  $T= \sqrt{\dfrac{r^3}{GM _e}}$    where $r$ and $M _e$  are the radius  of the orbit and mass of earth.

Thus time period doesn't depend on the mass of satellite.

$\implies $ for same time period, radius of the orbit will be same.
Hence  $r' : r= 1:1$  

The minimum number of satellites needed to be placed for world-wide communication between any two locations on earth's surface  is:

  1. 6

  2. 4

  3. 3

  4. 5


Correct Option: C
Explanation:

For a world-wide communication between any two locations of the earth, minimum 3 satellites must be placed such that each pair of  the satellites subtend angle of $120^o$ at the centre of earth. 

A body is dropped by a satellite in its geo -stationary orbit.

  1. it will burn on entering in to the atmosphere

  2. it will remain in the same place with respect to the earth

  3. it will reach the earth is $24$ hours

  4. it will perform uncertain motion


Correct Option: B
Explanation:

Geo-stationary satellite are those which have same angular velocity as that of earth, so they appear stationary with respect to earth. A body dropped off from satellite would also have same velocity as that of satellite, and so the angular velocity of the body would be same as that of earth also. So it would also seem stationary with respect to earth.

The earth satellite can move in an orbit the plane of which coincides with.

  1. The plane of any great circle round the earth

  2. The plane of any latitude circle of the earth

  3. Any plane not containing the centre of the earth

  4. The plane of tropic of cancer


Correct Option: A
Explanation:

Such is the case of a geo-stationary satellite, the angular velocity of satellite must be same as that of earth in any circular path around the earth.

If $R$ is the average radius of earth, $\omega $ is its angular velocity about its axis and $g$ is the gravitational acceleration on the surface of earth then the cube of the radius of orbit of a geostationary satellite will be equal to.

  1. $\dfrac {R^2g}{\omega }$

  2. $\dfrac {R^2\omega^2 }{g}$

  3. $\dfrac {Rg}{\omega^2 }$

  4. $\dfrac {R^2g}{\omega^2}$


Correct Option: D
Explanation:
As, $mr\omega^2=\dfrac{GMm}{r^2}$

$\implies r\omega^2=\dfrac{GM}{r^2}$

$\therefore r^3=\dfrac{GM}{\omega^2}=\dfrac{GM}{R^2}\dfrac{R^2}{\omega^2}=g\dfrac{R^2}{\omega^2}$

Motion of artificial earth satellites around the earth is powered by

  1. Liquid fuel

  2. Solar batteries

  3. Atomic energy

  4. None of the above


Correct Option: D
Explanation:

 A satellite orbits Earth when its speed is balanced by the pull of Earth gravity. Without this balance, the satellite would fly in a straight line off into space or fall back to Earth.

So, the the motion of satellites around the earth is powered by the gravitational pull of the earth, which provides the necessary centripetal force for the satellite to orbit the earth.

The moon waxes and wanes while going around the earth and hence it has circular motion

  1. True

  2. False


Correct Option: A
Explanation:

The moon waxes and wanes while going around the earth, because the terrestrial viewpoint of the moon changes relative to the moon's source of light. Hence it has a circular motion.

Height of geostationary satellite is

  1. $16000km$

  2. $22000km$

  3. $28000km$

  4. $36000km$


Correct Option: D
Explanation:

$Answer:-$ D

Approximately $35,786$ km.
A geostationary equatorial orbit (GEO) is a circular geosynchronous orbit in the plane of the Earth's equator with a radius of approximately $42,164$ km ($26,199$ mi) (measured from the center of the Earth). A satellite in such an orbit is at an altitude ofapproximately $35,786$ km ($22,236$ mi) above mean sea level.

Two planets, $X$ and $Y$, revolving in a orbit around star. Planet $X$ moves in an elliptical orbit whose semi-major axis has length $a$. Planet $Y$ moves in an elliptical orbit whose semi-major axis has a length of $9a$. If planet $X$ orbits with a period $T$, Find out the period of planet $Y$'s orbit?

  1. $729T$

  2. $27T$

  3. $3T$

  4. ${T}/{3}$

  5. ${T}/{27}$


Correct Option: B
Explanation:

Kepler's third law of planetary motion gives    $T^2\propto r^{3}$

    where $T=$ time period of revolution  ,  $r=$ length of semi major axis of elliptical orbit  

by this relation we get  ,  $\dfrac{T _{X}}{T _{Y}}=\dfrac{r _{X}^{3}}{r _{Y}^{3}}$

                                   $\dfrac{T _{X}^2}{T _{Y}^2}=\dfrac{a^{3}}{\left({9a}\right)^{3}}$

but given  $T _{X}=T$

                therefore   $\dfrac{T^2}{T _{Y}^2}=\dfrac{a^{3}}{729a^{3}}$

               or              $T _{Y}=27T$


An object is released from rest at a distance of ${2r} _{e}$ from the center of the Earth, where ${r} _{e}$ is the radius of the Earth. Find out the velocity of the object when it hits the Earth in terms of the gravitational constant $\left(G\right)$, the mass of the Earth $\left(M\right)$, and ${r} _{e}$.

  1. $\sqrt{{GM}/{{r} _{e}}}$

  2. ${GM}/{{r} _{e}}$

  3. $\sqrt{{GM}/{2{r} _{e}}}$

  4. ${GM}/{2{r} _{e}}$

  5. $2{GM}/{{r} _{e}}$


Correct Option: A
Explanation:

Given:

Initial velocity, $u=0$  ,  $h=2r _{e}$
Now, by using: $v^{2}=u^{2}+2gh$

                          $v^{2}=0+2g\times2r _{e}$

By putting  $g=GM/{r _{e}}^{2}$ in above equation 
We get, $v^{2}=2GM/{r _{e}}^{2}\times2r _{e}$
             $v^{2}=GM/2r _{e}$
             $v=\sqrt{GM/2r _{e}}$

A geosynchronous orbit is one in which the satellite makes one revolution around the Earth in 24 hrs.
How far above the surface of the Earth does this satellite have to orbit?
Assume the radius of the Earth is $6.37 \times {10}^{6} m$, and the mass of the Earth is $5.98 \times {10}^{24} kg$.

  1. $3.59 \times {10}^{7} m$

  2. $4.23 \times {10}^{7} m$

  3. $2.76 \times {10}^{6} m$

  4. $6.37 \times {10}^{6} m$

  5. $1.27 \times {10}^{7} m$


Correct Option: A
Explanation:

Given :   $T = 24$ hrs $ = 84400$  s               $M _e = 5.98 \times 10^{24}$ kg            $R _e  = 0.637 \times 10^7$  m

Time period of satellite moving in orbit of radius $r$            $T = 2\pi \sqrt{\dfrac{r^3}{GM _e}}$

$\therefore$    $86400 = 2\pi \sqrt{\dfrac{r^3}{(6.67 \times 10^{-11}) \times (5.98 \times 10^{24})}}$                         $\implies r  =4.23 \times 10^7$ m

Thus height of satellite above the surface        $h = r - R _e = (4.23 - 0.637) \times 10^7  \approx 3.59 \times 10^7$  m

Imagine a geostationary satellite of earth which is used as an inter continental telecast station. At what height will it have to be established?

  1. at $10^{3}m$

  2. at $6.4\times 10^{3}m$

  3. at $35.94\times 10^{6}m$

  4. at infinity


Correct Option: C
Explanation:

Time period for a geostationary satellite is 24 hour, so $t=24$

And $t=2\pi \sqrt{\dfrac{(R+h)^2}{GM}}$
So, $h=35.94\times 10^6m$

Which of the followings are correct uses of satellite placed in an orbit around Earth?

  1. For observing Earth from a distance

  2. For communication purposes

  3. To forecast weather

  4. All of the above


Correct Option: D
Explanation:
Satellite revolves around the earth with fixed a velocity in a circular orbit of certain radius. During its motion, satellite collects informative signals from earth and then sends those signal back to the other part of earth for interpretation. Satellites are used for weather forecasting, communication purposes and observing earth from a distance, GPS etc. 

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

  1. it must always be stationed above the equator

  2. it must be rotate from west to east

  3. it must be about $36,000 km$ above the earth surface

  4. it's orbit must be circular, and not elliptical


Correct Option: A,B,C,D
Explanation:

Geostationary satellites revolve in the equatorial plane.
It has to revolve along Earth's rotation direction that is from west to east.
For the time period to be close to $24$ hours, the height of the satellite has to be $36,000\ km$ from earth's surface.
And the orbit is circular.
All options.

Given that the universal gravitational constant, $G = 6.7 10^{-11} Nm^{2} kg^{-2}$ and that the mass,
M of the earth is $6.0 10^{24} kg$, find the speed of a satellite that is fixed to permanently
focus on the city of Abuja for broadcast of the 2010 IJSO competition.

  1. $ 3.08 \times 10^{3} ms^{-1}$

  2. $24 ms^{-1}$

  3. $40 ms^{-1}$

  4. $3.66 10^{3} ms^{-1}$


Correct Option: A
Explanation:

Centripetal force $=$ gravitational force
,$ \Rightarrow   mv _{2}r=GmMr _{2} , from     where    v _{2}=GMr$
where v is the velocity. Period, $T=2\pi rv$ giving
$r=Tv2\pi$
Hence $v _{3}=2\pi GMT   With   T = 24 hr = 8.64 \times  104 s$
$v=32\pi\times 6.7 10^{-11} \times 6.0 \times 10248.64\times 104 = 3.08 10^{3} ms$

For geo stationary satelites,

  1. Time period depends on the mass of the satelite

  2. The orbit radius is independent of the mass of earth.

  3. The period is equal to that of the rotation of earth about its axis

  4. None of these


Correct Option: C
Explanation:
For geostationary satellites is a circular geosynchronous orbit above earth's equator follows direction of earths rotation ship period is equal to rotation of earth about its axis.

A geo-stationary satellite orbits around the earth in a circular orbit of radius $36000\ km$. Then, the time period of a spy satellite orbiting a few $100\ km$ above the earth's surface ($R _{earth}=6400\ km$) will approximately be -

  1. $1/2\ hr$

  2. $1\ hr$

  3. $2\ hr$

  4. $4\ hr$


Correct Option: C
Explanation:
Satellite orbits in a radius of $36000㎞\quad \quad { R } _{ 1 }=36000㎞$
${ R } _{ earth }=6400㎞,\quad { R } _{ 2 }=6400$
By Lepler's third law, we know that
${ T }^{ 2 }\propto { R }^{ 3 }$
Also we know that time period of geostationary satellite is $24h$
${ T } _{ 1 }=24h$
$\therefore { \left( \cfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 2 }={ \left( \cfrac { { R } _{ 1 } }{ { R } _{ 2 } }  \right)  }^{ 3 }$
$\Rightarrow \cfrac { { \left( 24 \right)  }^{ 2 } }{ { \left( { T } _{ 2 } \right)  }^{ 2 } } ={ \left( \cfrac { 36000 }{ 6400 }  \right)  }^{ 3 }$
${ T } _{ 2 }=24{ \left( \cfrac { 6400 }{ 36000 }  \right)  }^{ 3/2 }$
${ T } _{ 2 }=2h$

A geostationary orbit will appear to move in

  1. Equitorial plane

  2. in planes other than equitorial plane

  3. in planes whose angular momentum is not conserved

  4. in planes whose angular momentum is conserved


Correct Option: B
Explanation:

A geostationary orbit will appear to move in in planes other than equitorial plane. The time period of such satellites will be 24 hrs only , but they will be at rest only with respect to the equitorial plane and hence such orbits are called parking orbits

The option (b) is the correct option

The mean radius of the earth is R, and its angular speed on its axis is $\omega $. What will be the radius of orbit of a geostationary satellite? 

  1. ${\left( {\frac{{Rg}}{{{\omega ^2}}}} \right)^{\frac{1}{3}}}$

  2. ${\left( {\frac{{{R^2}g}}{{{\omega ^2}}}} \right)^{\frac{1}{3}}}$

  3. ${\left( {\frac{{{R^2}g}}{\omega }} \right)^{\frac{1}{3}}}$

  4. ${\left( {\frac{{{R^2}{\omega ^2}}}{g}} \right)^{\frac{1}{3}}}$


Correct Option: B
Explanation:

$\dfrac{r^3}{T^2} = \dfrac{Gm}{R^2}. \dfrac{R^2}{4 \pi^2}$


$r^3 = T^2 . (\dfrac{Gm}{R^2}) \dfrac{R^2}{4 \pi^2}$

Since
$g = \dfrac{Gm}{R^2}$

$r^3 = (\dfrac{T^2}{2 \pi} )gR^2$

$r = (\dfrac{R^2g}{w^2})^{1/3}$

Hence (B) is correct answer

An instrument package is released from an orbiting earth satellite by simply detaching it from the outer. The package will :

  1. Go away from the earth and get lost in outer space

  2. Fall through a certain distance and then move in an orbit around the earth

  3. Fall towards the surface of earth

  4. Continue moving along with the satellite in the same orbit and with the same velocity


Correct Option: D
Explanation:

When the instrument package is released by simply detaching it from the satellite, then the velocity and acceleration of the package will be same as that of the satellite. So, it continues to moving along with the satellite in the same orbit and with the same velocity. 

A geostationary satellite is orbiting the earth at a height of $6R$ above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of $2.5R$ from the surface of the earth is $\underline{\hspace{0.5in}}$ hours.

  1. $6.45 h$

  2. $5.39 h$

  3. $6.23 h$

  4. $5.48 h$


Correct Option: A
Explanation:

According to Kepler's law,
$\dfrac {T _1^2}{T _2^2}=\dfrac {R _1^3}{R _2^3}$

$\Rightarrow \dfrac {24\times 24}{T _2^2}=\dfrac {6\times 6\times 6\times R^3}{2.5\times 2.5\times 2.5\times R^3}$

$\Rightarrow T _2^2=\dfrac {24\times 24\times 2.5\times 2.5\times 2.5}{6\times 6\times 6}=6.45 h$

Statement 1: Geostationary satellites may be setup in equatorial plane in orbits of any radius more than earth's radius.
Statement 2: Geostationary satellites have period of revolution of 24 hrs.

  1. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

  2. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

  3. Statement-1 is True, Statement-2 is False

  4. Statement-1 is False, Statement-2 is True


Correct Option: D
Explanation:

period of revolution must be 24 hours so radius will be find according to Kepler's law.

A geostationary satellite has an orbital speed of

  1. $2h$

  2. $6h$

  3. $12h$

  4. $24h$


Correct Option: A

A satellite revolves from east to west in a circular equatorial orbit of radius $R=1.00\times10^4:km$ around the Earth. Find the velocity ($v'$) of the satellite in the reference frame fixed to the Earth.

  1. $\displaystyle v^\prime = 49.0:km/s$

  2. $\displaystyle v^\prime = 7.0:km/s$

  3. $\displaystyle v^\prime = 21.0:km/s$

  4. $\displaystyle v^\prime = 14.0:km/s$


Correct Option: B

The orbital velocity of an artificial satellite in a circular orbit very close to Earth is $v$. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of $6R$ from Earth's surface will be

  1. $\displaystyle \cfrac {v}{\sqrt 7}$

  2. $\displaystyle \cfrac {v}{\sqrt 6}$

  3. $\displaystyle v$

  4. $\displaystyle \sqrt {6}v$


Correct Option: A
Explanation:

$\displaystyle v _1 \propto \dfrac {1}{\sqrt R}, v _2 \propto \dfrac {1}{\sqrt {7R}}$

$\displaystyle \dfrac {v _2}{v _1}=\dfrac {1}{\sqrt {7}} \Rightarrow v _2=\dfrac {v _1}{\sqrt {7}}=\dfrac {v}{\sqrt {7}}$

If the length of the day is $T$, the height of that TV satellite above the earth's surface which always appears stationary from earth, will be.

  1. $h=\left [ \dfrac {4 \pi^2GM}{T^2} \right ]^{1/3}$

  2. $h=\left [ \dfrac {4 \pi^2GM}{T^2} \right ]^{1/2}$

  3. $h=\left [ \dfrac {T^2GM}{4 \pi^2} \right ]^{1/3}$

  4. $h=\left [ \dfrac {4 \pi^2GM}{4 \pi^2} \right ]^{1/2}$


Correct Option: C
Explanation:

Since the satellite is stationary, it has same angular velocity as that of earth.

Thus $\omega=\dfrac{2\pi}{T}=\dfrac{v}{r}$

$\implies v=\dfrac{2\pi r}{T}$
The centripetal acceleration arises from the gravitational force earth exerts on satellite.
$\implies \dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$

Eliminating $v$ from above equations gives
$r=(\dfrac{GMT^2}{4\pi ^2})^{1/3}$

A geostationary satellite is revolving at a height $6R$ above the earth's surface, where $R$ is the radius of earth. The period of revolution of satellite orbiting at a height $2.5R$ above the earth's surface will be.

  1. $\text{24 hour}$

  2. $\text{12 hour}$

  3. $\text{6 hour}$

  4. $6 \sqrt 2\ \text{hour}$


Correct Option: D
Explanation:
$\text{The time period of satellite orbiting at a distance from the centre of the earth is given by,}$
$T^2=\dfrac{4\pi^2r^3}{GM^2}$
$\text{where M is the mass of the earth.}$
$\text{Therefore,the ratio of the time periods of two satellites at distance}\ r _1\ \text{and}  r _2$  
$\text{respectively from the centre of the earth is given by,}$ 
$\dfrac{T _1}{T _2}=(\dfrac{r _1}{r _2})^{3/2}$

$\text{or}\ T _2=T _1\left(\dfrac{r _2}{r _1}\right)^{3/2}$
$\text{For the geostationary satellite}\ T _1=1\ \text{day}=24  \text{hours   and} \ r _1 = 6R+R=7R$
$\text{For the other satellite}\quad r _2=2.5R+R=3.5R$
Therefore $T _2=24\times(\dfrac{3.5R}{7R})^{3/2}=24\times(\dfrac{1}{2})^{3/2}=6\sqrt{2}  $ hours
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