Tag: evs

Questions Related to evs

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:

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

  1. $10 m/s $

  2. $10^3 m/s$

  3. $10^5 m/s$

  4. $10^7 m/s$

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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 

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

  1. $3$ hrs

  2. $6$ hrs

  3. $4$ days

  4. $72$ days

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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 ?

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

  1. $4\ h$

  2. $1\ h$

  3. $2\ h$

  4. $1.5\ h$

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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 ?

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

  1. $8\ days$

  2. $4\ days$

  3. $2\ days$

  4. $16\ days$

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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

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

  1. 1 km/s

  2. 4 km/s

  3. 8 km/s

  4. 11.2 km/s

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Correct Option: A

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

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

  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

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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:

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

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

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

  3. $\sqrt{gR}$

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

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Correct Option: B

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

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

  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

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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

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

  1. 27 days

  2. 2.7 days

  3. 81 days

  4. 8.1 days

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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 :

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

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

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

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

  4. 4 times

Show answer
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.