Tag: artificial satellite

Questions Related to artificial satellite

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