Tag: artificial satellite

Questions Related to artificial satellite

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? 

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

Show answer
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 :

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  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

Show answer
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.

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  1. $6.45 h$

  2. $5.39 h$

  3. $6.23 h$

  4. $5.48 h$

Show answer
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.

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  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

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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

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  1. $2h$

  2. $6h$

  3. $12h$

  4. $24h$

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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.

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

Show answer
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

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

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

  3. $\displaystyle v$

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

Show answer
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.

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

Show answer
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.

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  1. $\text{24 hour}$

  2. $\text{12 hour}$

  3. $\text{6 hour}$

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

Show answer
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