Physics

Gravitation and Orbital Mechanics

522 Questions

Gravitation and orbital mechanics focus on planetary motion, elliptical orbits, and satellite deployment. Questions examine astrodynamics fundamentals, including geostationary orbits and perturbation theory. These topics are highly relevant for civil services and specialized technical examinations.

Planetary orbitsSatellite dynamicsGeostationary orbitsPerturbation theoryOrbital eccentricity

Gravitation and Orbital Mechanics Questions

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

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. $4\ hr$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $24h$
${ T } _{ 1 }=24h$
$\therefore { \left( \cfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 2 }={ \left( \cfrac { { R } _{ 1 } }{ { R } _{ 2 } }  \right)  }^{ 3 }$
$\Rightarrow \cfrac { { \left( 24 \right)  }^{ 2 } }{ { \left( { T } _{ 2 } \right)  }^{ 2 } } ={ \left( \cfrac { 36000 }{ 6400 }  \right)  }^{ 3 }$
${ T } _{ 2 }=24{ \left( \cfrac { 6400 }{ 36000 }  \right)  }^{ 3/2 }$
${ T } _{ 2 }=2h$
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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. ${\left( {\frac{{{R^2}{\omega ^2}}}{g}} \right)^{\frac{1}{3}}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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. 

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, 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. $5.48 h$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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

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

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

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. $\displaystyle v^\prime = 14.0\:km/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The velocity of a satellite in a circular orbit is v = sqrt(GM/R). Given R = 10,000 km, the calculation yields approximately 6.3 km/s. Option B is the closest approximation provided.

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

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. $\displaystyle \sqrt {6}v$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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. $h=\left [ \dfrac {4 \pi^2GM}{4 \pi^2} \right ]^{1/2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$r=(\dfrac{GMT^2}{4\pi ^2})^{1/3}$

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

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. $6 \sqrt 2\ \text{hour}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
Multiple choice

How often does the Space Station circle the Earth?

  1. 15 minutes

  2. 60 minutes

  3. 1 hour 30 minutes

  4. 3 hours 20 minutes
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The International Space Station orbits the Earth at an altitude of about 400 km, completing one revolution in approximately 90 minutes, or 1 hour and 30 minutes.

Multiple choice

What is it called when the Earth spins around on its axis?

  1. orbiting

  2. revolution

  3. circling

  4. rotation
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Rotation is the spinning of an object around its own internal axis, which takes about 24 hours for Earth. Revolution refers to the movement of one object around another, such as Earth orbiting the Sun.

Multiple choice

Why is the orbital period of Mercury 88 days while Earth's is 365 days?

  1. Earth has less mass

  2. Mercury has more mass

  3. Mercury is closer to the Sun

  4. Earth is closer to the Sun
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Kepler's third law of planetary motion, planets closer to the Sun have shorter orbital paths and experience stronger gravitational pull, resulting in faster orbital speeds and shorter orbital periods. Mercury is the closest planet to the Sun, so its year is much shorter than Earth's.