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

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

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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 $24 hrs$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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.

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

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. $\displaystyle\dfrac{\pi}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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   $6\omega _{1}=\displaystyle\dfrac{3\pi}{2}$

 $\implies \omega _{1}=\left (\displaystyle\dfrac{\pi}{4} \right )\displaystyle\dfrac{rad}{hr}$
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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.
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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

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

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

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

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. $1 : 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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  $r' : r= 1:1$  
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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

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

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

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

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. $\dfrac {R^2g}{\omega^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
As, $mr\omega^2=\dfrac{GMm}{r^2}$

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

$\therefore r^3=\dfrac{GM}{\omega^2}=\dfrac{GM}{R^2}\dfrac{R^2}{\omega^2}=g\dfrac{R^2}{\omega^2}$
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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. ${T}/{27}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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              $T _{Y}=27T$


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

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. $1.27 \times {10}^{7} m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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. 
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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.

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

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