Physics

Gravitation and Orbital Mechanics

500 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 geostationary satellite is at rest, relative to earth only for points in the 

  1. equitorial plane

  2. plane passing through the poles of the earth

  3. plane along magnetic north and south of the earth

  4. planes that are isoclinic
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A geostationary satellite is at rest, relative to earth only for points in the equitorial plane.

The correct option is (a)

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

Two satellites of masses $m _{1}$ and $m _{2} (m _{1} > m _{2})$ are revolving round the earth in circular orbits of radii $r _{1}$ and $r _{2}(r _{1} > r _{2})$ respectively. Which of the following statements is true regarding their speeds $v _{1}$ and $v _{2}$?

  1. $v _{1} = v _{2}$
  2. $v _{1} < v _{2}$
  3. $v _{1} > v _{2}$
  4. $(v _{1}/r _{1}) = (v _{2}/r _{2})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Velocity of satellite $=\sqrt { \dfrac { GM }{ R+h }  } $

Now, 
        ${ V } _{ 1 }=\sqrt { \dfrac { GM }{ { r } _{ 1 } }  } $
        ${ V } _{ 2 }=\sqrt { \dfrac { GM }{ { r } _{ 2 } }  } $
As ${ r } _{ 1 }>{ r } _{ 2 }$
So  ${ V } _{ 1 }<{ V } _{ 2 }$

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

An artificial satellite is moving around earth in a circular orbit with speed equal to one fourth the escape speed of a body from the surface of earth. The height of satellite above earth is : ($R$ is radius of earth)

  1. $3R$
  2. $5R$
  3. $7R$
  4. $8R$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let's consider,

$r=$ height  of the satellite above the earth

$R=$ radius of the earth

$G=$ gravitational constant

$g=$ acceleration due to gravity

The escape velocity of a body from the surface of earth,

$v _e=\sqrt{\dfrac{2GM}{R}}$

$\dfrac{1}{4}.v _e=\dfrac{1}{4}\sqrt{\dfrac{2GM}{R}}$. . . . . .(1)

The orbital velocity of the satellite is given by

$v _o=\sqrt{\dfrac{GM}{r}}$. . . . . .(2)

Equation equation (1) and (2), we get

$\sqrt{\dfrac{GM}{r}}=\dfrac{1}{4}\sqrt{\dfrac{2GM}{R}}$

$\dfrac{GM}{r}=\dfrac{1}{16}.\dfrac{2GM}{R}$

$r=8R$

The correct option is D.
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

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:

  1. $10 m/s $
  2. $10^3 m/s$
  3. $10^5 m/s$
  4. $10^7 m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Speed v = 2*pi*r / T. r = 3.84 * 10^8 m, T = 28.3 days = 28.3 * 24 * 3600 seconds approx 2.44 * 10^6 s. v = 2 * 3.14 * 3.84 * 10^8 / 2.44 * 10^6 approx 1000 m/s.

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

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. $16\ days$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

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

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

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

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

Orbital velocity v = sqrt(GM/r). G = 6.67 * 10^-11, M = 6 * 10^24, r = 8 * 10^6 m. v = sqrt(6.67 * 10^-11 * 6 * 10^24 / 8 * 10^6) = sqrt(40 * 10^13 / 8 * 10^6) = sqrt(5 * 10^7) approx 7071 m/s = 7.07 km/s.

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

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

Escape velocity v_e = sqrt(2GM/R) is a property of the planet and the starting point, not the speed at which it is fired (though it must be fired at least at that speed to escape). The other statements are physically incorrect.

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

The total energy of a satellite is-

  1. Always positive

  2. Always negative

  3. Always zero

  4. +ve or -ve depending upon radius of orbit.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For every bounded system the total energy is always negative because if $k=x$ then $u$ will be $-2r$ so that $E=-x.$

So, the total energy of a satellite is negative.
Hence, the answer is negative.

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

Two identical satellites are at distance R and 7R from the surface of the earth of radius R. Which is the wrong statement from the following ?

  1. The ration of their total energies will be 4 but the ration of their potential and kinetic energies will be 2

  2. The ration of their potential energies will be 4

  3. The ration of their kinetic energies will be 4

  4. The ration of their total energies will be 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total energy E = -GMm / (2r). For r1 = 2R and r2 = 8R (distances from center), the ratio of energies is 8/2 = 4. The potential energy U = -GMm/r, ratio is 8/2 = 4. Kinetic energy K = GMm/2r, ratio is 8/2 = 4. Option A is confusingly worded.