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

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.

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 R being the radius of the earth. What will be the time period of Another satellite at a height 2.5 R from the surface of the earth?

  1. 6 $\sqrt { 2 } $ hours
  2. 6 $\sqrt { 2.5 } $ hours
  3. 6 $\sqrt { 3 } $ hours

  4. 12 hours
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Kepler's Third Law: T^2 is proportional to r^3. r1 = 7R, r2 = 3.5R. (T2/T1)^2 = (3.5R/7R)^3 = (1/2)^3 = 1/8. T2 = T1 / sqrt(8) = 24 / (2*sqrt(2)) = 12 / sqrt(2) = 6*sqrt(2) hours.

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

What is the nature of relation betweenthe kinetic energy $\left( \mathrm { E } _ { \mathrm { k } } \right)$ and their orbitalradius $( \mathrm { r } )$ of the satellites revolvingaround the Earth?

  1. $E _ { k } \propto 1$
  2. $E _ { k } \propto \frac { 1 } { r }$
  3. $E _ { k } \propto r ^ { 2 }$
  4. $E _ { k } \propto \frac { 1 } { r ^ { 2 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} \dfrac { { GMm } }{ { { r^{ 2 } } } } =\dfrac { { m{ v^{ 2 } } } }{ r }  \ \Rightarrow \dfrac { { m{ v^{ 2 } } } }{ 2 } =\dfrac { { GMm } }{ { 2r } }  \ \therefore K _E\propto \dfrac { 1 }{ r }  \end{array}$

$\therefore $ Option $B$ is correct .

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

Two artificial satellite of masses $ m _1 $ and $ m _2 $ are moving with speed $ v _1 $ and $ v _2 $ in orbits of radii$ r _1 $ and $ r _2 $ respectively. if $ r _ 1>r _2 $ then which of the following statements in true:-

  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
C Correct answer
Explanation

Orbital velocity v = sqrt(GM/r). As r increases, v decreases. Since r1 > r2, v1 < v2.

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 mu always be stationed above the equator.

  2. It must rotate from west to east.

  3. It must be about 36,000 km above the earth.

  4. Its orbit must be circular, and not elliptical.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Since the satellite rotates in a plane which passes through the centre of earth, for it to be stationary relative to the earth, its angular velocity must be same as that of earth (direction also). And hence it must rotate in the equitorial plane.


Since the angular velocity is same as that of earth, its direction must be west to east.

Balancing forces,

$\dfrac { GMm }{ { (R+h) }^{ 2 } } =m{ \omega  }^{ 2 }(R+h)$

$\omega =\sqrt { \dfrac { GM }{ R+h }  } =\dfrac { 2\pi  }{ 3600X24 } $

This gives $h=36000km$

For constant $\omega$, orbit must be circular.

Answer is ABCD.

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

Orbital decay, a process of prolonged reduction in the attitude of a satellites orbit is caused by 
A) Atmospheric drag   B) Gravitational Pull      C) Tides

  1. A only

  2. C only

  3. A and C only

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

Orbital decay is primarily caused by atmospheric drag. Gravitational perturbations and tidal forces can also affect orbits over long periods.

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. $h = \left[ \dfrac { 4 x ^ { 2 } G m } { T ^ { 2 } } \right] ^ { - 6 }$
  2. $h = \left[ \dfrac { 4 x ^ { 2 } G M } { T ^ { 2 } } \right] ^ { - 1 / 2 } - R$
  3. $h = \left[ \dfrac { G M T ^ { 2 } } { 4 \pi ^ { 2 } } \right] ^ { 1/3 } - R$
  4. $h = \left[ \dfrac { G M T ^ { 2 } } { 4 \pi ^ { 2 } } \right] ^ { 2 } + R$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a geostationary satellite, the orbital period T = 2*pi*sqrt(r^3/GM). Solving for r: r^3 = GMT^2 / (4*pi^2). r = (GMT^2 / 4*pi^2)^(1/3). The height h = r - R_e.

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

A planet of small mass m moves around the sun of mass M along an elliptical orbit such that its minimum and maximum distance from the sun are r and R respectively. Its period of revolution will be: 

  1. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{6GM}}} $
  2. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{3GM}}} $
  3. $\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{2GM}}} $
  4. $2\pi \sqrt {\dfrac{{{{\left( {r + R} \right)}^3}}}{{GM}}} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Kepler's Third Law states T^2 is proportional to a^3, where a is the semi-major axis. For an elliptical orbit, a = (r + R) / 2. Thus, T = 2*pi*sqrt(a^3 / GM) = 2*pi*sqrt(((r+R)/2)^3 / GM) = 2*pi*sqrt((r+R)^3 / 8GM). None of the options match exactly, but D is the standard form for T^2 proportional to a^3.

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

A geostationary 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 $\displaystyle { R } _{ earth }={ 6400 } \quad km$ will approximately be

  1. $\cfrac { 1 }{ 2 } { h }$
  2. ${ 1h }$
  3. ${ 2h }$
  4. ${ 4h }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given :   $R _{G}= 36000    km                              R _{S}= 6400+ 100= 6500     km$

Time period, $T= 2\pi\sqrt{\dfrac{r}{GM}}$
Thus   $(\dfrac{T _S}{T _G})^2= (\dfrac{R _S}{R _G})^3$ 
 $(\dfrac{T _S}{24 })^2= (\dfrac{6500}{36000})^3$ 
$\implies     T _S= 1.84    h        \approx  2  h$

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

A space shuttle is revolving around the earth in circular orbit. A certain point pilot fires forward pointing thruster to decrease shuttle's mechanical energy. Then orbital time period $T$ of shuttle

  1. Will increase

  2. Will decrease

  3. Will remain constant

  4. Will first decrease and then increase.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Decreasing mechanical energy in a circular orbit forces the satellite into a lower orbit. According to Kepler's Third Law, the orbital period T is proportional to r^(3/2), so a smaller radius results in a smaller orbital period.