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 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.

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, $r= 36000+6400   km=  42400    km$ 
$d= \dfrac{42400}{6400}  R  \approx  7R$