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 physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

The radius and mean density of the earth are $R$ & $p$ respectively . the critical orbital speed of a satellite for a low altitude orbit is 

  1. $ 2 \sqrt {\dfrac { \pi G p}{3R}} $
  2. $ 2R \sqrt {\dfrac { \pi Gp}{3}} $
  3. $ 2R \sqrt {\dfrac {2 \pi G p}{3}} $
  4. $ 2R \sqrt {\dfrac { \pi Gp}{2}} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Orbital speed v = sqrt(GM/R). Mass M = density * volume = p * (4/3) * pi * R^3. Substituting M: v = sqrt(G * p * 4/3 * pi * R^3 / R) = sqrt(4/3 * pi * G * p * R^2) = 2R * sqrt(pi * G * p / 3).

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

A body is suspended from a spring balance kept in a satellite The reading of the balance is $\displaystyle W _{1}$ when the satellite goes in an orbit of radius $R$ and is $\displaystyle W _{2}$ when it goes in an orbit of radius $2R$ Then

  1. $\displaystyle W _{1}=W _{2}$
  2. $\displaystyle W _{1}< W _{2}$
  3. $\displaystyle W _{1}>W _{2}$
  4. $\displaystyle W _{1}\neq W _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The reading on the spring balance is independent of the radius and thus both weight will be the same. Also, there is no gravitational force acting on a satellite.
 
Hence, ${ W } _{ 1 }={ W } _{ 2 }$.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

Consider a satellite going round the earth in a circular orbit. Which of the following statements is wrong?

  1. It is a freely falling body

  2. It is a moving with constant speed.

  3. It is acted upon by a force directed away from the centre of the earth which counter- balances the gravitational pull.

  4. Its angular momentum remains constant.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Satellite going around the earth in circular orbit is in state of free fall, and its speed is constant. speed depends upon the radius of orbit of satellite.

So Its angular velocity($\omega=v\times r$) is also constant and thus angular momentum $m\omega$ is also constant.
Apart from gravitational pull of the earth, there is no other force on the satellite. So option C is incorrect.

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

An Earth's satellite is moving in a circular orbit with a uniform speed $v$. If the gravitational force of the Earth suddenly disappears, the satellite will

  1. vanish into outer space.
  2. continue to move with velocity $v$ in original orbit.

  3. fall down with increasing velocity.
  4. fly off tangentially from the orbit with velocity $v$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The satellite revolves around the Earth in a circular orbit when centrifugal force acting outward on the satellite is balanced by the gravitational force of the Earth. Thus, when gravitational force suddenly disappears, then only centrifugal force will be acting and velocity is tangential to the orbit and hence, the satellite will fly off tangentially. It is similar to an object tied with a string, fly off when string suddenly breaks.

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

The orbit of the earth is an ellipse is an and not a circle. The distance between earth sun thus varies. On January 3, earth is closet to the sun (Perihelion). Similarly, earth is said to be at Aphelion, when it is farthest from the sun on ________.

  1. March, 23

  2. July, 4

  3. December, 23

  4. April, 21
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Earth is about 147.1 million kilometres (91.4 million miles) from the Sun at perihelion around January 3, in contrast to about 152.1 million kilometres (94.5 million miles) at aphelion around July 4, a difference of about 5.0 million kilometres (3.1 million miles).

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

A satellite of the earth is revolving in circular orbit with a uniform velocity V. If the gravitational force suddenly disappears, the satellite will

  1. continue to move with the same velocity in the same orbit.

  2. move tangentially to the original orbit with velocity V.

  3. fall down with increasing velocity.

  4. come to a stop somewhere in its original orbit.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The satellite is revolving around earth because the centripetal force is balanced by earth's gravitational pull.If the gravitational pull disappears, the satellite free of centripetal force. So, it will travel with its instantaneous velocity i.e. in the direction tangential to the circular path.

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

If an orbiting satellite comes to a standstill suddenly,

  1. the satellite will move along the tangent.

  2. the satellite will move radically towards centre of the orbit.

  3. the satellite will go to outer space and will be lost.

  4. the satellite will continue to move in the same orbit.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the satellite is brought to standstill $v=0$ which gives the net force on the satellite=
$F=\dfrac{GMm}{{a}^{2}} $ in the direction of the centre of the earth which earlier was providing centripetal acceleration. So the satellite will fall on the earth

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

The moon revolves round the earth  $13$  times in one year. If the ratio of sun-earth distance to earth-moon distance is  $392,$  then the ratio of masses of sun and earth will be

  1. $365$
  2. $356 \times 10 ^ { - 12 }$
  3. $3.56 \times 10 ^ { 5 }$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Kepler's Third Law and the relationship between orbital periods and distances, the ratio of the masses can be derived from the given orbital data.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

The mean radius of the earth is $R$, its angular speed about its own axis $\omega$ and the acceleration due to gravity at the earth surface is $g$. The cube of radius of orbit of 'geostationary satellite' will be

  1. $(R^{2}g/ \omega)$
  2. $(R^{2}\omega/ g)$
  3. $(Rg/ \omega^{2})$
  4. $(R^{2}g/ \omega^{2})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$mr\omega^{2} = \dfrac {GMm}{r^{2}}$
$\Rightarrow r\omega^{2} = \dfrac {GM}{r^{2}}$
$\Rightarrow r^{3} = \dfrac {GM}{\omega^{2}} = \dfrac {GM}{R^{2}} . \dfrac {R^{2}}{\omega^{2}}$
$\Rightarrow r^{3} = g \dfrac {R^{2}}{\omega^{2}}$
Hence (D) is correct.

Multiple choice physics motion of system of particles and rigid bodies centre of gravity turning effects of forces forces - vectors and moments

Given that there is a relationship between the orbital radius of a planet and its period of revolution and that the periods of revolution of Mercury, Earth, Jupiter and Neptune and nearly 0.24, 1, 11.8 and 165 years. It follows that the period of revolution of
1. Venus is less than 0.24 years
2. Mars is less than 12 years
3. Uranus is more than 165 years
4. Uranus is less than 165 years but more than 12 years.

  1. 1 and 3

  2. 4 only

  3. 3 o nly

  4. 2 and 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis (distance from the Sun). Therefore, planets further from the Sun have longer orbital periods. Mars (between Earth and Jupiter) must have a period between 1 and 11.8 years. Uranus (further than Neptune) would have a period longer than 165 years. The logic in option 2 and 4 fits the observed orbital distances.

Multiple choice social studies importance of transport system importance of transportation system introduction to transport and communication transportation services language, writing and great books

How many geosynchronous communication satellites are required to cover all inhabited regions of the Earth?

  1. 4

  2. 3

  3. 5

  4. 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • A single geostationary satellite is on a line of sight with about 40 percent of the earth's surface. 
  • Three such satellites, each separated by 120 degrees of longitude, can provide coverage of the entire plane the north and south geographic poles.
Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

Assertion (A): Even though a planet revolve around the sun in an elliptical orbit, the angular momentum of planet is constant.
Reason (R) : Any force other than mutual gravitational force is absent between the planet and the sun.

  1. Both A and R are true and R is correct explanation of A

  2. Both A and R are true and R is not correct explanation of A

  3. A is true and R is false

  4. A is false and R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Angular momentum of planet revolving around the sun is constant because the line of force of the gravitational force passes through the  sun . Therefore torque about the sun is zero.
$\therefore$  Angular momentum will remain constant.  And there is no other force acting on the planet, which results into zero tangential force.