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 satellite revolves from east to west in a circular equatorial orbit of radius $R=1.00\times10^4:km$ around the Earth. Find the velocity ($v'$) of the satellite in the reference frame fixed to the Earth.

  1. $\displaystyle v^\prime = 49.0\:km/s$
  2. $\displaystyle v^\prime = 7.0\:km/s$
  3. $\displaystyle v^\prime = 21.0\:km/s$
  4. $\displaystyle v^\prime = 14.0\:km/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The velocity of a satellite in a circular orbit is v = sqrt(GM/R). Given R = 10,000 km, the calculation yields approximately 6.3 km/s. Option B is the closest approximation provided.

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

The orbital velocity of an artificial satellite in a circular orbit very close to Earth is $v$. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of $6R$ from Earth's surface will be

  1. $\displaystyle \cfrac {v}{\sqrt 7}$
  2. $\displaystyle \cfrac {v}{\sqrt 6}$
  3. $\displaystyle v$
  4. $\displaystyle \sqrt {6}v$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle v _1 \propto \dfrac {1}{\sqrt R}, v _2 \propto \dfrac {1}{\sqrt {7R}}$

$\displaystyle \dfrac {v _2}{v _1}=\dfrac {1}{\sqrt {7}} \Rightarrow v _2=\dfrac {v _1}{\sqrt {7}}=\dfrac {v}{\sqrt {7}}$

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. $h=\left [ \dfrac {4 \pi^2GM}{T^2} \right ]^{1/3}$
  2. $h=\left [ \dfrac {4 \pi^2GM}{T^2} \right ]^{1/2}$
  3. $h=\left [ \dfrac {T^2GM}{4 \pi^2} \right ]^{1/3}$
  4. $h=\left [ \dfrac {4 \pi^2GM}{4 \pi^2} \right ]^{1/2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the satellite is stationary, it has same angular velocity as that of earth.

Thus $\omega=\dfrac{2\pi}{T}=\dfrac{v}{r}$

$\implies v=\dfrac{2\pi r}{T}$
The centripetal acceleration arises from the gravitational force earth exerts on satellite.
$\implies \dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$

Eliminating $v$ from above equations gives
$r=(\dfrac{GMT^2}{4\pi ^2})^{1/3}$

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

A geostationary satellite is revolving at a height $6R$ above the earth's surface, where $R$ is the radius of earth. The period of revolution of satellite orbiting at a height $2.5R$ above the earth's surface will be.

  1. $\text{24 hour}$
  2. $\text{12 hour}$
  3. $\text{6 hour}$
  4. $6 \sqrt 2\ \text{hour}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\text{The time period of satellite orbiting at a distance from the centre of the earth is given by,}$
$T^2=\dfrac{4\pi^2r^3}{GM^2}$
$\text{where M is the mass of the earth.}$
$\text{Therefore,the ratio of the time periods of two satellites at distance}\ r _1\ \text{and}  r _2$  
$\text{respectively from the centre of the earth is given by,}$ 
$\dfrac{T _1}{T _2}=(\dfrac{r _1}{r _2})^{3/2}$

$\text{or}\ T _2=T _1\left(\dfrac{r _2}{r _1}\right)^{3/2}$
$\text{For the geostationary satellite}\ T _1=1\ \text{day}=24  \text{hours   and} \ r _1 = 6R+R=7R$
$\text{For the other satellite}\quad r _2=2.5R+R=3.5R$
Therefore $T _2=24\times(\dfrac{3.5R}{7R})^{3/2}=24\times(\dfrac{1}{2})^{3/2}=6\sqrt{2}  $ hours
Multiple choice

How often does the Space Station circle the Earth?

  1. 15 minutes

  2. 60 minutes

  3. 1 hour 30 minutes

  4. 3 hours 20 minutes
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The International Space Station orbits the Earth at an altitude of about 400 km, completing one revolution in approximately 90 minutes, or 1 hour and 30 minutes.

Multiple choice

What is it called when the Earth spins around on its axis?

  1. orbiting

  2. revolution

  3. circling

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

Rotation is the spinning of an object around its own internal axis, which takes about 24 hours for Earth. Revolution refers to the movement of one object around another, such as Earth orbiting the Sun.

Multiple choice

One astronomical unit (AU) is equal to:

  1. The circumference of Earth's orbit around the sun

  2. The mean distance of Earth from the sun

  3. The diameter of the sun

  4. The time required for Earth to complete one revolution
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

One astronomical unit (AU) is defined as the average distance between the Earth and the Sun, approximately 150 million kilometers.

Multiple choice

A planet is orbiting the sun. Which of the following could cause the planet to leave orbit?

  1. An increase in orbital velocity due to a collision

  2. Molecules on the surface reaching escape velocity

  3. An increase in mass due to planetary accretion

  4. The planet is losing interest in the sun
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An orbiting planet maintains its path due to the balance between gravitational pull and its orbital velocity. If a collision significantly increases the planet's orbital velocity, it can exceed the escape velocity required to maintain a closed orbit, causing it to leave the system.

Multiple choice physics measurements and units measuring mass measurement of mass measuring instruments

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

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

A satellite in orbit is in a state of continuous free fall toward the Earth. Because the body inside the satellite is also in free fall, it experiences weightlessness regardless of the orbital radius. Therefore, the reading of the spring balance will be zero in both cases, making W1 equal to W2.