Satellites and Orbital Mechanics - class-XI

LANDSAT 1 through $3$ operated in a near-polar orbit at an altitude of $920 km$ with an $18-day$ repeat coverage cycle. These satellites circled the earth every 103 min completing $14$ orbits a day

The energy required to rocket an orbiting satellite out of earth's gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.
Hence, option (a) is incorrect statement.

Geostationary satellites are launched to the equitorial plane, so the it could have a time period of 24 hours exact and can have maximum reach, i.e on both heminpheres for information transfer.

Geostationary satellite appears to be stationary above particular point on the Earth. We can have geostationary satellites only above equator. If it is not above the equator then it will change its position relative to particular position of the Earth which we do not want. 

The time period of a satellite orbiting close to a spherical body is proportional to the square root of the inverse of density. Specifically, T equals 2 times pi times the square root of (R cubed over G M), which can be rewritten using density as T being inversely proportional to the square root of the density of the earth. Therefore, option C is correct while the other proportionality relations are incorrect.

The sense of synchronous satellite must be same as the sense of rotation of earth i.e., from west to east

Answer is B.

A geostationary orbit can only be achieved at an altitude very close to $35,786$ km $(22,236 mi)$, and directly above the Equator. This equates to an orbital velocity of $3.07 km/s (1.91 mi/s)$ or an orbital period of $1,436$ minutes, which equates to almost exactly one sidereal day or $23.934461223$ hours, which is approximately $24$ hours.

A geostationary satellite is an earth-orbiting satellite, placed at an altitude of approximately 35,800 kilometers (22,300 miles) directly over the equator, that revolves in the same direction the earth rotates (west to east). 

For the satellite to be 'synchronous' with the Earth, its period of revolution should be equal to Earth's rotation. For this, the satellite has to revolve along the direction of Earth's rotation thus their angular velocity vector always point in the same direction.

For the satellite to be 'synchronous' with the Earth, its period of revolution should be equal to Earth's rotation. For this, the satellite has to revolve along the direction of Earth's rotation that is from West to East.

For the satellite to be geostationary at equatorial plane, its time period should be equal to Earth's rotation time period i.e 24 hours.

Communication satellites are also referred as Polar satellites. Polar satellites are those satellites that passes above or nearly above both the poles of the body orbited on each revolution and hence it has very close inclination of 90 degree's to the body's equator.

Satellite revolves around the earth with fixed a velocity in a circular orbit of certain radius. During its motion, satellite collects informative signals from earth and then sends those signal back to the other part of earth for interpretation. Satellites are generally used for navigation, live telecast of programmer, weather forecasting, communication purposes, GPS etc. 

The centrifugal force acting on the satellite, revolving around the earth with certain velocity, is balanced by the gravitational force acting on it due to earth. The satellite keeps revolving around the earth in circular path of certain radius as seen from the earth.

Moon revolves around earth with certain fixed velocity in a circular orbit of certain radius. Moon is a natural satellite of earth whose time period of revolution is $27.3$ days.

A geostationary satellite must remain directly above a fixed point on the earth's equator, which requires its orbital plane to be the equatorial plane. Since New Delhi is located at a northern latitude and not on the equator, no satellite can remain directly above it at all times.

Parking orbit for a geostationary satellite is in equatorial plane of earth, since it has to be at rest always with respect to equator

The correct option is option (b)

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

The correct option is (a)

A geostationary satellite is a satellite which always appears to be stationary from an earth's surface.

The correct option is option (a)

As we know, $T\alpha \dfrac { 1 }{ \sqrt { g }  } $

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

Using conservation of mechanical energy, the initial energy at distance 3R from the center equals the final energy at the surface of the moon (distance R). Setting initial kinetic energy to zero, we get minus G M m over 3R equals half m v squared minus G M m over R. Solving for v yields the square root of (4 G M over 3R), making option B correct.

Speed v = 2pir / 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.

Orbital speed v = sqrt(GM/r). G = 6.67 * 10^-11, M = 6 * 10^24, r = 3.84 * 10^8 m. v = sqrt(6.67 * 10^-11 * 6 * 10^24 / 3.84 * 10^8) = sqrt(40.02 * 10^13 / 3.84 * 10^8) = sqrt(10.4 * 10^5) approx 1020 m/s = 1 km/s.

Escape velocity v_e = sqrt(2) * v_o, where v_o is orbital velocity. v_e = 1.414 * v_o. To escape, the speed must increase by 0.414 * v_o, which is 41.4%.

$M=5.98 \times 10^{24}kg$

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.

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.

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

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.

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 / (2sqrt(2)) = 12 / sqrt(2) = 6sqrt(2) hours.

We have,

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

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

$\begin{array}{l} V=\sqrt { \dfrac { { GM } }{ R }  }  \ \frac { { -GMm } }{ R } +\dfrac { 1 }{ 2 } m\left( { \dfrac { { GM } }{ R }  } \right) =\dfrac { { -GMm } }{ { \left( { R+h } \right)  } } +0 \ \Rightarrow \dfrac { { -GMm } }{ R } +\dfrac { { GMm } }{ { 2R } } =\dfrac { { -GMm } }{ { \left( { R+h } \right)  } }  \ \Rightarrow \dfrac { { -GMm } }{ { 2R } } =\dfrac { { -GMm } }{ { \left( { R+h } \right)  } }  \ \Rightarrow 2R=R+h \ \Rightarrow r=R \ Hence, \ option, , C, , is, , correct, , answer. \end{array}$

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.

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

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

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 = 2pisqrt(a^3 / GM) = 2pisqrt(((r+R)/2)^3 / GM) = 2pisqrt((r+R)^3 / 8GM). None of the options match exactly, but D is the standard form for T^2 proportional to a^3.

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

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.

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.

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

A geostationary satellite must satisfy the following requirements:

A geostationary satellite is placed at an altitude of about 36000 km  from the earth surface.

For a world-wide communication between any two locations of the earth, minimum 3 satellites must be placed such that each pair of  the satellites subtend angle of $120^o$ at the centre of earth. 

Geo-stationary satellite are those which have same angular velocity as that of earth, so they appear stationary with respect to earth. A body dropped off from satellite would also have same velocity as that of satellite, and so the angular velocity of the body would be same as that of earth also. So it would also seem stationary with respect to earth.

Such is the case of a geo-stationary satellite, the angular velocity of satellite must be same as that of earth in any circular path around the earth.

 A satellite orbits Earth when its speed is balanced by the pull of Earth gravity. Without this balance, the satellite would fly in a straight line off into space or fall back to Earth.

The moon waxes and wanes while going around the earth, because the terrestrial viewpoint of the moon changes relative to the moon's source of light. Hence it has a circular motion.

$Answer:-$ D

Kepler's third law of planetary motion gives    $T^2\propto r^{3}$

    where $T=$ time period of revolution  ,  $r=$ length of semi major axis of elliptical orbit  

by this relation we get  ,  $\dfrac{T _{X}}{T _{Y}}=\dfrac{r _{X}^{3}}{r _{Y}^{3}}$

                                   $\dfrac{T _{X}^2}{T _{Y}^2}=\dfrac{a^{3}}{\left({9a}\right)^{3}}$

but given  $T _{X}=T$

                therefore   $\dfrac{T^2}{T _{Y}^2}=\dfrac{a^{3}}{729a^{3}}$

               or              $T _{Y}=27T$

Given:

Given :   $T = 24$ hrs $ = 84400$  s               $M _e = 5.98 \times 10^{24}$ kg            $R _e  = 0.637 \times 10^7$  m

Time period for a geostationary satellite is 24 hour, so $t=24$

Geostationary satellites revolve in the equatorial plane.
It has to revolve along Earth's rotation direction that is from west to east.
For the time period to be close to $24$ hours, the height of the satellite has to be $36,000\ km$ from earth's surface.
And the orbit is circular.
All options.

Centripetal force $=$ gravitational force
,$ \Rightarrow   mv _{2}r=GmMr _{2} , from     where    v _{2}=GMr$
where v is the velocity. Period, $T=2\pi rv$ giving
$r=Tv2\pi$
Hence $v _{3}=2\pi GMT   With   T = 24 hr = 8.64 \times  104 s$
$v=32\pi\times 6.7 10^{-11} \times 6.0 \times 10248.64\times 104 = 3.08 10^{3} ms$

A geostationary orbit will appear to move in in planes other than equitorial plane. The time period of such satellites will be 24 hrs only , but they will be at rest only with respect to the equitorial plane and hence such orbits are called parking orbits

The option (b) is the correct option

$\dfrac{r^3}{T^2} = \dfrac{Gm}{R^2}. \dfrac{R^2}{4 \pi^2}$

When the instrument package is released by simply detaching it from the satellite, then the velocity and acceleration of the package will be same as that of the satellite. So, it continues to moving along with the satellite in the same orbit and with the same velocity. 

According to Kepler's law,
$\dfrac {T _1^2}{T _2^2}=\dfrac {R _1^3}{R _2^3}$

period of revolution must be 24 hours so radius will be find according to Kepler's law.

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.

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

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

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