Earth satellite MCQ Quiz in मल्याळम - Objective Question with Answer for Earth satellite - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Let us consider the time period of the first satellite is T₁ and that of second satellite is T₂.

Given,

\(\frac{T_1}{T_2}=\frac 18\)

We know that angular velocity is given by- 

\(\omega =\frac{2 \pi}{T}\)

⇒ \(\omega \propto \frac 1T\)

Therefore,

\(\frac {\omega_1}{\omega_2} = \frac{T_2}{T_1}\)

\(\frac {\omega_1}{\omega_2} = \frac{T_2}{T_1}=\frac 81\)

Hence Option 1 is the correct choice.

CONCEPT:

• A satellite is an object that has been placed into the orbit. This is also known as an artificial satellite.
  • There is a natural satellite also. eg. Moon
• Satellites orbits around the planet because they have enough speed that is fast enough to defeat the downward pull of gravity.

EXPLANATION:

• In the case of a moving satellite, the gravitational pull provides the necessary centripetal force to the satellite to move the satellite on the circular orbit.
• If the gravitational pull disappears/becomes zero suddenly then there will be no centripetal force acting on the satellite. So the satellite will not move in the circular orbit.

• The satellite will move with the same speed and path is tangential to the circular orbit at that point where the gravitational pull disappears. So option 2 is correct.

The correct answer is Artificial satellite.

 Key Points

• Artificial satellites are human-built objects orbiting the Earth and other planets in the Solar System.
  • Artificial satellites are used to study the Earth, other planets, to help us communicate, and even to observe the distant Universe.
  • The first artificial satellite was the Soviet Sputnik 1 mission, launched in 1957.

 Important Points

• A satellite is a body that orbits around another body in space.
• There are two different types of satellites – natural and man-made.
• Examples of natural satellites are the Earth and Moon.
• The Earth rotates around the Sun and the Moon rotates around the Earth.
• The Moon is a natural satellite of the Earth.
• ​Space shuttle
  • Space shuttle, also called Space Transportation System, partially reusable rocket-launched vehicle designed to go into orbit around Earth, to transport people and cargo to and from orbiting spacecraft, and to glide to a runway landing on its return to Earth’s surface.
  • It was developed by the U.S. National Aeronautics and Space Administration (NASA).

Concept:

Second Pendulum: A pendulum which has a time period for 2 seconds is called second pendulum.

Time Period of Pendulum (T) is given by

\(T = 2\pi\sqrt{\frac{l}{g}}\)   ........................... (i)

where l = length of Pendulum, g = Acceleration Due to Gravity

and, Acceleration Due to Gravity (g) is formulated as

\(g = \frac{GM_p}{R_p^2}\) .................................. (ii)

where G = Gravitational Constant, M_p = Mass of Planet, R_p = Radius of Planet

Hence from equation (i) and (ii), 

\(T = 2\pi\sqrt{\frac{lR_p^2}{GM_p}}\)  ................................. (iii)

\(T \propto \frac{R_p}{\sqrt{M_p}}\) ......................................... (iv)

Calculation:

Given:

M_p = 2M_e, R_p = 3R_e

From the equation (iv)

\(\frac{T_p}{T_e} = (\frac{R_p}{R_e})(\sqrt{\frac{M_e}{M_p}})\)

\(\frac{T_p}{2} = (\frac{3R_e}{R_e})(\sqrt{\frac{M_e}{2M_e}})\)

\(\frac{T_p}{2} = \frac{3}{\sqrt{2}}\)

T_p = 4.242 seconds

The correct answer is 1) 35000 km. 
Concept: 

• A geostationary satellite is one that orbits the Earth at the same rate that the Earth rotates.
• This means it stays in the same position relative to the surface of the Earth, specifically over the equator.
• To achieve this, the satellite must be placed in a circular orbit directly above the equator at a very specific altitude where its orbital period matches the Earth's rotational period of about 24 hours.
• This special orbit is known as a geostationary orbit.

Explanation:

• The altitude required for a satellite to remain geostationary is precisely calculated to ensure that the gravitational forces and the orbital velocity balance in such a way that the satellite's orbital period is the same as the Earth’s rotation period.
• This is critical because if the satellite is too low or too high, it will orbit too quickly or too slowly, respectively, to maintain a stationary position relative to the surface.
• The altitude for a geostationary orbit is approximately 35,786 kilometers (about 22,236 miles) above mean sea level.
• At this altitude, the satellite's orbital speed aligns perfectly with the Earth's rotation, allowing it to hover over the same spot on the Earth's surface at all times.

Key Points

• Orbital period: The time it takes for a satellite to complete one orbit around the Earth. For a geostationary satellite, this is 24 hours  the same as the Earth's rotation period.
• Equatorial orbit: Geostationary satellites must orbit along the Earth's equator. Any deviation from this path would cause the satellite to appear to move north and south over the Earth’s surface.
• Specific altitude: The altitude is crucial for maintaining geostationary orbit. If it's incorrect, the satellite will not remain fixed over one spot.

CONCEPT:

Orbital Velocity of Satellite:

• Satellites are natural or artificial bodies describing an orbit around a planet under its gravitational attraction.
• The orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth.
• For the revolution of a satellite around the earth, the gravitational pull provides the required centripetal force.

• It is given as,

\(\Rightarrow v_{o} = \sqrt{\frac{GM}{r}}\)

Where vo = orbital speed, G = gravitational constant, M = mass of the planet, and r = radius of the orbit

• Orbital speed is independent of the mass of the orbiting satellite.

Kinetic Energy:

• The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,

​\(⇒ KE=\frac{1}{2}m× v^{2}\)

Where KE = kinetic energy, m = mass and v = speed

CALCULATION:

Given mA = mB = m and \(\frac{r_A}{r_B}=\frac{1}{2}\)

• The orbital velocity of satellites A is,

\(\Rightarrow v_{oA} = \sqrt{\frac{GM}{4r_A}}\)​

• The kinetic energy of satellite A is,

\(⇒ KE_A=\frac{1}{2}m_A× v_{oA}^{2}\)

\(⇒ KE_A=\frac{1}{2}m\times\frac{GM}{4r_A}\)     -----(1)

• The orbital velocity of satellites B is,

\(\Rightarrow v_{oB} = \sqrt{\frac{GM}{4r_B}}\)

• The kinetic energy of satellite B is,

\(⇒ KE_B=\frac{1}{2}m_B× v_{oB}^{2}\)

\(⇒ KE_B=\frac{1}{2}m\times\frac{GM}{4r_B}\)     -----(2)

By equation 1 and equation 2,

\(⇒ \frac{KE_A}{KE_B}=\frac{GMm}{8r_A}\times\frac{8r_B}{GMm}\)

\(⇒ \frac{KE_A}{KE_B}=\frac{2}{1}\)

• Hence, option 3 is correct.

The correct answer is option 2) i.e. 12100 km

CONCEPT:

• The time period of the satellite: It is the time taken by the satellite to complete one revolution around the Earth.
• Consider a satellite orbiting the earth at a height h from the surface of the earth of radius R.

The circumference of orbit of satellite = 2π(R+h)

• The orbital velocity of the satellite is given by,

\(\Rightarrow v_0 =\sqrt{\frac{GM}{R+h}}\)

• Time period of the satellite is, 

\(\Rightarrow T = \frac{circumference}{orbital \: velocity} = \frac{2\pi(R+h)}{\sqrt{\frac{GM}{R+h}}} = 2\pi\sqrt{\frac{(R+h)^3}{GM}}\)

CALCULATION:

Given that:

Radius of Earth, R = 6300 km

Ratio of time period = \(\frac{T_A}{T_B} = \frac{2}{3}\)

The height at which satellite B is orbiting, h_B = 2h_A

• Time period of the satellite is, 

\(\Rightarrow T = 2\pi\sqrt{\frac{(R+h)^3}{GM}}\)

⇒ T ∝ (R + h)^3/2

\(⇒ \frac{T_A}{T_B} = \frac{(R+h_A)^{3/2}}{(R+h_B)^{3/2} }\)

\(⇒ \frac{2}{3} = ( \frac{R+h_A}{R+2h_A })^{3/2} ⇒ (\frac{2}{3})^{2/3} =\frac{(R+h_A)}{(R+2h_A) } \)

⇒ 0.76(R + 2h_A) = R + h_A

⇒ 0.52hA = 0.24R

⇒ 0.52hA = 0.24 × 6300

⇒ hA = 2907 km

⇒ Distance of B from the centre of Earth = R + h_B = R + 2hA = 6300 + (2 × 2907) = 12114 km ≈ 12100 km

CONCEPT:

• Kepler’s First law: Every planet revolves around the Sun in an elliptical orbit and Sun is situated at one of its two foci. It is also termed as ‘the Law of Orbits’.
• Conservation of Angular Momentum: The velocity and distance from the Sun both change as the planet moves in an elliptical orbit, but the product of the velocity times the distance stays constant.

L = m v r,

• Thus angular momentum is conserved for a planet rotating around the sun in an elliptical orbit.
• Kepler’s Second Law: The line that joins any planet to the sun sweeps equal areas in equal intervals of time. So the area velocity is constant.
• Kepler’s third law or the law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semi-major axis of the orbit i.e. T2 ∝ r3

EXPLANATION:

• As from the law of conservation of angular momentum; we can say that the speed of the mercury will be the maximum when its distance from the sun is the minimum;

∴ m v r = constant

• So, the speed (v) of the planet will be maximum at ‘P1 because the distance between the sun and the planet is the lowest at ‘P1S’ in the figure.
• Thus the kinetic energy will be maximum at P₁ and minimum at P₃. So the kinetic energy is not constant. Hence option 1 is correct.

CONCEPT:

• Orbital velocity is defined as the velocity required to maintain a satellite into its orbit around earth.
• Mathematically orbital velocity is given by

\(V= \sqrt{\frac{GM}{R}} \)

Where G = Gravitational constant, M = Mass, R = Distance from earth to satellite

CALCULATION:

Given-  R_A = 4R, R_B = R, V_A = 3V, V_B = V, and M₁ = M₂ = M

• The velocity of a satellite in a circular orbit is given by

\(⇒ V= \sqrt{\frac{GM}{R}} \)  

• Velocities of Sataelites A can be written as

\(⇒ V_A= \sqrt{\frac{GM}{R_A}} = \sqrt{\frac{GM}{4R}} \)      -------(1)

• Velocities of Sataelites B can be written as

\(⇒ V_B= \sqrt{\frac{GM}{R_B}} = \sqrt{\frac{GM}{R}} \)       --------(2)

On dividing equations 1 and 2, we get

\(⇒ \frac{V_A}{V_B} =\sqrt{\frac{R}{4R}} \)

\(⇒ \frac{V_A}{V_B} =\frac{1}{2}\)

⇒ V_B = 2V_A

⇒ V_B = 2 × 3 V = 6V

• The velocity of Satelite B is V_B = 6 V

• Escape velocity is the minimum amount of velocity required for a body projected vertically upwards to escape from the earth's gravitational attraction.
• Mathematical expression for escape velocity is given by 

\(V= √{\frac{2GM}{R}}\)

• Escape velocity is√ 2 times orbital velocity.

A geostationary satellite orbits at a specific altitude and position above the Earth's equator in such a way that it appears to remain fixed relative to the Earth's surface. This fixed height and position allow it to match the rotation of the Earth, making it appear stationary from the perspective of an observer on the ground.