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Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - NEET MCQ


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10 Questions MCQ Test - Test: Earth Satellites and Energy of an Orbiting Satellite (August 12)

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Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 1

The total energy of a circularly orbiting satellite is-

Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 1

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 at a height h is given by:


Where M is the mass of the Earth.

  • The kinetic energy of the satellite of mass m in a circular orbit is:

  • The potential energy at distance (R +h) from the centre of the earth is given by:

  • The total energy of a circularly orbiting satellite ​is

Explanation:

  • A negative sign in the value of energy shows that the nature of force between satellite and Earth is attractive.
  • Thus, the total energy of a circularly orbiting satellite is negative.
Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 2

Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,

Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 2
  • The total energy of the earth satellite bounded system is negative = (−GM/ 2R)
    where R is the radius of the satellite and M is the mass of the earth.
  • The negative sign indicates that the force of attraction between satellite and earth is continuously reduced due to atmospheric friction (viscous force).
  • As a result, the radius of the orbit or height decreases gradually, and ultimately it comes back to earth with increasing speed and burns in the atmosphere.

Hence, option (3) is the correct answer.

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Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 3

The ratio of kinetic energy and the potential energy of the satellite rotating around the earth is:

Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 3

Satellite:

  • Satellites are the natural or artificial bodies that describe an orbit around a planet by its gravitational force of attraction.
  • Moon is a natural satellite.

Kinetic energy

  • The energy possessed by a body due to the virtue of its motion is called kinetic energy.

Gravitational potential energy:

  • The work done in bringing anybody from infinity to a point in the gravitational field is called the gravitational potential energy of that body at that point.
  • The gravitational potential energy at a height h from the earth's surface is given as,

Where U = gravitational potential energy, KE = kinetic energy, M = mass of the earth, m = mass of the body, R = radius of the earth, v = velocity, and h = height from the surface of the earth

Calculations:

  • We know that the kinetic energy of the satellite is given as,

  • We know that the potential energy of the satellite is given as,

Where r = distance of the satellite from the center of the planet

By equation 1 and equation 2,


Hence, option 1 is correct.

Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 4

The relationship between total energy (E) of an orbiting satellite and its kinetic energy (KE) is given by-

Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 4

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 at a height h is given by:


Where M is the mass of the Earth.

  • The kinetic energy of the satellite of mass m in a circular orbit is:

  • The potential energy at distance (R +h) from the centre of the earth is given by:

  • The total energy of a circularly orbiting satellite ​is

Explanation:

The relationship between the total energy (E) of an orbiting satellite and its kinetic energy (KE) is given by:

E = -KE.

Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 5

A ball is dropped from a satellite revolving around the earth at a height of 120 km. The ball will

Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 5

Orbital velocity of the satellite is given by

  • From the above equation, it is clear that the orbital velocity of the satellite is independent of the mass of the satellite. 
  • Speed of the ball will be the same as that of the satellite, so when the ball is dropped from a satellite revolving around the earth at height of 120 Km, it will continue to move with the same speed along the original orbit of the satellite. Therefore option 2 is correct.
Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 6

A satellite is orbiting the Earth in a circular orbit of radius R. Choose the correct statement from the following.

Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 6

Angular momentum is the rotational equivalent of linear momentum.
Angular momentum (L) for a point object accelerating with respect to a fixed point at a distance R is given by:
L = R x mv

Where m is the mass of the body, and v is the velocity.

So, for a satellite rotating about the Earth, angular momentum ∝ R
The potential energy of an orbiting satellite is inversely proportional to the radial distance from the centre of the earth i.e. (R+h).
So, potential energy ∝ R-1

​Therefore, none of the given options is correct.

Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 7
The escape velocity from the surface of the Earth is approximately:
Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 7
The escape velocity from the surface of the Earth is approximately 11.2 km/s.
Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 8
Which of the following quantities remains constant for a satellite in a circular orbit around the Earth?
Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 8
For a satellite in a circular orbit around the Earth, the total mechanical energy remains constant.
Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 9
If a satellite is in a geostationary orbit, its orbital period is:
Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 9
A satellite in a geostationary orbit has an orbital period of 24 hours, which is the same as the Earth's rotation period.
Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 10
The total energy of a satellite cannot be zero because:
Detailed Solution for Test: Earth Satellites and Energy of an Orbiting Satellite (August 12) - Question 10
The total energy of a satellite cannot be zero because it would escape to infinity if its energy is zero.
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