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Mark the correct statement (i) Escape ...

Mark the correct statement (i) Escape velocity does not depend on mass of body
(ii) If total energy of a satellite becomes positive , it escape from earth .
(iii) Orbit of geostationary orbit is called parking orbit

A

(i) only

B

(i),(ii) only

C

(i),(ii) and (iii)

D

(i),(iii) only

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to evaluate each statement provided and determine which one is correct. ### Step 1: Analyze Statement (i) **Statement (i): Escape velocity does not depend on mass of body.** - The formula for escape velocity \( v_e \) is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius from the center of the planet to the point of escape. - From the formula, we can see that the escape velocity depends only on the mass of the planet \( M \) and the radius \( R \). It does not depend on the mass of the object trying to escape. **Conclusion:** Statement (i) is correct. ### Step 2: Analyze Statement (ii) **Statement (ii): If total energy of a satellite becomes positive, it escapes from Earth.** - The total mechanical energy \( E \) of a satellite in orbit is given by: \[ E = K + U \] where \( K \) is the kinetic energy and \( U \) is the potential energy. For a bound system (like a satellite in orbit), the total energy is negative. - If the total energy becomes positive, it indicates that the satellite has enough energy to escape the gravitational pull of the Earth. **Conclusion:** Statement (ii) is also correct. ### Step 3: Analyze Statement (iii) **Statement (iii): Orbit of geostationary orbit is called parking orbit.** - A geostationary orbit is a specific type of orbit where a satellite appears to be stationary relative to a point on the Earth's surface. This orbit is at an altitude of approximately 35,786 kilometers above the equator. - The term "parking orbit" typically refers to a temporary orbit used to hold a spacecraft before it maneuvers to its final orbit. A geostationary orbit is not referred to as a parking orbit. **Conclusion:** Statement (iii) is incorrect. ### Final Conclusion From the analysis, we find that: - Statement (i) is correct. - Statement (ii) is correct. - Statement (iii) is incorrect. Thus, the correct statements are (i) and (ii).
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Knowledge Check

  • If v_(e) is escape velocity and v_(0) , is orbital velocity of satellite for orbit close to the earth's surface. Then are related by

    A
    `v_(o)=sqrt(2)v_(e )`
    B
    `v_(o)=v_(e )`
    C
    `v_(e )=(v_(o))/(2)`
    D
    `v_(e )=sqrt(2)v_(o)`
  • The escape velocity of a body form the earth depends on (i) the mass of the body. (ii) the location from where it is projected. (iii) the direction of projection. (iv) the height of the location form where the body is launched.

    A
    (i) and (ii)
    B
    (ii) and (iv)
    C
    (i) and (iii)
    D
    (iii) and (iv)
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