Home
Class 11
PHYSICS
Compute the additional velocity required...

Compute the additional velocity required by a satellite orbiting around earth with radius 2R to become free from earth's gravitational field. Mass of earth is M.

A

`sqrt((2GM)/(R))(sqrt(2)-1)`

B

`sqrt((GM)/(2R))(sqrt(2)-1)`

C

`sqrt((GM)/(R))(sqrt(3)-1)`

D

`sqrt((GM)/(R))(sqrt(2)+1)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of computing the additional velocity required by a satellite orbiting around Earth at a radius of \(2R\) to escape Earth's gravitational field, we can follow these steps: ### Step 1: Understand the Escape Velocity The escape velocity (\(v_e\)) from a distance \(r\) from the center of the Earth is given by the formula: \[ v_e = \sqrt{\frac{2GM}{r}} \] where \(G\) is the gravitational constant and \(M\) is the mass of the Earth. ### Step 2: Calculate the Escape Velocity at Radius \(2R\) For a satellite at a distance of \(2R\) from the center of the Earth, we can substitute \(r = 2R\) into the escape velocity formula: \[ v_e = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}} \] ### Step 3: Calculate the Orbital Velocity at Radius \(2R\) The orbital velocity (\(v_0\)) of a satellite in circular orbit at a distance \(r\) is given by: \[ v_0 = \sqrt{\frac{GM}{r}} \] Substituting \(r = 2R\): \[ v_0 = \sqrt{\frac{GM}{2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}} \] ### Step 4: Find the Additional Velocity Required The additional velocity (\(\Delta v\)) required for the satellite to escape Earth's gravitational field is the difference between the escape velocity and the orbital velocity: \[ \Delta v = v_e - v_0 \] Substituting the values we calculated: \[ \Delta v = \sqrt{\frac{GM}{R}} - \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}} \] Factoring out \(\sqrt{\frac{GM}{R}}\): \[ \Delta v = \sqrt{\frac{GM}{R}} \left(1 - \frac{1}{\sqrt{2}}\right) \] ### Step 5: Simplify the Expression To simplify \(1 - \frac{1}{\sqrt{2}}\): \[ 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \] Thus, we have: \[ \Delta v = \sqrt{\frac{GM}{R}} \cdot \frac{\sqrt{2} - 1}{\sqrt{2}} \] ### Final Answer The additional velocity required by the satellite to escape Earth's gravitational field is: \[ \Delta v = \sqrt{\frac{GM}{R}} \cdot \frac{\sqrt{2} - 1}{\sqrt{2}} \]

To solve the problem of computing the additional velocity required by a satellite orbiting around Earth at a radius of \(2R\) to escape Earth's gravitational field, we can follow these steps: ### Step 1: Understand the Escape Velocity The escape velocity (\(v_e\)) from a distance \(r\) from the center of the Earth is given by the formula: \[ v_e = \sqrt{\frac{2GM}{r}} \] where \(G\) is the gravitational constant and \(M\) is the mass of the Earth. ...
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    DC PANDEY ENGLISH|Exercise (B) Chapter Exercises|31 Videos
  • GRAVITATION

    DC PANDEY ENGLISH|Exercise (C) Chapter Exercises|45 Videos
  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Check Point 10.6|20 Videos
  • GENERAL PHYSICS

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|2 Videos
  • KINEMATICS

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|10 Videos

Similar Questions

Explore conceptually related problems

If the kinetic energy of a satellite orbiting around the earth is doubled then

By how much percent does the speed of a satellite orbiting in circular orbit be increased so that it will escape from the gravitational field of the earth ?

By how much percent does the speed of a satellite orbiting in circular orbit be increased so that it will escape from the gravitational field of the earth ?

What will be velocity of a satellite revolving around the earth at a height h above surface of earth if radius of earth is R :-

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R). The minimum increase in its orbital velocity required, So that the satellite could escape from the earth's gravitational field, is close to :(Neglect the effect of atomsphere.)

Assertion : Orbital velocity of a satellite is greater than its escape velocity. Reason : Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.

A satellite is orbiting the earth in a circular orbit of radius r . Its

A spaceship is launched into a circular orbit close to the earth's surface . What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth = 6400 km , g = 9.8m//s^(2) .

A spaceship is launched into a circular orbit close to the earth's surface . What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth = 6400 km , g = 9.8m//s^(2) .

The relative uncertainty in the period of a satellite orbiting around the earth is 10^(-2) . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

DC PANDEY ENGLISH-GRAVITATION-(A) Chapter Exercises
  1. Different points in the earth are at slightly different distances from...

    Text Solution

    |

  2. Two satellite of same mass are launched in the same orbit of radius r ...

    Text Solution

    |

  3. Compute the additional velocity required by a satellite orbiting aroun...

    Text Solution

    |

  4. Particles of masses 2M, m and M are respectively at points A, B and C ...

    Text Solution

    |

  5. Earth orbiting satellite will escape if

    Text Solution

    |

  6. The energy required to move a satellite of mass m from an orbit of rad...

    Text Solution

    |

  7. A body attains a height equal to the radius of the earth. The velocity...

    Text Solution

    |

  8. Suppose the gravitational attraction varies inversely as the distance ...

    Text Solution

    |

  9. The rotation of the earth about its axis speeds up such that a man on ...

    Text Solution

    |

  10. The rotation of the earth radius R about its axis speeds upto a value ...

    Text Solution

    |

  11. Assume the radius of the earth to be 6.4xx10^(6)m a. Calculate the t...

    Text Solution

    |

  12. If gravitational attraction between two points masses be given by F=G(...

    Text Solution

    |

  13. Suppose the gravitational force varies inversely as the n^(th) power o...

    Text Solution

    |

  14. A body is projected vertically upwards from the surface of the earth w...

    Text Solution

    |

  15. A particle takes a time t(1) to move down a straight tunnel from the s...

    Text Solution

    |

  16. If the earth were to spin faster, acceleration due to gravity at the p...

    Text Solution

    |

  17. A body which is initially at rest at a height R above the surface of t...

    Text Solution

    |

  18. A planet of mass m moves around the Sun of mass Min an elliptical orbi...

    Text Solution

    |

  19. A rocket is launched vertical from the surface of the earth of radius ...

    Text Solution

    |

  20. Two particles of equal mass go around a circle of radius R under the a...

    Text Solution

    |