Maharashtra State BoardHSC Science (General) 11th

Answer the following question in detail. Obtain an expression for binding energy of a satellite revolving around the Earth at certain altitude. - Physics

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Answer in Brief

Answer the following question in detail.

Obtain an expression for the binding energy of a satellite revolving around the Earth at a certain altitude.

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Solution

An expression for the binding energy of satellite revolving in a circular orbit around the Earth:

  1. Consider a satellite of mass m revolving at height h above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy.
  2. Let M be the mass of the Earth, R be the Radius of the Earth, vc be the critical velocity of satellite, r = (R + h) be the radius of the orbit.
  3. Kinetic energy of satellite = `1/2 "mv"_"c"^2 = 1/2 "GMm"/"r"`
  4. The gravitational potential at a distance r from the centre of the Earth is `- "GM"/"r"`
    ∴ The potential energy of a satellite
    = Gravitational potential × mass of a satellite
    = `- "GMm"/"r"`
  5. The total energy of satellite is given as
    T.E. = K.E. + P.E.
    `= 1/2 "GMm"/"r" - "GMm"/"r" = - 1/2 "GMm"/"r"`
  6. The total energy of a circularly orbiting satellite is negative. The negative sign indicates that the satellite is bound to the Earth, due to the gravitational force of attraction. For the satellite to be free from the Earth’s gravitational influence its total energy should become zero or positive.
  7. Hence the minimum energy to be supplied to unbind the satellite is `+ 1/2 "GMm"/"r"`. This is the binding energy of a satellite.
Concept: Earth Satellites
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APPEARS IN

Balbharati Physics 11th Standard Maharashtra State Board
Chapter 5 Gravitation
Exercises | Q 3. (xi) | Page 98
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