Home
Class 11
PHYSICS
Derive an expression for energy of satel...

Derive an expression for energy of satellite.

Text Solution

Verified by Experts

The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
`U = - (GM_(s)M_(E))/((R_(E)+h)) " " ......(1)`
Here `M_(s)` - mass of the satellite,`M_(E)` - mass of the Earth, `R_(E)` - radius of the Earth.
The Kinetic energy of the satellite is
`K.E = 1/2 m_(s)v^(2)` ...(2)
Here v is the orbital speed of the satellite and is equal to
` v = sqrt((GM_(E))/((R_(E)+h)))`
Substituting the value of v in (2) the kinetic energy of the satellite becomes,
`K.E = 1/2 (GM_(E)M_(s))/((R_(R)+h))`
Therefore the total energy of the satellite is
` E = 1/2 (GM_(E)M_(s))/((R_(E)+h)) - (GM_(s)M_(E))/((R_(E)+h))`
`E = -(GM_(s)M_(E))/(2(R_(E)+h))`
The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.
Note: As h approaches the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth.s gravity and is not bound to Earth at large distance.
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    FULL MARKS|Exercise O|2 Videos
  • GRAVITATION

    FULL MARKS|Exercise TEXTUAL EVALUATION SOLVED (IV. Conceptual Questions)|6 Videos
  • GRAVITATION

    FULL MARKS|Exercise TEXTUAL EVALUATION SOLVED (II. Short Answer Questions)|15 Videos
  • EXAMINATION QUESTION PAPER MARCH 2019

    FULL MARKS|Exercise PART-IV|10 Videos
  • HEAT AND THERMODYNAMICS

    FULL MARKS|Exercise ADDITIONAL QUESTIONS SOLVED (Numerical Problems)|16 Videos

Similar Questions

Explore conceptually related problems

Derive an expression for electrostatic potential energy of the dipole in a uniform electric field .

Derive an expression for potenstial energy of bar magnet in a uniform magnetic field.

Derive an expression for escape speed.

Derive an expression for power and velocity.