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Q8.6(a):
Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.

Solution :

Step 1: Establishing the Gravitational Potential Energy

For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center, the gravitational potential energy $U$ is defined as the work done by an external agent in bringing the mass $m$ from infinity to distance $r$. With the convention that potential energy is zero at infinity, the expression is given by:

$U = -\frac{GMm}{r}$ [Per the definition of gravitational potential in a central force field]

Step 2: Deriving the Kinetic Energy of an Orbiting Satellite

For a satellite to maintain a stable circular orbit, the gravitational force must provide the necessary centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$ [By Newton’s Second Law and Universal Law of Gravitation]

Multiplying both sides by $r/2$, we derive the expression for kinetic energy ($K$):

$K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$

Step 3: Calculating Total Mechanical Energy ($E$)

The total mechanical energy $E$ of the satellite is the sum of its kinetic and potential energies:

$E = K + U$

$E = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right)$

$E = \frac{GMm - 2GMm}{2r} = -\frac{GMm}{2r}$

Step 4: Comparative Analysis

Comparing the total energy $E$ with the expressions for $K$ and $U$ derived in Steps 1 and 2:

  • $K = \frac{GMm}{2r} \implies E = -K$
  • $U = -\frac{GMm}{r} \implies E = \frac{1}{2}U$

Thus, the total energy of an orbiting satellite is the negative of its kinetic energy.

r M m

Conclusion

Given the derivation $E = -K$, the correct alternative is that the total energy is the negative of the kinetic energy.

Final Solution: The total energy of an orbiting satellite is the negative of its kinetic energy.


More Questions from Class 11 Physics Gravitation EXERCISES


CBSE Solutions for Class 11 Physics Gravitation


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